## Chemistry - Chemical Kinetics: Solved Example Problems | 12th Chemistry : UNIT 7 : Chemical Kinetics

Chapter: 12th chemistry : unit 7 : chemical kinetics, chemical kinetics: solved example problems, molecularity: solved example problems.

Consider the oxidation of nitric oxide to form NO 2

2NO(g) + O 2 (g) → 2NO 2 (g)

(a). Express the rate of the reaction in terms of changes in the concentration of NO,O 2 and NO 2 .

(b). At a particular instant, when [O 2 ] is decreasing at 0.2 mol L −1 s −1 at what rate is [NO2 ] increasing at that instant?

Evaluate yourself 1

1) Write the rate expression for the following reactions, assuming them as elementary reactions.

i) 3A + 5B 2 →4CD

ii) X 2 + Y 2 →2XY

2). Consider the decomposition of N 2 O 5 (g) to form NO 2 (g) and O 2 (g) . At a particular instant N 2 O 5 disappears at a rate of 2.5x10 -2 mol dm -3 s -1 . At what rates are NO 2 and O 2 formed? What is the rate of the reaction?

1. What is the order with respect to each of the reactant and overall order of the following reactions?

(a). 5Br − ( aq ) + BrO 3 − (aq ) + 6H + (aq)

→ 3Br 2 (l ) + 3H 2 O(l)

The experimental rate law is

Rate = k [Br − ][BrO 3 ][H + ] 2

(b). CH 3 CHO(g ) → Δ → CH 4 (g) + CO(g)

the experimental rate law is

Rate = k [CH 3 CHO] 3/2

a) First order with respect to Br − , first order with respect to BrO 3 − and second order with respect to H+ . Hence the overall order of the reaction is equal to 1 + 1 + 2 = 4

b) Order of the reaction with respect to acetaldehyde is 3/2 and overall order is also 3/2

2. The rate of the reaction x + 2y → product is 4 x 10 −3 mol L −1 s −1 if [x]=[y]=0.2 M and rate constant at 400K is 2 x 10 -2 s -1 , What is the overall order of the reaction.

Rate = k [x] n [y] m

4 x 10 -3 mol L -1 s -1 = 2 x 10 -2 s -1 (0.2 mol L -1 ) n (0.2mol L -1 ) m

Comparing the powers on both sides

The overall order of the reaction n + m = 1

Evaluate yourself 2

1). For a reaction, X + Y → product ; quadrupling [x] , increases the rate by a factor of 8. Quadrupling both [x] and [y], increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall order of the reaction?

2). Find the individual and overall order of the following reaction using the given data.

2NO(g) + Cl 2 (g) → 2NOCl(g)

## Half life period of a reaction: Solved Example Problems

A first order reaction takes 8 hours for 90% completion. Calculate the time required for 80% completion. (log 5 = 0.6989 ; log10 = 1)

For a first order reaction,

Let [A 0 ] = 100M

t = t 90% ; [A]=10M (given that t 90 % =8hours)

t = t 80% ; [A]=20M

Find the value of k using the given data

Substitute the value of k in equation (2)

t 80% = 8hours x 0.6989

t 80% = 5.59hours

(ii) The half life of a first order reaction x → products is 6.932 x 10 4 s at 500K . What percentage of x would be decomposed on heating at 500K for 100 min. (e 0.06 = 1.06)

Given t 1/2 = 0.6932 x 10 4 s

To solve :2 when t=100 min,

[ [A 0 ] −[A] / [A 0 ] ] x 100 = ?

We know that

For a first order reaction, t 1/2 = 0.6932 / k

k = 10 −5 s −1

Show that in case of first order reaction, the time required for 99.9% completion is nearly ten times the time required for half completion of the reaction.

Evaluate yourself:

1. In a first order reaction A → products 60% of the given sample of A decomposes in 40 min. what is the half life of the reaction?

2. The rate constant for a first order reaction is 2.3 X 10 −4 s −1 If the initial concentration of the reactant is 0.01M . What concentration will remain after 1 hour?

3. Hydrolysis of an ester in an aqueous solution was studied by titrating the liberated carboxylic acid against sodium hydroxide solution. The concentrations of the ester at different time intervals are given below.

Show that, the reaction follows first order kinetics.

## Arrhenius equation - The effect of temperature on reaction rate: Solved Example Problems

The rate constant of a reaction at 400 and 200K are 0.04 and 0.02 s-1 respectively. Calculate the value of activation energy.

According to Arrhenius equation

Ea = 2305 J mol−1

Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation

Where Ea is the activation energy. When a graph is plotted for log k Vs 1/T a straight line with a slope of - 4000K is obtained. Calculate the activation energy

E a = − 2.303 R m

E a = − 2.303 x 8.314 J K −1 mol −1 x (− 4000K )

E a = 76,589J mol −1

E a = 76.589 kJ mol −1

## Evaluate yourself

For a first order reaction the rate constant at 500K is 8 X 10 −4 s −1 . Calculate the frequency factor, if the energy of activation for the reaction is 190 kJ mol -1 .

## Chemical Kinetics - Example : Solved Example Problems

1. The rate law for a reaction of A, B and C hasbeenfoundtobe rate = k [ A ] 2 [ B ][ L ] 3/2

How would the rate of reaction change when

(i) Concentration of [L] is quadrupled

(ii) Concentration of both [A] and [B] are doubled

(iii) Concentration of [A] is halved

(iv) Concentration of [A] is reduced to(1/3) and concentration of [L] is quadrupled.

2. The rate of formation of a dimer in a second order reaction is 7.5 × 10 − 3 mol L − 1 s − 1 at 0.05 mol L − 1 monomer concentration. Calculate the rate constant.

Let us consider the dimerisation of a monomer M

Rate= k [M] n

Given that n=2 and [M] = 0.05 mol L -1

Rate = 7.5 X 10 -3 mol L -1 s -1

k = Rate/[M] n

k = 7.5x10 -3 /(0.05) 2 = 3mol -1 Ls -1

3. For a reaction x + y + z → products the rate law is given by rate = k [ x ] 3/2 [ y ] 1/2 what is the overall order of the reaction and what is the order of the reaction with respect to z.

i.e., second order reaction.

Since the rate expression does not contain the concentration of z , the reaction is zero order with respect to z.

15. The decomposition of Cl 2 O 7 at 500K in the gas phase to Cl 2 and O 2 is a first order reaction. After 1 minute at 500K, the pressure of Cl 2 O 7 falls from 0.08 to 0.04 atm. Calculate the rate constant in s -1 .

4. A gas phase reaction has energy of activation 200 kJ mol -1 . If the frequency factor of the reaction is 1.6 × 10 13 s − 1 Calculate the rate constant at 600 K. ( e − 40 .09 = 3.8 × 10 − 18 )

20. For the reaction 2 x + y → L find the rate law from the following data.

5. The rate constant for a first order reaction is 1.54 × 10 -3 s -1 . Calculate its half life time.

We know that, t 1/2 = 0.693/ k

t 1/2 = 0.693/1.54 x 10 -3 s -1 = 450 s

6. The half life of the homogeneous gaseous reaction SO 2 Cl 2 → SO 2 + Cl 2 which obeys first order kinetics is 8.0 minutes. How long will it take for the concentration of SO 2 Cl 2 to be reduced to 1% of the initial value?

We know that, k = 0.693/ t 1/2

k = 0.693/ 8.0 minutes= 0.087 minutes -1

7. The time for half change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?

8. A zero order reaction is 20% complete in 20 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?

i) Let A = 100M, [A 0 ]–[A] = 20M,

For the zero order reaction

k=([A 0 ]-[A] / t)

k=(20M / 20min) = 1 Mmin -1

Rate constant for a reaction = 1Mmin -1

ii) To calculate the time for 80% of completion

k = 1Mmin -1 , [A 0 ] = 100M, [A 0 ]-[A] = 80M, t = ?

t=([A 0 ]-[A] / k) = (80M / 1Mmin -1 ) = 80min

9. The activation energy of a reaction is 225 k Cal mol -1 and the value of rate constant at 40°C is 1.8 × 10 − 5 s − 1 . Calculate the frequency factor, A.

Here, we are given that

E a = 22.5 kcal mol -1 = 22500 cal mol -1

T = 40 ° C = 40 + 273 = 313 K

k = 1.8 × 10 -5 sec -1

Substituting the values in the equation

log A = log (1.8) −5 + (15.7089)

log A = 10.9642

A = antilog (10.9642)

A = 9.208 × 10 10 collisions s −1

10. Benzene diazonium chloride in aqueous solution decomposes according to the equation C 6 H 5 N 2 Cl → C 6 H 5 Cl + N 2 . Starting with an initial concentration of 10 g L − 1 , the volume of N 2 gas obtained at 50 °C at different intervals of time was found to be as under:

Show that the above reaction follows the first order kinetics. What is the value of the rate constant?

For a first order reaction

In the present case, V∞ = 58.3 ml.

The value of k at different time can be calculated as follows:

Since the value of k comes out to be nearly constant, the given reaction is of the first order. The mean value of k = 0.0676 min -1

11. From the following data, show that the decomposition of hydrogen peroxide is a reaction of the first order:

Where t is the time in minutes and V is the volume of standard KMnO4 solution required for titrating the same volume of the reaction mixture.

In the present case, V o = 46.1 ml.

The value of k at each instant can be calculated as follows:

Thus, the value of k comes out to be nearly constant. Hence it is a reaction of the first order.

12. Where t is the time in minutes and V is the volume of standard KMnO 4 solution required for titrating the same volume of the reaction mixture.

i) For the first order reaction k = 2.303/t log [A 0 ]/[A]

Assume, [A 0 ] = 100 %, t = 50 minutes

Therefore, [A] = 100 – 40 = 60

k = (2.303 / 50) log (100 / 60)

k = 0.010216 min -1

Hence the value of the rate constant is 0.010216 min -1

ii) t = ?, when the reaction is 80% completed,

[A] = 100 – 80 = 20%

From above, k = 0.010216 min -1

t = (2.303 / 0.010216) log (100 / 20)

t = 157.58 min

The time at which the reaction will be 80% complete is 157.58 min.

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Chemical Kinetics

Chemical Reactions

Activation Energy

Chemical Reactions and Kinetics

The Mechanisms of Chemical Reactions

What happens when the first step in a multi-step reaction is not the rate-limiting step? Consider the reaction between NO and O 2 to form NO 2 , for example.

2 NO( g ) + O 2 ( g ) 2 NO 2 ( g )

This reaction involves a two-step mechanism. The first step is a relatively fast reaction in which a pair of NO molecules combine to form a dimer, N 2 O 2 . The product of this step then undergoes a much slower reaction in which it combines with O 2 to form a pair of NO 2 molecules.

The net effect of these reactions is the transformation of two NO molecules and one O 2 molecule into a pair of NO 2 molecules.

In this reaction, the second step is the rate-limiting step. No matter how fast the first step takes place, the overall reaction cannot proceed any faster than the second step in the reaction. As we have seen, the rate of any step in a reaction is directly proportional to the concentrations of the reactants consumed in that step. The rate law for the second step in this reaction is therefore proportional to the concentrations of both N 2 O 2 and O 2 .

Because the first step in the reaction is much faster, the overall rate of reaction is more or less equal to the rate of this rate-limiting step.

Rate k (N 2 O 2 )(O 2 )

This rate law is not very useful because it is difficult to measure the concentrations of intermediates, such as N 2 O 2 , that are simultaneously formed and consumed in the reaction. It would be better to have an equation that related the overall rate of reaction to the concentrations of the original reactants.

Let's take advantage of the fact that the first step in this reaction is reversible.

The rate of the forward reaction in this step depends on the concentration of NO raised to the second power.

The rate of the reverse reaction depends only on the concentrations of N 2 O 2 .

Because the first step in this reaction is very much faster than the second, the first step should come to equilibrium. When that happens, the rate of the forward and reverse reactions for the first step are the same.

k f (NO) 2 = k r (N 2 O 2 )

Let's rearrange this equation to solve for one of the terms that appears in the rate law for the second step in the reaction.

(N 2 O 2 ) = ( k f /k r ) (NO) 2

Substituting this equation into the rate law for the second step gives the following result.

Rate 2nd = k ( k f /k r ) (NO) 2 (O 2 )

Since k , k f , and k r are all constants, they can be replaced by a single constant, k' , to give the experimental rate law for this reaction described in Exercise 22.6.

The Relationship Between the Rate Constants and the Equilibrium Constant for a Reaction

There is a simple relationship between the equilibrium constant for a reversible reaction and the rate constants for the forward and reverse reactions if the mechanism for the reaction involves only a single step . To understand this relationship, let's turn once more to a reversible reaction that we know occurs by a one-step mechanism.

The rate of the forward reaction is equal to a rate constant for this reaction, k f , times the concentrations of the reactants, ClNO 2 and NO.

Rate forward = k f (ClNO 2 )(NO)

The rate of the reverse reaction is equal to a second rate constant, k r , times the concentrations of the products, NO 2 and ClNO.

Rate reverse = k r (NO 2 )(ClNO)

This system will reach equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction.

Rate forward = Rate reverse

Substituting the rate laws for the forward and reverse reactions when the system is at equilibrium into this equation gives the following result.

k f [NO][ClNO 2 ] = k r [ClNO][NO 2 ]

This equation can be rearranged to give the equilibrium constant expression for the reaction.

Thus , the equilibrium constant for a one-step reaction is equal to the forward rate constant divided by the reverse rate constant.

Determining the Order of a Reaction from Rate of Reaction Data

The rate law for a reaction can be determined by studying what happens to the initial instantaneous rate of reaction when we start with different initial concentrations of the reactants. To show how this is done, let's determine the rate law for the decomposition of hydrogen peroxide in the presence of the iodide ion.

Data on initial instantaneous rates of reaction for five experiments run at different initial concentrations of H 2 O 2 and the I - ion are given in the table below.

Rate of Reaction Data for the Decomposition of H 2 O 2 in the Presence of the I - Ion

The only difference between the first three trials is the initial concentration of H 2 O 2 . The difference between Trial 1 and Trial 2 is a two-fold increase in the initial H 2 O 2 concentration, which leads to a two-fold increase in the initial rate of reaction.

The difference between Trial 1 and Trial 3 is a three-fold increase in the initial H 2 O 2 concentration, which produces a three-fold increase in the initial rate of reaction.

The only possible conclusion is that the rate of reaction is directly proportional to the H 2 O 2 concentration.

Experiments 1, 4, and 5 were run at the same initial concentration of H 2 O 2 but different initial concentrations of the I - ion. When we compare Trials 1 and 4 we see that doubling the initial I - concentration leads to a twofold increase in the rate of reaction.

Trials 1 and 5 show that tripling the initial I - concentration leads to a three-fold increase in the initial rate of reaction. We therefore conclude that the rate of the reaction is also directly proportional to the concentration of the I - ion.

The results of these experiments are consistent with a rate law for this reaction that is first-order in both H 2 O 2 and I - .

Rate = k (H 2 O 2 )(I - )

The Integrated Form of First-Order and Second-Order Rate Laws

The rate law for a reaction is a useful way of probing the mechanism of a chemical reaction but it isn't very useful for predicting how much reactant remains in solution or how much product has been formed in a given amount of time. For these calculations, we use the integrated form of the rate law.

Let's start with the rate law for a reaction that is first-order in the disappearance of a single reactant, X .

When this equation is rearranged and both sides are integrated we get the following result.

In this equation, ( X ) is the concentration of X at any moment in time, ( X ) 0 is the initial concentration of this reagent, k is the rate constant for the reaction, and t is the time since the reaction started.

To illustrate the power of the integrated form of the rate law for a reaction, let's use this equation to calculate how long it would take for the 14 C in a piece of charcoal to decay to half of its original concentration. We will start by noting that 14 C decays by first-order kinetics with a rate constant of 1.21 x 10 -4 yr -1 .

The integrated form of this rate law would be written as follows.

We are interested in the moment when the concentration of 14 C in the charcoal is half of its initial value.

Substituting this relationship into the integrated form of the rate law gives the following equation.

We now simplify this equation

and then solve for t .

It therefore takes 5730 years for half of the 14 C in the sample to decay. This is called the half-life of 14 C. In general, the half-life for a first-order kinetic process can be calculated from the rate constant as follows.

Let's now turn to the rate law for a reaction that is second-order in a single reactant, X .

The integrated form of the rate law for this reaction is written as follows.

Once again, ( X ) is the concentration of X at any moment in time, ( X ) 0 is the initial concentration of X , k is the rate constant for the reactio n, and t is the time since the reaction started.

The half-life of a second-order reaction can be calculated from the integrated form of the second-order rate law.

We start by asking: "How long it would take for the concentration of X to decay from its initial value, ( X ) 0 , to a value half as large?"

The first step in simplifying this equation involves multiplying the top and bottom halves of the first term by 2.

Subtracting one term on the left side of this equation from the other gives the following result.

We can now solve this equation for the half-life of the reaction.

There is an important difference between the equations for calculating the half-life of first order and second-order reactions. The half-life of a first-order reaction is a constant, which is proportional to the rate constant for the reaction.

The half-life for a second-order reaction is inversely proportional to both the rate constant for the reaction and the initial concentration of the reactant that is consumed in the reaction.

Discussions of the half-life of a reaction are therefore usually confined to first-order processes.

Determining the Order of a Reaction with the Integrated form of Rate Laws

The integrated form of the rate laws for first- and second-order reactions provides another way of determining the order of a reaction. We can start by assuming, for the sake of argument, that the reaction is first-order in reactant X .

Rate = k ( X )

We then test this assumption by checking concentration versus time data for the reaction to see whether they fit the first-order rate law.

To see how this is done, let's start by rearranging the integrated form of the first-order rate law as follows.

ln ( X ) - ln ( X ) 0 = - kt

We then solve this equation for the natural logarithm of the concentration of X at any moment in time.

ln ( X ) = ln ( X ) 0 - kt

ln ( X ) = - kt + ln ( X ) 0

If the reaction is first-order in X , a plot of the natural logarithm of the concentration of X versus time will be a straight line with a slope equal to - k , as shown in the figure below.

If the plot of ln ( X ) versus time is not a straight line, the reaction can't be first-order in X . We therefore assume, for the sake of argument, that it is second-order in X .

Rate = k ( X ) 2

We then test this assumption by checking whether the experimental data fit the integrated form of the second-order rate law.

If the reaction is second-order in X , a plot of the reciprocal of the concentration of X versus time will be a straight line with a slope equal to k , as shown in the figure below.

Reactions That are First-Order in Two Reactants

What about reactions that are first-order in two reactants, X and Y, and therefore second-order overall?

Rate = k ( X )( Y )

A plot of 1/( X ) versus time won't give a straight line because the reaction is not second-order in X . Unfortunately, neither will a plot of ln ( X ) versus time, because the reaction is not strictly first-order in X . It is first-order in both X and Y .

One way around this problem is to turn the reaction into one that is pseudofirst-order by making the concentration of one of the reactants so large that it is effectively constant. The rate law for the reaction is still first-order in both reactants. But the initial concentration of one reactant is so much larger than the other that the rate of reaction seems to be sensitive only to changes in the concentration of the reagent present in limited quantities.

Assume, for the moment, that the reaction is studied under conditions for which there is a large excess of Y . If this is true, the concentration of Y will remain essentially constant during the reaction. As a result, the rate of the reaction will not depend on the concentration of the excess reagent. Instead, it will appear to be first order in the other reactant, X . A plot of ln ( X ) versus time will therefore give a straight line.

Rate = k '( X )

If there is a large excess of X , the reaction will appear to be first-order in Y . Under these conditions, a plot of log ( Y ) versus time will be linear.

Rate = k '( Y )

In our discussion of acid-base equilibria, we argued that the concentration of water is so much larger than any other component of these solutions that we can build it into the equilibrium constant expression for the reaction.

We now understand why this is done. Because the concentration of water is so large, the reaction between an acid or a base and water is a pseudo-first-order reaction that only depends on the concentration of the acid or base.

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## Solved Examples

Example : 1.

The experimental data for the reaction 2A + B2 → 2AB is as follows:

From an examination of above data, it is clear that when the concentration of B 2 is doubled, the rate is doubled. Hence the order of reaction with respect to B 2 is one. Further when concentration of A is doubled, the rate remain unaltered. So, order of reaction with respect to A is zero. The probable rate law for the reaction will be -dx/dt = k[B 2 ][A] 0 = k[B 2 ] Alternatively Rate = k[B 2 ] α 1.6 × 10-4 = k[0.5] α 3.2 × 10-4 = k[1] α On dividing we get α = 1 .·. Rate = k[A] 0 [B 2 ] = k[B 2 ]

__________________________________________________________________________

## Example : 2

For the reaction A + 2B → 2C the following data were obtained.

Write down the rate law for the reaction.

Let the rate law be -dx/dt = k[A] x [B] y By keeping the concentration of B constant in experiments (1), (2) and (3) and increasing concentration uniformly, the rate also increases uniformly, thus, Rate ∞ [A], i.e., x = 1 By keeping the concentration of A constant in experiments (1), (4) and (5) and increasing the concentration of B, the rate remains the same. Hence, y = 0 The rate law is -dx/dt = k[A]

Alternatively method: From Expt. (1), k[1.0] x [1.0] y = 0.15 ....(i) From Expt. (2), k[2.0] x [1.0] y = 0.30 ... (ii) Dividing Eq. (ii) by Eq. (i), [2.0] x /[1.0] x = 0.30/0.15 = 2

So , x = 1

From Expt. (1), k[1.0] x [1.0] y = 0.15 ....(i) From Expt. (4), k[1.0] x [2.0] y = 0.15 .... (iii) Dividing Eq. (iii) by Eq. (i), [2.0] y /[1.0] y = 1 So , y = 0 Hence, the rate law is -dx/dt = k[A].

___________________________________________________________________________________

## Example : 3

For the reaction: 2NO + Cl2 → 2NOCl at 300 K following data are obtained.

Write rate of law for the reaction. What is the order of the reaction? Also calculate the specific rate constant. .

Let the rate law for the reaction be

Rate = k[NO] x [Cl 2 ] y

From Expt. (1),

1.2×10-4 = k[0.010] x [0.010] y ... (i)

From Expt. (2),

2.4×10 -4 = k[0.010] x [0.020] y ... (ii)

Dividing Eq. (ii) by Eq. (i),

(9.6×10-4)/(2.4×10-4) = ([0.020] x )/[0.010] x or 2 = (2) y y = 1 From Expt. (2), 2.4×10-4 = k[0.010] x [0.020] y .... (ii) From Expt. (3), 9.6×10-4 = k[0.020] x [0.020] y .... (ii) Dividing Eq. (ii) by Eq. (ii), (9.6 × 10-4)/(2.4 × 10-4) = ([0.020]x )/[0.010]x or 4 = 2x x = 2 Order of reaction = x + y = 2 + 1 = 3 Rate law for the reaction is Rate = k[NO] 2 [Cl 2 ] Considering Eq. (i) again, 1.2 × 10 -4 = k[0.010] 2 [0.010] k = (1.2×10-4)/[0.010] 3 = 1.2×10 2 mol -2 litre 2 sec-1 ________________________________________________________________________________________________________

For the hypothetical reaction 2A + B → products the following data are obtained.

Find out how the rate of the ration depends upon the concentration of A and B and fill in the blanks.

## Solution :

From Expt. (2) and (3), it is clear that when concentration of A is kept constant and that of B is doubled, the rate increases four times. This shows that the reaction is of second order with respect to B. Similarly, from Expt. (1) and (2), it is observed that when concentration of A is increased three times and that of B two times, the rate becomes twelve times. Hence, the reaction is first order with respect to A. Thus the rate law for the reaction is Rate = k[A][B]2

From Expt. (1), 1.75×10-4 = k[0.0017] x [0.0017] y [1.0] z From Expt. (4), 3.50×10-4 = k[0.0017] x [0.0017] y [0.5] z (1.75 × 10-4)/(3.50 × 10-4) = [1.0] z /[0.5] z or 1/2 = 2z or 2-1 = 2z z = -1, i.e., order w.r.to OH - is -1. Rate law = k[OCI- ][I- ]/[OH-] From Expt.(1)k = (1.75×10 -4 [OH - ])/[OCI - ][I - ] = (1.75 × 10 -4 ×1.0)/(0.0017×0.0017) = 60.55 s -1 ____________________________________________________________________________

## Example : 5

The rate law for the reaction, 2Cl 2 O → 2Cl 2 + O 2 at 200oC is found to be : rate = k[Cl 2 O] 2 (a) How would the rate change if [Cl 2 O] is reduced to one-third of its original value? (b) How should the [Cl 2 O] be changed in order to double the rate? (c) How would the rate change if [Cl 2 O] is raised to threefold of its original value?

## Solution :

(a) Rate equation for the reaction, r = k[Cl 2 O] 2 Let the new rate be r'; so r' = k[(Cl 2 O)/3] 2 = 1/9 r (b) In order to have the rate = 2r, let the concentration of Cl 2 O be x. So 2r = kx 2 .... (i) We know that r = k[Cl 2 O] 2 .... (ii) Dividing Eq. (i) by (ii), 2r/r = (kx 2 )/(k[Cl 2 O] 2 ) or 2 = x 2 /[Cl 2 O] 2 or x 2 = 2[Cl 2 O] 2 or x = √√2 [Cl 2 O] (d) New rate = k[3Cl 2 O] 2 = 9k[Cl 2 O] 2 = 9r _____________________________________________________________________________________________________

## Example : 6

For a reaction in which A and B from C, the following data were obtained from three experiments:

What is the rate equation of the reaction and what is the value of rate constant?

## Solution:

Let the rate equation be k[A] x [B] y From Expt. (1), 0.3×10 -4 = k[0.03] x [0.03] y ... (i) From Expt. (2), 1.2×10 -4 = k[0.06] x [0.06] y ... (ii) (1.2 × 10-4)/(0.3×10-4) = ([0.06] x [0.06] y )/([0.03] x [0.03] y ) = 2 x × 2y = 4 ... (iii) Similarly from Expt. (1) and (3), 2x ×3y = 9 ....(iv) Solving Eq. (iii) and (iv), x = 0, y = 2 Rate equation, Rate = k[B] 2 Considering Eq. (i) again, k = (0.3×10 -4 )/[0.03] 2 = 3.33 × 10-2 mol L -1 s -1 ________________________________________________________________________________

## Example : 7

For a first order reaction when log k was plotted against , a straight 1/T line with a slope of -6000 was obtained. Calculate the activation energy of the reaction

Slope = -Ea/2.303R = –6000

= 27483.935Cal = 27.48 Kcal

## Example : 8

Solution: .

___________________________________________________________________________________________

## Example : 9

The time required for 10% completion of first order reaction at 298 K is equal to that required for its 25% completion at 308K. If the preexponential factor for the reaction is 3.56 ´ 10 9 s –1 , calculate the energy of activation

_________________________________________________________________________________

## Question : 10

At 380°C, the half-life period for the first order decomposition of H 2 O 2 is 360 min. the energy of activation of the reaction is 200 kJ mol –1 , Calculate the time required for 75% decomposition at 450°C.

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## Latest articles from Blog

## Chemical Kinetics and Rate of Reaction

## Core Concepts – Kinetics and Rate of Reaction

In this article, we will learn about chemical kinetics and rate of reactions, including what properties affect them, conditions needed for reactions to occur, and how to calculate reaction rates.

## Topics Covered in Other Articles

- Collision Theory
- Reaction Rates and How To Determine Rate Laws
- Activation Energy
- Understanding Kinetic Molecular Theory
- Solving Stoichiometric Problems
- Introduction to Energy

## What is the Rate of a Reaction?

The rate of a reaction measures the speed at which a chemical reaction takes place. This is done by dividing the change in concentration of reactants by the change in time. So it measures either the rate of disappearance of the reactants or the rate of appearance of the products. To calculate the rate let’s first look at the general formula:

A few notes:

- The rate for reactants A and B are negative because they are being consumed. This means that C and D are positive for the opposite reason—they are being produced.
- The change in concentration over the change in time for each species (reactants and products) is divided by its stoichiometric equivalent.
- The concentration of each species is measured in molarity as denoted by the brackets [ ].

## What is Kinetics?

The general definition of kinetics is the study of the effects of forces on mechanisms. In terms of chemistry (chemical kinetics), it is the study of rates of reaction. Using our knowledge of reaction rates now, we can say it is the study of how fast chemical reactions go from reactants to products or how fast reactants are consumed and products are produced. A useful measure in studying the kinetics of a reaction is kinetic energy, the energy a particle has due to its motion.

## What is Collision Theory?

For us to discuss reaction rates and kinetics more in depth, we must first understand collision theory . Reacting particles must collide for a reaction to occur, but not every collision results in a reaction. Collision theory outlines the necessary conditions for a reaction to take place. There are three conditions that satisfy collision theory:

- Particles must collide
- Collisions must have sufficient energy
- Collisions must have proper orientation

## Particles Must Collide

The first condition of collision theory—particles must collide—is the basis for reactions. Without colliding, particles do not interact, and therefore reactions do not occur.

## Collisions Must Have Sufficient Energy

As stated before, the kinetic energy of a particle is the energy due to its motion. So a particle’s kinetic energy is proportional to its speed. The kinetic energy of reacting particles is important because reactions have specific activation energies , aka the energy needed for a reaction to occur (like a threshold). The kinetic energy of colliding particles must match the activation energy for the reaction to occur.

## Collisions Must Have Proper Orientation

If particles collide and with sufficient energy, a reaction still isn’t guaranteed to occur. The direction at which the reactants collide will determine if the reaction occurs. Even further, a reaction may occur but the orientation of the collision can determine what products are formed. This is because different individual atoms will interact depending on which angle two molecules collide.

## Effects on Kinetics and the Rate of Reactions

Knowing this about collision theory, we can now discuss properties that affect reaction rates. There are four properties than can affect the rate of a reaction :

- Concentration
- Temperature

## Physical State

- Presence of a Catalyst or Inhibitor

## Concentration

Concentration and reaction rate are positively related, meaning that if the concentration of particles increases, the rate of the reaction will increase. The opposite is also true. Using collision theory we can explain this. The higher concentration of particles increases the amount of collisions and therefore increases the reaction rate.

## Temperature

Temperature and reaction rate are also positively related. This is because temperature is a measure of the average kinetic energy of particles. The higher the temperature, the higher the kinetic energy the particles have. This not only increases the amount of collisions (because particles are moving faster) but it also increases the likelihood that collisions are occurring at sufficient energy which will increase the reaction rate.

The physical state of a reactant can affect the rate of a reaction because of surface area. The particles in solids can not move freely like that of liquids and gasses which lowers the reactants surface area. The smaller the surface area, the less opportunity for reacting particle interaction which will slow the reaction rate. Conversely, gas particles move most freely opposed to other physical states which can also slow reaction rates. Unlike the other physical states, gas particles fill up the space of their containers, but imagine we are comparing the same amount of particles for each physical state. In the gaseous state there would be the same amount of particles taking up more space. This can make collisions more of a chance occurrence and slow the reaction rate.

## Presence of a Catalyst or Inhibitor

Catalysts increase reaction rates while inhibitors decrease reaction rates. Catalysts lower the activation energy of a reaction which allows more collisions to have sufficient energy for a reaction to occur. Inhibitors act oppositely. They force a reaction to take a path with a higher activation energy. So, less collisions meet the sufficient energy and the rate of the reaction slows.

## For More Help, Watch Our Interactive Video Introducing Kinetics!

Chemical kinetics and rate of reaction practice problems.

Consider the combustion of butadiene:

How does the rate of butadiene consumption relate to the production of CO 2 ? By what factor is butadiene consumption faster or slower than CO 2 production?

Between the four properties affecting reaction rate (reactant concentration, temperature, reactant physical state, and use of catalyst/inhibitor), which properties increase or decrease the proportion of a given amount of reactant capable of performing the reactant?

## Kinetics and Rate of Reaction Practice Problem Solutions

1: CO 2 production is four times as fast as butadiene consumption.

2: Temperature (increases or decreases proportion of reactants above activation energy) and Physical State (increases or decreases proportion of available reactant based on surface area)

## Related Articles to Chemical Kinetics and Rate of Reaction

- Reaction Order
- Integrated Rate Laws
- What is the Equilibrium Constant
- Concentration Units: Normality, PPM, and Volume Percent

- Chemistry Concept Questions and Answers
- Kinetics Questions

## Chemical Kinetics Questions

Chemical Kinetics deals with the rates of the reactions. The study of Kinetics involves the factors affecting the rate of the chemical reactions, the mechanisms and the transitions states involved, if any.

## Chemical Kinetics Chemistry Questions with Solutions

Q1: What is the difference between the average rate and instantaneous rate?

Answer: Average rate is the rate measured for a long period of time. While the instantaneous rate is the rate measured for an infinitesimally small period of time.

Q2. What does the given graph represent about the nature of reaction? Which of the following expressions are in favour of the graph?

- Δ[A]/Δt = Δ[B]/Δt
- -Δ[B]/Δt = Δ[A]/Δt
- -Δ[A]/Δt = Δ[B]/Δt
- None of the above

Answer: (c.)

Explanation: The given graph represents that with the decrease in the concentration of A, the concentration of B increases. This implies that A is the reactant and B is the product of the reaction. The reaction can be represented as: A → B.

Since in this reaction, A is the reactant and B is the product; the change in concentration of either of A and B with time gives the rate of the reaction.

Hence, the average rate of formation of B can be written as: Δ[B]/Δt (with a + sign as the concentration of B increases through the reaction).

While the average rate of consumption of A can be written as: -Δ[A]/Δt (with a – sign as the concentration of A decreases through the reaction)

The correct expression for the reaction curves shown in the graph is -Δ[A]/Δt = Δ[B]/Δt.

Q3. The reaction rate of a substance is directly proportional to its_____.

Answer: Active Mass

The active mass is the concentration of the reacting substance in mol l -1 . Hence, the rate of the reaction of the substance is directly proportional to its active mass.

Q4. In the reaction rate expression, the change in concentration of each of the reactants and products are divided by the respective stoichiometric number present in the reaction equation. Why is this division done?

Answer: To understand the above fact, we must look at the following balanced equation:

2N 2 O 5 → 4NO 2 + O 2

The rate of decomposition of N 2 O 5 is -d[N 2 O 5 ]/dt. The rate of formation of NO 2 and O 2 are d[NO 2 ]/dt and d[O 2 ]/dt respectively.

However, these rates are not equal. This is because when 2 moles of N 2 O 5 decompose, 4 moles of NO 2 and 1 mole of O 2 are formed. This implies that the rate of decomposition of N 2 O 5 is twice the rate of formation of O 2 and the rate of formation of NO 2 is 4 times the rate of formation of O 2 .

Hence, the rate of formation of products and the rate of decomposition of the reactants are divided by their stoichiometric coefficients in the reaction in order to get an identical value for the rate of the reaction.

Therefore, the rate of reaction in terms of each of the reactants and the products is :

Rate of the reaction =

Q5. Calculate the rate of the reaction in terms of the different reactants and products for the following reaction.

4NH 3 (g) + 5O 2 (g) → 4NO (g) + 6H 2 O (g)

Given the rate of formation of NO is 3.6 x 10 -3 molL -1 s -1 , calculate the rate of disappearance of NH 3 and the rate of formation of H 2 O.

Answer: The rate of the reaction in terms of each of the reactants and products is given as:

Rate of Reaction =

Since, the coefficients of NO and NH 3 in the balanced chemical equation are the same, both of them have the same rate of formation and disappearance respectively.

Hence, rate of formation of NO = rate of disappearance of NH 3 = 3.6 x 10 -3 mol L -1 s -1

Now, the rate of reaction :

Hence, the rate of formation of water :

Hence, the rate of formation of water is 5.4 x 10 -3 mol L -1 s -1 .

Q6. Calculate the overall order from the given rate expressions.

- Rate = k[A] ½ [B] 3/2
- Rate = k[A] 3/2 [B] -1

Answer: The overall orders are calculated as:

- The order w.r.t. A = 1/2

The order w.r.t. B = 3/2

Overall Order = ½ + 3/2 = 2

- The order w.r.t. A = 3/2

The order w.r.t. B = -1

Overall Order = 3/2 – 1 = 1/2

Q7. The rate constants of 3 reactions are given. Identify the order of each of the given reactions.

- k = 2.3 x 10 -5 L mol -1 s -1
- k = 3.1 x 10 -4 s -1
- k = 9.3 x 10 -4 mol L -1 s -1

Answer: Since only rate constants for each reaction are given, the order of the reactions will be determined based on the units of the rate constants.

Q8. What are the differences between the rate of the reaction and the reaction rate constant?

Answer: The differences between the two are:

Q9. Give an example of the reaction of the 4th order.

Answer: The dissociation of potassium chlorate to form potassium perchlorate is an example of the reaction of 4th order.

4KClO 3 → 3KClO 4 + KCl

Q10. Derive the general expression of the time taken by the reactant to reduce to its nth fraction in the first order reaction.

Answer: Let us assume the initial amount of the reactant (A) = a

Hence, the nth fraction decreased from “a” in time t = a/n

From the first order reaction,

Hence, the general expression of the time taken by the reactant to reduce to its nth fraction in the first order reaction is represented by t = (2.303/k) logn.

Q11. The half-life period of a 1st order reaction is 60 min. What percentage of the substance will be left after 240 min?

Answer: 1 half-life period = 60 min

Number of half-lives after 240 min = 240/60 = 4 hal-lives i.e. n = 4

Amount of substance left after n half-lives = A o /2 n

Percentage of the amount of substance left after 4 half-lives = A o /2 4 x 100 = A o /16 x 100 = 6.25% of A.

Q12. The dissociation of N 2 O 5 in CCl 4 takes place by the 1st order rate law. The table below shows the concentration of N 2 O 5 measured at different times.

From the given observations, calculate k at t = 410 s and t = 1130 s.

Answer: From the first order rate law:

So, the value of k for the given reaction at t = 410 s is 7.768 x 10 -4 s -1 and that at t = 1130 s is 7.341 x 10 -4 s -1 .

Q13. A 1st order reaction gets 40% completed in 50 minutes. Calculate:

(i.) the rate constant

(ii.) the time in which the reaction will get 80% completed.

Answer: (i.) From the 1st order:

Where 𝑥 = (40/100)a = 0.4a, t= 50 min

The rate constant for the reaction is 0.010216 min -1 .

(ii.)To determine t = ?, 𝑥 = 0.8a

As k is constant for a given reaction, k = 0.010216 min -1

Hence, the time in which the reaction will get 80% completed is 157.58 min.

Q14. Determine the order of a reaction whose rate constant has the same unit as the rate of the reaction.

Answer: Zero Order reaction.

Q15. A reaction is 2nd ordered w.r.t. a reactant. Determine the change in the rate of the reaction when the amount of the reactant is:

- Reduced to its half

Answer: For a 2nd order reaction, Rate = k[A] 2 = ka 2

- When A = 2a, Rate = k(2a) 2 = 4.ka 2 , the reaction rate increases 4 times.
- When A = ½ a, Rate = k(½ a) 2 = ¼ times, the reaction rate decreases by ¼ times.

## Practise Questions on Chemical Kinetics

Q1. The given reaction is carried out in a closed vessel:

2N 2 O 5 (g) ⇌ 4NO 2 (g) + O 2 (g)

It was observed that the concentration of NO 2 increased by 2.0 x 10 -2 mol L -1 within 5 seconds of the reaction. Calculate:

(i.) the rate of the reaction.

(ii.) the rate of change of concentration of N 2 O 5 .

Q2. For A → Products, k = 2.0 x 10 -2 s -1 . Calculate the concentration of A after 100 s.

Given [A] o = 1.0 mol L -1

Q3. The equation followed by the composition of a hydrocarbon is:

k = (4.5 x 10 11 s -1 )e -28000 K/T

Calculate the E a .

Q4. A 1st order reaction gets 30% decomposed in 40 min. Calculate its t 1/2 .

Q5. The rate law for a reaction is: Rate = k = [A][B] ½ . Can this be an elementary reaction?

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## Chemical Kinetics

Chemical Kinetics is a part of physical chemistry that focuses on the rate of chemical reactions and the reasons behind them. It involves figuring out the speed and details of chemical reactions, studying the factors that affect how fast they happen, and how molecules interact during these changes.

This field is important in various scientific areas, from improving industrial processes to understanding complicated reactions in biological systems. By focusing on how molecules interact, chemical kinetics helps us to understand the complex movements of atoms and molecules in the world of chemistry.

## What is Chemical Kinetics?

The term “kinetics” originates from the Greek word ‘kinesis,’ meaning movement. Unlike thermodynamics, which addresses the feasibility of a reaction, chemical kinetics focuses on explaining the rate of a reaction.

The study of chemical kinetics is important because Factors such as concentration, temperature, pressure, and catalysts play pivotal roles in controlling reaction rates. Chemical reactions are often categorized based on their rates.

Examples include fast reactions, such as the reaction of sodium with water (Na + H 2 O), moderate reactions, like the reaction of magnesium with water (Mg + H 2 O), and slow reactions, as observed in esterification processes.

## What is the Rate of Reaction?

The rate of a reaction, or its speed, is defined as the change in concentration of a reactant or product per unit of time.

Specifically, it can be defined in terms of either the rate of decrease in the concentration of a reactant or the rate of increase in the concentration of a product. In every chemical reaction, as the reaction proceeds, the quantity of reactants decreases while the amount of products rises.

By plotting the concentration of reactants and products against time, one can readily compute the rates of product formation and reactant disappearance based on the slopes of their respective curves. Notably, the overall reaction rate may or may not be equal to the formation and disappearance rates.

## Factors Affecting Reaction Rates

The rate of a reaction can be modified by changing certain parameters. Here are the key factors:

Concentration of Reactants: According to collision theory, an increase in the concentration of reactants leads to more collisions between molecules, consequently elevating the reaction rate.

- Example: In a reaction A + B → C, if the concentration of A and B is increased, the rate of the reaction will likely increase due to a higher frequency of collisions.

Nature of the Reactants: The reaction rate is influenced by the types of substances involved. Acid/base reactions and salt formation tend to be fast, while reactions resulting in the formation of larger molecules through covalent bonds are usually slower. The nature and strength of bonds in reactant molecules play a significant role in the rate of transformation into products.

- Example: Acid/base reactions and salt formation often exhibit faster rates. Conversely, the formation of larger molecules through covalent bonds, like in the reaction C 6 H 12 O 6 → 2C 2 H 5 OH + 2CO 2 , tends to be slower.

Physical State of Reactants: The physical state (solid, liquid, or gas) of a reactant affects the rate of change. Homogeneous phases (e.g., both in an aqueous solution) allow for increased collisions, while different phases limit reactions to interfaces between reactants.

- Example: In the reaction between solid magnesium (Mg) and hydrochloric acid (HCl), the physical state influences the rate. When Mg is in a powdered form, the reaction proceeds faster than when it is in larger solid pieces.

Surface Area of Reactants: Increasing the surface area, such as by crushing a solid into smaller particles, enhances the frequency of collisions between reactant particles, leading to a faster reaction.

- Example: For the reaction 2H 2 (g) + O 2 (g) → 2H 2 O(g), finely divided powdered iron as a catalyst increases the surface area, promoting a faster reaction.

Temperature: An increase in temperature results in more collisions between reactant molecules per second, thereby boosting the reaction rate. Depending on whether the reaction is endothermic or exothermic, temperature influences the rates of forward or backward reactions.

- Example: In the combustion of methane (CH 4 ) with oxygen (O 2 ), increasing the temperature leads to a higher reaction rate, as more collisions occur between the molecules.

Effect of Solvent: The nature of the solvent influences the reaction rate of solute particles. For instance, reactions may occur faster in organic solvents like DMF compared to solvents like methanol due to specific bonding characteristics.

- Example: The reaction of sodium acetate with methyl iodide occurs faster in organic solvents like DMF (dimethylformamide) than in CH 3 OH (methanol) due to specific bonding characteristics.

Catalyst: Catalysts modify the reaction rate by altering the reaction mechanism. Two types of catalysts, promoters, and poisons, respectively increase and decrease reaction rates.

- Example: The decomposition of hydrogen peroxide (2H 2 O 2 ) is catalyzed by manganese dioxide (MnO 2 ), increasing the rate of the reaction without being consumed in the process.

## Average and Instantaneous Rate

Average rate of reaction.

The rate of a reaction is categorized into average and instantaneous rates based on the period considered. When the period is finite, it is referred to as the average rate and is denoted as:

r avg = ΔC/Δt

Where r avg average rate, ΔC change in concentration, Δt change in time.

## Instantaneous Rate of Reaction

The average rate often needs to provide precise information regarding the completion of a reaction. For instance, consider the hydrolysis of esters into acid and alcohol in the presence of water

RCOOR’ → RCOOH + R’OH

Assuming a 2M solution of ester at time t = 0 which becomes 1M in 30 minutes, one might logically infer that the completion time will be 1 hour. However, in reality, the reaction requires more than 3 hours to complete. To obtain a more comprehensive understanding of the time needed for completion and for other analytical purposes, the “Instantaneous rate” is employed, represented as

r inst = lim Δt→0 ΔC/Δt = d[C]/dt

Where r inst is the instantaneous rate.

## Unit of Rate of Reaction

The unit of rate is commonly expressed as Molarity per second (Mol s -1 ) or mol per liter per second (mol/L/s), representing the ratio of concentration to time (mol L -1 divided by seconds).

The units of the rate constant, k, depend on the overall reaction order, and for a zero-order reaction, the units of k are M/s, for a first-order reaction, the units are 1/s, and for a second-order reaction, the units are 1/(M·s)

Rate Law or Law of Mass Action is a principle stating that the rate of a chemical reaction is directly proportional to the product of the masses of the reactants, each raised to a power equal to its coefficient in the balanced chemical equation.

This law was developed by Norwegian scientists Cato M. Guldberg and Peter Waage between 1864 and 1879. This fundamental principle that explains the dynamic equilibrium behavior observed in solutions. Moreover, it also states that the ratio of reactant to product concentrations remains constant at a state of chemical equilibrium.

## Equilibrium Constant (K c )

At equilibrium, the concentrations of reactants and products remain constant at a given temperature. Consider a simple reversible reaction where A and B are reactants, and C and D are products:

aA + bB → cC + dD

An equilibrium mixture, comprising both products and reactants, maintains a specific relationship between their concentrations. This relationship is expressed through the equilibrium constant, Kc, as follows

K c =[C] c [D] d /[A] a [B] b

Here, K c represents the equilibrium constant measured in moles per litre.

In this equation, the concentrations of A, B, C, and D at equilibrium are denoted by [A],[B],[C], and[D] respectively, with stoichiometric coefficients for each species.

## Order of Reaction

In the rate equation, we know,

Rate = k [A] x [B] y

Where x and y indicate how sensitive the rate is to the change in concentration of A and B.

The sum of these exponents, i.e., x + y gives the overall order of a reaction whereas x and y represent the order concerning the reactants A and B respectively.

Read More about Order of Reaction .

Different types of chemical reactions are categorized based on how the rate of the reaction depends on the concentration of the reactants:

- Zeor Order Reactions

## First-Order Reactions

Pseudo-first order reactions.

- Second Order Reactions

Let’s discuss these types in brief as follows:

## Zero Order Reactions

The rate of reaction in these reactions is unaffected by the concentration of the reactants . Changes in the concentration of the reactants do not impact the reaction speed.

- Example: Enzyme-catalyzed oxidation of ethanol (CH 3 CH 2 OH) to acetaldehyde (CH 3 CHO).

The rates of these reactions depend on the concentration of only one reactant (order of reaction is 1). Even if multiple reactants are present, only one reactant has a first-order concentration, while the others have zero-order concentration.

- Example: 2H 2 O 2 → 2H 2 O + O 2 .

Read more about First Order Reaction .

In pseudo-first-order reactions, one reactant’s concentration remains constant and is included in the rate constant in the rate expression. This constant concentration may be due to an excess compared to other reactants or because it acts as a catalyst.

- Example: CH 3 COOCH 3 + H 2 O → CH 3 COOH + CH 3 OH (pseudo-first order kinetics due to excess water).

Read more about Pseudo First Order Reaction .

## Second-Order Reaction

A second-order reaction has an order of 2. The rate of these reactions can be expressed as the square of the concentration of one reactant or as the product of the concentrations of two separate reactants.

- Example: r = k[A] 2 or r = k[A][B], e.g., NO 2 + CO → NO + CO 2

## Integrated Rate Law

Integrated rate law is a fundamental concept in chemical kinetics that describes the concentration of reactants or products over time during a chemical reaction. It provides a mathematical relationship between the concentration of a reactant or product and time, thereby allowing for the determination of reaction kinetics and rate constants.

Zero-Order Reaction Rate

In a zero-order reaction, the reaction rate is independent of the concentration of the reactant. The integrated rate equation for a zero-order reaction is:

[A] t = −k.t+[A] 0

First-Order Reaction Rate

In a first-order reaction, the reaction rate is directly proportional to the concentration of the reactant. The integrated rate equation for a first-order reaction is:

ln[A] t = −k.t +ln[A] 0

Second-Order Reaction Rate

In a second-order reaction, the reaction rate is proportional to the square of the concentration of the reactant. The integrated rate equation for a second-order reaction is:

1/[A] t = k.t+1/[A] 0

- [A] t is the concentration of the reactant at time t,
- k is the rate constant,
- [A] 0 is the initial concentration of the reactant.

Read More about Integrated Rate Laws .

## Half-Life of a Reaction

In chemical kinetics, the concept of half-life represents the time required for the concentration of a reactant to decrease to half of its initial value. The half-life is particularly useful in understanding the kinetics of first-order reactions, where the reaction rate is directly proportional to the concentration of the reactant.

For a first-order reaction, the relationship between the half-life (t 1/2 ), the rate constant k, and the initial concentration of the reactant [A 0 ]can be expressed as

t 1/2 =0.693/k

This equation indicates that the half-life of a first-order reaction is inversely proportional to the rate constant k. As the rate constant increases, the half-life decreases, indicating a faster rate of reaction.

## Some Other Terms Related to Chemical Kinetics

Some of the other common related terms to chemical kinetics are:

## Activation Energy

Collision theory.

Let’s discuss these terms in detail as follows.

In chemical kinetics, activation energy (E a ) defines the energy barrier that must be overcome for a chemical reaction to occur. It represents the minimum amount of energy required for reactant molecules to undergo a successful collision and transform into products. The concept of activation energy is central to the understanding of reaction rates.

The Arrhenius equation is a mathematical formula that explains the influence of temperature on the pace of a chemical reaction, serving as the basis for predictive models used to determine reaction-rate constants. It is expressed as an exponential function:

k = A. exp(- E/RT)

Where k denotes the reaction-rate constant, A signifies the collision frequency leading to the reaction, E represents the activation energy , R denotes the ideal gas constant (8.314 joules per kelvin per mole), and T is the absolute temperature.

Catalysis is a fundamental concept in chemical kinetics that involves the acceleration of chemical reactions by substances known as catalysts.

Catalysts are substances that facilitate chemical reactions by providing an alternative reaction pathway with a lower activation energy, thereby reducing the energy barrier required for the reaction to occur. There are two main types of catalysis:

- Homogeneous Catalysis: In homogeneous catalysis, the catalyst and the reactants are in the same phase (usually liquid or gas). The catalyst interacts with the reactants to form an intermediate complex, which then proceeds to form the products. The catalyst is typically present in small quantities relative to the reactants.
- Heterogeneous Catalysis: In heterogeneous catalysis, the catalyst is in a different phase from the reactants. For example, the reactants may be in the gas phase while the catalyst is a solid. The reactant molecules adsorb onto the surface of the catalyst, where the reaction takes place. The products then desorb from the catalyst surface.

Catalysis plays a crucial role in various industrial processes, including petroleum refining, chemical synthesis, and environmental remediation.

Collision Theory is a framework utilized to forecast the rates of chemical reactions, particularly in the context of gases. It states that for a reaction to occur, the reacting entities—be they atoms or molecules—must converge or collide. However, not all collisions result in chemical transformation.

An effective collision leading to chemical change necessitates that the colliding entities possess a minimum level of internal energy, equivalent to the activation energy of the reaction. Additionally, the colliding entities must be arranged in a manner conducive to the required rearrangement of atoms and electrons.

## Conclusion: Chemical Kinetics

In conclusion, chemical kinetics stands as a fundamental discipline within the realm of physical chemistry, shedding light on the intricacies of reaction rates and mechanisms. From industrial applications to understanding biological processes, chemical kinetics plays a pivotal role in diverse scientific endeavors. Its significance extends beyond the laboratory, impacting areas such as material synthesis, environmental studies, and pharmaceutical development.

Factors Affecting Rate of a Chemical Reaction Temperature Dependence of the Rate of a Reaction Rate of Reaction

## Solved Example on Chemical Kinetics

Let’s solve some example problems on Chemical Kinetics

Example 1 : Identify the reaction order from each of the following rate constants.

- k = 2.3 × 10 -5 L mol -1 s -1
- k = 3 × 10 -4 s -1

(1) The unit of second order rate constant is L mol -1 s -1 ,therefore k = 2.3 × 10 -5 L mol -1 s -1 represents a second order reaction. (2) The unit of a first-order rate constant is s -1 therefore k = 3 × 10 -4 s -1 represents a first-order reaction.

Example 2: A first-order reaction is found to have a rate constant, k = 5.5 × 10 -14 s -1 . Find the half-life of the reaction.

The half-life for a first-order reaction is t 1/2 =0.693/k So here, t 1/2 = 0.693/ (5.5 × 10 -14 ) s -1 = 1.26 × 10 13 s

Example 3: Calculate the overall order of a reaction which has the rate expression

- Rate = k [A] 1/2 [B] 3/2
- Rate = k [A] 3/2 [B] -1

Rate = k [A] x [B] y Order = x + y 1. Order = 1/2 + 3/2 = 2, i.e., hence the reaction is of second order 2. Order = 3/2 + (–1) = 1/2, i.e., hence the reaction is of half order.

## Chemical Kinetics FAQs

Chemical kinetics is the branch of physical chemistry focused on understanding the rates of chemical reactions.

## Who is the Father of Chemical Kinetics?

Ludwig Ferdinand Wilhelminy, commonly known as Ludwig Wilhelmy, is considered the Father of Chemical Kinetics due to his pioneering work studying the rate of inversion of sucrose in 1850.

## Why do we Study Chemical Kinetics?

Chemical kinetics is studied to understand reaction mechanisms, improve control over reactions, optimize industrial processes, construct mathematical models, and contribute to fundamental scientific curiosity.

## What is K in Chemical Kinetics?

In chemical kinetics, the symbol K represents the reaction rate constant, which is a proportionality constant that quantifies the rate of a chemical reaction.

## What is the Basic Theory of Chemical Kinetics?

The basic theory of chemical kinetics involves understanding the rates of chemical reactions and the factors that influence them.

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## Solved Objective Problems: Chemical Kinetics | Chemistry Class 12 - NEET PDF Download

Problem : 1

In gaseous reactions important for understanding of the upper atmosphere H 2 O and O react bimolecularly to form two OH radicals. DH for this reaction is 72kJ/mol at 500 K and E a = 77 kJ/mol, then E a for two bimolecular recombination of 2OH radicals to form H 2 O & O is

(A) 3 kJ mol -1 (B) 4 kJ mol -1 (C) 5 kJ mol -1 (D) 7 kJ mol -1

Sol. As ΔH is positive, therefore reaction is endothermic

This is the energy profile diagram for an endothermic reaction.

Now when the products is converted back to reactant the energy of activation is x as shown in fig.

Evidently x = E a - ΔH

= (77 - 72) = 5 kJ mol -1 Therefore, (C)

Problem : 2

In a certain reaction 10% of the reactant decomposes in the first hour, 20% is second hour, 30% in third hour and so on. What are the dimensions of rate constant.

(A) hour -1 (B) mol lit -1 sec -1 (C) lit mol -1 sec -1 (D) mol sec -1

Sol. If the amount of products formed which is 10%, 20% and 30% is plotted against time i.e., 1 hr, 2 hr and 3 hr respectively, it is a straight line passing through the origin.

Therefore, dimensions of k = moles lit -1 sec -1

Therefore, (B)

Problem : 3

Two substances A(t½ = 5 mins) and B(t½ = 15 mins) are taken is such a way that initially [A] = 4[B]. The time after which the conentration of both will be equal is

(A) 5 min

(B) 15 min

(D) concentration can never be equal

Sol. t½ of A is 5 min

Therefore, in 15 mins it will become 1/8 of initial and t½ of B is 15 mins

Therefore, in 15 mins it will become ½ of initial

Therefore, ratio of [A] : [B] after 15 min is 4 : 1

But given [A] = 4[B]

Therefore, [A] = [B] after 15 min

Therefore, [B]

Problem : 4

The reaction A(g) 2B(g) → C(g) D(g) is an elementary process. In an experiment, the initial partial pressure of A & B are P A = 0.60 and P B = 0.80 atm. When P c = 0.2 atm the rate of reaction relative to the initial rate is

(A) 1/48

(B) 1/24

(C) 9/16

(D) 1/6

(Rate) i = k[A] [B] 2 = k[0.6][0.8] 2

(Rate) t = k[0.4] [0.4] 2

Problem : 5

For a hypothetical reaction A B → C D, the rate = k[A] -1/2 [B] 3/2 . On doubling the concentration of A and B the rate will be

(A) 4 times

(B) 2 times

(C) 3 times

(D) none of these

Sol. k = k[2] -1/2 [2] 3/2 = k[2] 3/2-1/2 = k = [2] 1 = 2k

Problem : 6

An organic compound A decomposes by following two parallel first order mechanisms :

Select the correct statement(s)

(A) If three moles of A are completely decomposed then 2 moles of B and 1 mole of C will be formed.

(B) If three moles of A are completely decomposed then 1 moles of B and 2 mole of C will be formed.

(C) half life for the decomposition of A is 20 min

(D) half life for the decomposition of B is 0.33 min

Sol. BC

Problem : 7

For any I st order gaseous reaction A → 2B Pressure devoloped after 20 min is 16.4 atm and after very long time is 20 atm. What is the total pressure developed after 10 min.

(A) 12

(B) 13

(C) 14

(D) 15

t = 0 P O -

t = 10 P O - P 2P

t = 20 P O - P' 2P'

16.4 = P O - P' 2P'

16.4 = 10 P'

for first order reaction at equal time conc/pr is in G. P.

3.6 × 10 = (10 - P) 2

P 10 min = P O P = 10 4 = 14 atm

Problem : 8

For any reaction, 2A → B, rate constant of reaction is 0.231 min -1 . Time (in sec) when 25% of A will remain unreacted.

(A) 150

(B) 180

(C) 200

(D) 140

Sol. B

Problem : 9

For any reaction A(g) → B(g), rate constant k = 8.21 × 10 -2 atm/min at 300 K. If initial concentration of A is 2M then what is the half life (in hr.)?

(A) 5

(B) 6

(C) 7

(D) 8

Problem : 10

The mechanism of the above reaction given as

E 1 = Activation energy for K 1

E 2 = Activation energy for K 2

E 3 = Activation energy for K 3

Calculate E reaction

(given : E 1 = 10 kJ, E 3 = 5 kJ, E 2 = 12 kJ)

(D) 3

rate = K 3 [x][B]

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## Chapter 15.3: Solving Equilibrium Problems

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## Learning Objective

- To solve quantitative problems involving chemical equilibriums.

There are two fundamental kinds of equilibrium problems: (1) those in which we are given the concentrations of the reactants and the products at equilibrium (or, more often, information that allows us to calculate these concentrations), and we are asked to calculate the equilibrium constant for the reaction; and (2) those in which we are given the equilibrium constant and the initial concentrations of reactants, and we are asked to calculate the concentration of one or more substances at equilibrium. In this section, we describe methods for solving both kinds of problems.

## Calculating an Equilibrium Constant from Equilibrium Concentrations

We saw in the exercise in Example 6 in Section 15.2 that the equilibrium constant for the decomposition of CaCO 3 (s) to CaO(s) and CO 2 (g) is K = [CO 2 ]. At 800°C, the concentration of CO 2 in equilibrium with solid CaCO 3 and CaO is 2.5 × 10 −3 M. Thus K at 800°C is 2.5 × 10 −3 . (Remember that equilibrium constants are unitless.)

A more complex example of this type of problem is the conversion of n -butane, an additive used to increase the volatility of gasoline, to isobutane (2-methylpropane). This reaction can be written as follows:

\( n-butane \left ( g \right ) \rightleftharpoons isobutane \left ( g \right ) \tag{15.3.1} \)

and the equilibrium constant K = [isobutane]/[ n -butane]. At equilibrium, a mixture of n -butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n -butane. Substituting these concentrations into the equilibrium constant expression,

\( K= \dfrac{isobutane}{n-butane}=\dfrac{0.041\;\cancel{M}}{0.016 \; \cancel{M}} = 2.6 \tag{15.3.2} \)

Thus the equilibrium constant for the reaction as written is 2.6.

## Example 15.3.1

The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid:

\( 2SO_{2}\left ( g \right ) + O_{2}\left ( g \right ) \rightleftharpoons 2SO_{3}\left ( g \right ) \)

A mixture of SO 2 and O 2 was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained 5.0 × 10 −2 M SO 3 , 3.5 × 10 −3 M O 2 , and 3.0 × 10 −3 M SO 2 . Calculate K and K p at this temperature.

Given: balanced equilibrium equation and composition of equilibrium mixture

Asked for: equilibrium constant

Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain K .

Substituting the appropriate equilibrium concentrations into the equilibrium constant expression,

\( K=\dfrac{\left [ SO_{3} \right ]^{2}}{\left [ SO_{2} \right ]^{2}\left [ O_{2} \right ]}=\dfrac{\left ( 5.0\times 10^{-2} \right )^{2}}{\left ( 3.0\times 10^{-3} \right )^{2}\left ( 3.5\times 10^{-3} \right )}=7.9\times 10^{4} \)

To solve for K p , we use Equation 15.2.17 , where Δ n = 2 − 3 = −1:

\( K_{p}= K\left ( RT \right )^{\Delta n} \) \( =7.9\times 10^{4}\left [ \left (0.082606\; L\cdot atm/mol\cdot \cancel{K} \right ) \left ( 800 \; \cancel{K} \right )\right ] \) \( =1.2\times 10^{3}\)

Hydrogen gas and iodine react to form hydrogen iodide via the reaction

\( H_{2}\left ( g \right ) + I_{2}\left ( g \right ) \rightleftharpoons 2HI\left ( g \right ) \)

A mixture of H 2 and I 2 was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained 1.37 × 10 −2 M HI, 6.47 × 10 −3 M H 2 , and 5.94 × 10 −4 M I 2 . Calculate K and K p for this reaction.

Answer: K = 48.8; K p = 48.8

Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example 9 shows one way to do this.

## Example 15.3.2

A 1.00 mol sample of NOCl was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of Cl 2 . Calculate K at this temperature. The equation for the decomposition of NOCl to NO and Cl 2 is as follows:

\( 2NOCl \left ( g \right ) \rightleftharpoons 2NO\left ( g \right ) + Cl_{2}\left ( g \right ) \)

Given: balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium

Asked for: K

A Write the equilibrium constant expression for the reaction. Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations).

B Calculate all possible initial concentrations from the data given and insert them in the table.

C Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Insert those concentration changes in the table.

D Obtain the final concentrations by summing the columns. Calculate the equilibrium constant for the reaction.

A The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows:

\( K=\dfrac{\left [ NO_{2} \right ]^{2}\left [ Cl_{2} \right ]}{\left [ NOCl \right ]^{2}} \)

To obtain the concentrations of NOCl, NO, and Cl 2 at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations.

B Initially, the system contains 1.00 mol of NOCl in a 2.00 L container. Thus [NOCl] i = 1.00 mol/2.00 L = 0.500 M. The initial concentrations of NO and Cl 2 are 0 M because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of Cl 2 in a 2.00 L container, so [Cl 2 ] f = 0.056 mol/2.00 L = 0.028 M. We insert these values into the following table:

C We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of Cl 2 , the substance for which initial and final concentrations are known:

Δ[Cl 2 ] = [0.028 M (final) − 0.00 M (initial)] = +0.028 M

According to the coefficients in the balanced chemical equation, 2 mol of NO are produced for every 1 mol of Cl 2 , so the change in the NO concentration is as follows:

\( \Delta \left [ NO \right ] = \left ( \dfrac{0.028 \; \cancel{mol\;Cl_{2}}}{L} \right )\left ( \dfrac{2\; mol\; NO}{1\;\cancel{mol\;Cl_{2}}} \right )=0.056\; M \)

Similarly, 2 mol of NOCl are consumed for every 1 mol of Cl 2 produced, so the change in the NOCl concentration is as follows:

\( \Delta \left [ NOCl \right ] = \left ( \dfrac{0.028 \; \cancel{mol\;Cl_{2}}}{L} \right )\left ( \dfrac{-2\; mol\; NO}{1\;\cancel{mol\;Cl_{2}}} \right )=-0.056\; M \)

We insert these values into our table:

D We sum the numbers in the [NOCl] and [NO] columns to obtain the final concentrations of NO and NOCl:

[NO] f = 0.000 M + 0.056 M = 0.056 M [NOCl] f = 0.500 M + (−0.056 M) = 0.444 M

We can now complete the table:

We can now calculate the equilibrium constant for the reaction:

\( K=\dfrac{\left [ NO_{2} \right ]^{2}\left [ Cl_{2} \right ]}{\left [ NOCl \right ]^{2}}=\dfrac{\left ( 0.056 \right )^{2}\left ( 0.028 \right )}{0.444}^{2}=4.5\times 10^{-4} \)

The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia (NH 3 ) by reacting 0.1248 M H 2 and 0.0416 M N 2 at about 500°C. At equilibrium, the mixture contained 0.00272 M NH 3 . What is K for the reaction N 2 + 3 H 2 ⇌ 2NH 3 at this temperature? What is K p ?

Answer: K = 0.105; K p = 2.61 × 10 −5

The original laboratory apparatus designed by Fritz Haber and Robert Le Rossignol in 1908 for synthesizing ammonia from its elements. A metal catalyst bed, where ammonia was produced, is in the large cylinder at the left. The Haber-Bosch process used for the industrial production of ammonia uses essentially the same process and components but on a much larger scale. Unfortunately, Haber’s process enabled Germany to prolong World War I when German supplies of nitrogen compounds, which were used for explosives, had been exhausted in 1914.

## Calculating Equilibrium Concentrations from the Equilibrium Constant

To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of n -butane to isobutane (Equation 15.26), for which K = 2.6 at 25°C. If we begin with a 1.00 M sample of n -butane, we can determine the concentration of n -butane and isobutane at equilibrium by constructing a table showing what is known and what needs to be calculated, just as we did in Example 9.

The initial concentrations of the reactant and product are both known: [ n -butane] i = 1.00 M and [isobutane] i = 0 M. We need to calculate the equilibrium concentrations of both n -butane and isobutane. Because it is generally difficult to calculate final concentrations directly, we focus on the change in the concentrations of the substances between the initial and the final (equilibrium) conditions. If, for example, we define the change in the concentration of isobutane (Δ[isobutane]) as + x , then the change in the concentration of n -butane is Δ[ n -butane] = − x . This is because the balanced chemical equation for the reaction tells us that 1 mol of n -butane is consumed for every 1 mol of isobutane produced. We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone.

Substituting the expressions for the final concentrations of n -butane and isobutane from the table into the equilibrium equation,

\( K=\dfrac{\left [ isobutane \right]}{\left [ n-butane \right ]}=\dfrac{x}{1.00-x}=2.6 \)

Rearranging and solving for x ,

\( x = 2.6\left ( 1.00-x \right )=2.6-2.6x \) \( x + 2.6x =2.6 \) \( x = 0.72 \)

We obtain the final concentrations by substituting this x value into the expressions for the final concentrations of n -butane and isobutane listed in the table:

[ n -butane] f = (1.00 − x ) M = (1.00 − 0.72) M = 0.28 M [isobutane] f = (0.00 + x ) M = (0.00 + 0.72) M = 0.72 M

We can check the results by substituting them back into the equilibrium constant expression to see whether they give the same K that we used in the calculation:

\( K=\dfrac{\left [ isobutane \right]}{\left [ n-butane \right ]}=\dfrac{0.72 \; \cancel{M}}{0.28 \; \cancel{M}}=2.6 \)

This is the same K we were given, so we can be confident of our results.

Example 10 illustrates a common type of equilibrium problem that you are likely to encounter.

## Example 15.3.3

The water–gas shift reaction is important in several chemical processes, such as the production of H 2 for fuel cells. This reaction can be written as follows:

\( H_{2}\left ( g \right ) + CO_{2}\left ( g \right ) \rightleftharpoons H_{2}O\left ( g \right ) + CO\left ( g \right )\)

K = 0.106 at 700 K. If a mixture of gases that initially contains 0.0150 M H 2 and 0.0150 M CO 2 is allowed to equilibrate at 700 K, what are the final concentrations of all substances present?

Given: balanced equilibrium equation, K , and initial concentrations

Asked for: final concentrations

A Construct a table showing what is known and what needs to be calculated. Define x as the change in the concentration of one substance. Then use the reaction stoichiometry to express the changes in the concentrations of the other substances in terms of x . From the values in the table, calculate the final concentrations.

B Write the equilibrium equation for the reaction. Substitute appropriate values from the table to obtain x .

C Calculate the final concentrations of all species present. Check your answers by substituting these values into the equilibrium constant expression to obtain K .

A The initial concentrations of the reactants are [H 2 ] i = [CO 2 ] i = 0.0150 M. Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of H 2 O as x , then Δ[H 2 O] = + x . We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of x . For example, 1 mol of CO is produced for every 1 mol of H 2 O, so the change in the CO concentration can be expressed as Δ[CO] = + x . Similarly, for every 1 mol of H 2 O produced, 1 mol each of H 2 and CO 2 are consumed, so the change in the concentration of the reactants is Δ[H 2 ] = Δ[CO 2 ] = − x . We enter the values in the following table and calculate the final concentrations.

B We can now use the equilibrium equation and the given K to solve for x :

\( K=\dfrac{\left [ H_{2}O] \right ] \left [ CO \right ]}{\left [ H_{2} \right ]\left [ CO_{2} \right ]}=\dfrac{\left (x \right )\left ( x \right ) }{\left ( 0.0150-x \right )\left ( 0.0150-x \right )}=\dfrac{x^{2}}{\left ( 0.0150-x \right )^{2}}=0.160 \notag \)

We could solve this equation with the quadratic formula, but it is far easier to solve for x by recognizing that the left side of the equation is a perfect square; that is,

\[\dfrac{x^2}{(0.0150−x)^2}=\left(\dfrac{x}{0.0150−x}\right)^2=0.106 \notag \]

(The quadratic formula is presented in Essential Skills 7 in Section 15.7 .) Taking the square root of the middle and right terms,

\[\dfrac{x^2}{(0.0150−x)^2} =(0.106)^{1/2}=0.326 \notag \]

\[x =(0.326)(0.0150)−0.326x \notag \]

\[1.326x=0.00489 \notag \]

\[x =0.00369=3.69 \times 10^{−3} \notag \]

C The final concentrations of all species in the reaction mixture are as follows:

- \([H_2]_f=[H_2]_i+Δ[H_2]=(0.0150−0.00369) \;M=0.0113\; M\)
- \([CO_2]_f =[CO_2]_i+Δ[CO_2]=(0.0150−0.00369)\; M=0.0113\; M\)
- \([H_2O]_f=[H_2O]_i+Δ[H_2O]=(0+0.00369) \;M=0.00369\; M\)
- \([CO]_f=[CO]_i+Δ[CO]=(0+0.00369)\; M=0.00369 \;M\)

We can check our work by inserting the calculated values back into the equilibrium constant expression:

\[K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.00369)^2}{(0.0113)^2}=0.107 \notag \]

To two significant figures, this K is the same as the value given in the problem, so our answer is confirmed.

Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation:

\[H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)} \notag \]

K = 54 at 425°C. If 0.172 M H 2 and I 2 are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture?

Answer: [HI] f = 0.270 M; [H 2 ] f = [I 2 ] f = 0.037 M

In Example 10, the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. Such a case is described in Example 11.

## Example 15.3.4

In the water–gas shift reaction shown in Example 10, a sample containing 0.632 M CO 2 and 0.570 M H 2 is allowed to equilibrate at 700 K. At this temperature, K = 0.106. What is the composition of the reaction mixture at equilibrium?

Given: balanced equilibrium equation, concentrations of reactants, and K

Asked for: composition of reaction mixture at equilibrium

A Write the equilibrium equation. Construct a table showing the initial concentrations of all substances in the mixture. Complete the table showing the changes in the concentrations ( x ) and the final concentrations.

B Write the equilibrium constant expression for the reaction. Substitute the known K value and the final concentrations to solve for x .

C Calculate the final concentration of each substance in the reaction mixture. Check your answers by substituting these values into the equilibrium constant expression to obtain K .

A [CO 2 ] i = 0.632 M and [H 2 ] i = 0.570 M. Again, x is defined as the change in the concentration of H 2 O: Δ[H 2 O] = + x . Because 1 mol of CO is produced for every 1 mol of H 2 O, the change in the concentration of CO is the same as the change in the concentration of H 2 O, so Δ[CO] = + x . Similarly, because 1 mol each of H 2 and CO 2 are consumed for every 1 mol of H 2 O produced, Δ[H 2 ] = Δ[CO 2 ] = − x . The final concentrations are the sums of the initial concentrations and the changes in concentrations at equilibrium.

B We can now use the equilibrium equation and the known K value to solve for x :

\[K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{x^2}{(0.570−x)(0.632−x)}=0.106 \notag \]

In contrast to Example 10, however, there is no obvious way to simplify this expression. Thus we must expand the expression and multiply both sides by the denominator:

\[x^2 = 0.106(0.360 − 1.20x + x^2) \notag \]

Collecting terms on one side of the equation,

\[0.894x^2 + 0.127x − 0.0382 = 0 \notag \]

This equation can be solved using the quadratic formula:

\[ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{−0.127 \pm \sqrt{(0.127)^2−4(0.894)(−0.0382)}}{2(0.894)} \notag \]

\[x =0.148 \text{ and } −0.290 \notag \]

Only the answer with the positive value has any physical significance, so Δ[H 2 O] = Δ[CO] = +0.148 M, and Δ[H 2 ] = Δ[CO 2 ] = −0.148 M.

- \([H_2]_f[ = [H_2]_i+Δ[H_2]=0.570 \;M −0.148\; M=0.422 M\)
- \([CO_2]_f =[CO_2]_i+Δ[CO_2]=0.632 \;M−0.148 \;M=0.484 M\)
- \([H_2O]_f =[H_2O]_i+Δ[H_2O]=0\; M+0.148\; M =0.148\; M\)
- \([CO]_f=[CO]_i+Δ[CO]=0 M+0.148\;M=0.148 M\)

We can check our work by substituting these values into the equilibrium constant expression:

\[K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.148)^2}{(0.422)(0.484)}=0.107 \notag \]

Because K is essentially the same as the value given in the problem, our calculations are confirmed.

The exercise in Example 8 showed the reaction of hydrogen and iodine vapor to form hydrogen iodide, for which K = 54 at 425°C. If a sample containing 0.200 M H 2 and 0.0450 M I 2 is allowed to equilibrate at 425°C, what is the final concentration of each substance in the reaction mixture?

Answer: [HI] f = 0.0882 M; [H 2 ] f = 0.156 M; [I 2 ] f = 9.2 × 10 −4 M

In many situations it is not necessary to solve a quadratic (or higher-order) equation. Most of these cases involve reactions for which the equilibrium constant is either very small ( K ≤ 10 −3 ) or very large ( K ≥ 10 3 ), which means that the change in the concentration (defined as x ) is essentially negligible compared with the initial concentration of a substance. Knowing this simplifies the calculations dramatically, as illustrated in Example 12.

## Example 15.3.5

Atmospheric nitrogen and oxygen react to form nitric oxide:

\[N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)} \notag \]

K p = 2.0 × 10 −31 at 25°C. What is the partial pressure of NO in equilibrium with N 2 and O 2 in the atmosphere (at 1 atm, P{N 2 } = 0.78 atm and P{O 2 } = 0.21 atm

Given: balanced equilibrium equation and values of K p , P{O 2 } and P{N 2 }

Asked for: partial pressure of NO

A Construct a table and enter the initial partial pressures, the changes in the partial pressures that occur during the course of the reaction, and the final partial pressures of all substances.

B Write the equilibrium equation for the reaction. Then substitute values from the table to solve for the change in concentration ( x ).

C Calculate the partial pressure of NO. Check your answer by substituting values into the equilibrium equation and solving for K .

A Because we are given K p and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of O 2 is 0.21 atm and that of N 2 is 0.78 atm. If we define the change in the partial pressure of NO as 2 x , then the change in the partial pressure of O 2 and of N 2 is − x because 1 mol each of N 2 and of O 2 is consumed for every 2 mol of NO produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium.

B Substituting these values into the equation for the equilibrium constant,

\[K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(2x)^2}{(0.78−x)(0.21−x)}=2.0 \times 10^{−31} \notag \]

In principle, we could multiply out the terms in the denominator, rearrange, and solve the resulting quadratic equation. In practice, it is far easier to recognize that an equilibrium constant of this magnitude means that the extent of the reaction will be very small; therefore, the x value will be negligible compared with the initial concentrations. If this assumption is correct, then to two significant figures, (0.78 − x ) = 0.78 and (0.21 − x ) = 0.21. Substituting these expressions into our original equation,

\[\dfrac{(2x)^2}{(0.78)(0.21)} = 2.0 \times 10^{−31} \notag \]

\[\dfrac{4x^2}{0.16} =2.0 \times10^{−31} \notag \]

\[x^2=\dfrac{0.33 \times 10^{−31}}{4} \notag \]

\[x^=9.1 \times 10^{−17} \notag \]

C Substituting this value of x into our expressions for the final partial pressures of the substances,

- \(P_{NO}=2x \; atm=1.8 \times 10^{−16} \;atm \)
- \(P_{N_2}=(0.78−x) \;atm=0.78 \;atm \)
- \(P_{O_2}=(0.21−x) \;atm=0.21\; atm\)

From these calculations, we see that our initial assumption regarding x was correct: given two significant figures, 2.0 × 10 −16 is certainly negligible compared with 0.78 and 0.21. When can we make such an assumption? As a general rule, if x is less than about 5% of the total, or 10 −3 > K > 10 3 , then the assumption is justified. Otherwise, we must use the quadratic formula or some other approach. The results we have obtained agree with the general observation that toxic NO, an ingredient of smog, does not form from atmospheric concentrations of N 2 and O 2 to a substantial degree at 25°C. We can verify our results by substituting them into the original equilibrium equation:

\[K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(1.8 \times 10^{−16})^2}{(0.78)(0.21)}=2.0 times 10^{−31} \notag \]

The final K p agrees with the value given at the beginning of this example.

Under certain conditions, oxygen will react to form ozone, as shown in the following equation:

\[H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)} \notag \]

K p = 2.5 × 10 −59 at 25°C. What ozone partial pressure is in equilibrium with oxygen in the atmosphere P(O 2 ) =0.21 atm ?

Answer: 4.8 × 10 −31 atm

Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large ( K ≥ 10 3 ). A large equilibrium constant implies that the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. When we solve this type of problem, we view the system as equilibrating from the products side of the reaction rather than the reactants side. This approach is illustrated in Example 13.

## Example 15.3.6

The chemical equation for the reaction of hydrogen with ethylene (C 2 H 4 ) to give ethane (C 2 H 6 ) is as follows:

K = 9.6 × 10 18 at 25°C. If a mixture of 0.200 M H 2 and 0.155 M C 2 H 4 is maintained at 25°C in the presence of a powdered nickel catalyst, what is the equilibrium concentration of each substance in the mixture?

Given: balanced chemical equation, K , and initial concentrations of reactants

Asked for: equilibrium concentrations

A Construct a table showing initial concentrations, concentrations that would be present if the reaction were to go to completion, changes in concentrations, and final concentrations.

B Write the equilibrium constant expression for the reaction. Then substitute values from the table into the expression to solve for x (the change in concentration).

C Calculate the equilibrium concentrations. Check your answers by substituting these values into the equilibrium equation.

A From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. If the reaction were to go to completion, the concentration of ethane would be 0.155 M and the concentration of ethylene would be 0 M. Because the concentration of hydrogen is greater than what is needed for complete reaction, the concentration of unreacted hydrogen in the reaction mixture would be 0.200 M − 0.155 M = 0.045 M. The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example 12. If we define − x as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is + x . The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions.

B Substituting values into the equilibrium constant expression,

\[K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155−x}{(0.045+x)x}=9.6 \times 10^{18} \notag \]

Once again, the magnitude of the equilibrium constant tells us that the equilibrium will lie far to the right as written, so the reverse reaction is negligible. Thus x is likely to be very small compared with either 0.155 M or 0.045 M, and the equation can be simplified [(0.045 + x ) = 0.045 and (0.155 − x ) = 0.155] as follows:

\[K=\dfrac{0.155}{0.045x} = 9.6 \times 10^{18} \notag \]

\[x=3.6 \times 10^{−19} \notag \]

C The small x value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table:

- \([C_2H_6]_f = (0.155 − x)\; M = 0.155 \; M\)
- \([C_2H_4]_f = x\; M = 3.6 \times 10^{−19} M \)
- \([H_2]_f = (0.045 + x) \;M = 0.045 \;M\)

We can verify our calculations by substituting the final concentrations into the equilibrium constant expression:

\[K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155}{(0.045)(3.6 \times 10^{−19})}=9.6 \times 10^{18} \notag \]

This K value agrees with our initial value at the beginning of the example.

Hydrogen reacts with chlorine gas to form hydrogen chloride:

\[H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)} \notag \]

K p = 4.0 × 10 31 at 47°C. If a mixture of 0.257 M H 2 and 0.392 M Cl 2 is allowed to equilibrate at 47°C, what is the equilibrium composition of the mixture?

- \([H_2]_f = 4.8 \times 10^{−32}\; M\)
- \([Cl_2]_f = 0.135\; M\)
- \([HCl]_f = 0.514\; M\)

When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium.

## Key Takeaway

- Various methods can be used to solve the two fundamental types of equilibrium problems: (1) those in which we calculate the concentrations of reactants and products at equilibrium and (2) those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture.

## Conceptual Problems

Describe how to determine the magnitude of the equilibrium constant for a reaction when not all concentrations of the substances are known.

Calculations involving systems with very small or very large equilibrium constants can be dramatically simplified by making certain assumptions about the concentrations of products and reactants. What are these assumptions when K is (a) very large and (b) very small? Illustrate this technique using the system A + 2B ⇌ C for which you are to calculate the concentration of the product at equilibrium starting with only A and B. Under what circumstances should simplifying assumptions not be used?

## Numerical Problems

Please be sure you are familiar with the topics discussed in Essential Skills 7 ( Section 15.7 ) before proceeding to the Numerical Problems.

In the equilibrium reaction A + B ⇌C, what happens to K if the concentrations of the reactants are doubled? tripled? Can the same be said about the equilibrium reaction A ⇌B + C?

The following table shows the reported values of the equilibrium P{O 2 } at three temperatures for the reaction Ag 2 O(s) ⇌ 2 Ag(s) + 1/2 O 2 (g) for which Δ H ° = 31 kJ/mol. Are these data consistent with what you would expect to occur? Why or why not?

Given the equilibrium system N 2 O 4 (g) ⇌ 2 NO 2 (g), what happens to K p if the initial pressure of N 2 O 4 is doubled? If K p is 1.7 × 10 −1 at 2300°C, and the system initially contains 100% N 2 O 4 at a pressure of 2.6 × 10 2 atm, what is the equilibrium pressure of each component?

At 430°C, 4.20 mol of HI in a 9.60 L reaction vessel reaches equilibrium according to the following equation: H 2 (g) + I 2 (g) ⇌2HI(g) At equilibrium, [H 2 ] = 0.047 M and [HI] = 0.345 M. What are K and K p for this reaction?

Methanol, a liquid used as an automobile fuel additive, is commercially produced from carbon monoxide and hydrogen at 300°C according to the following reaction: CO(g) + 2H 2 (g) ⇌ CH 3 OH(g) and K p = 1.3 × 10 −4 . If 56.0 g of CO is mixed with excess hydrogen in a 250 mL flask at this temperature, and the hydrogen pressure is continuously maintained at 100 atm, what would be the maximum percent yield of methanol? What pressure of hydrogen would be required to obtain a minimum yield of methanol of 95% under these conditions?

Starting with pure A, if the total equilibrium pressure is 0.969 atm for the reaction A (s ⇌ 2 B(g) + C(g), what is K p ?

The decomposition of ammonium carbamate to NH 3 and CO 2 at 40°C is written as NH 4 CO 2 NH 2 (s) ⇌ 2NH 3 (g) + CO 2 If the partial pressure of NH 3 at equilibrium is 0.242 atm, what is the equilibrium partial pressure of CO 2 ? What is the total gas pressure of the system? What is K p ?

At 375 K, K p for the reaction SO 2 Cl 2 (g) ⇌ SO 2 (g) + Cl 2 g) is 2.4, with pressures expressed in atmospheres. At 303 K, K p is 2.9 × 10 −2 .

- What is K for the reaction at each temperature?
- If a sample at 375 K has 0.100 M Cl 2 and 0.200 M SO 2 at equilibrium, what is the concentration of SO 2 Cl 2 ?
- If the sample given in part b is cooled to 303 K, what is the pressure inside the bulb?

For the gas-phase reaction aA ⇌ bB, show that K p = K ( RT ) Δ n assuming ideal gas behavior.

For the gas-phase reaction I 2 ⇌2I, show that the total pressure is related to the equilibrium pressure by the following equation:

\[P_T=\sqrt{K_pP_{I_2}} + P_{I_2} \notag \]

Experimental data on the system Br2(l) ⇌ Br 2 (aq) are given in the following table. Graph [Br 2 ] versus moles of Br 2 (l) present; then write the equilibrium constant expression and determine K .

Data accumulated for the reaction n- butane(g) ⇌ isobutane(g) at equilibrium are shown in the following table. What is the equilibrium constant for this conversion? If 1 mol of n -butane is allowed to equilibrate under the same reaction conditions, what is the final number of moles of n -butane and isobutane?

Solid ammonium carbamate (NH 4 CO 2 NH 2 ) dissociates completely to ammonia and carbon dioxide when it vaporizes:

\[ NH_4CO_2NH_{2(s)} \rightleftharpoons 2NH_{3(g)}+CO_{2(g)} \notag \]

At 25°C, the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium partial pressure of each gas? What is K p ? If the concentration of CO 2 is doubled and then equilibrates to its initial equilibrium partial pressure + x atm, what change in the NH 3 concentration is necessary for the system to restore equilibrium?

The equilibrium constant for the reaction COCl 2 (g) ⇌ CO(g) + Cl 2 (g) is K p = 2.2 × 10 −10 at 100°C. If the initial concentration of COCl 2 is 3.05 × 10 −3 M, what is the partial pressure of each gas at equilibrium at 100°C? What assumption can be made to simplify your calculations?

Aqueous dilution of IO 4 − results in the following reaction:

\[IO^−_{4(aq)}+2H_2O_{(l)} \rightleftharpoons H_4IO^−_{6(aq)} \notag \]

and K = 3.5 × 10 −2 . If you begin with 50 mL of a 0.896 M solution of IO 4 − that is diluted to 250 mL with water, how many moles of H 4 IO 6 − are formed at equilibrium?

Iodine and bromine react to form IBr, which then sublimes. At 184.4°C, the overall reaction proceeds according to the following equation:

\[I_{2(g)}+Br_{2(g)} \rightleftharpoons 2IBr_{(g)} \notag \]

K p = 1.2 × 10 2 . If you begin the reaction with 7.4 g of I 2 vapor and 6.3 g of Br 2 vapor in a 1.00 L container, what is the concentration of IBr(g) at equilibrium? What is the partial pressure of each gas at equilibrium? What is the total pressure of the system?

For the reaction

\[C_{(s)} + 12N_{2(g)}+\frac{5}{2}H_{2(g)} \rightleftharpoons CH3NH2(g) \notag \]

K = 1.8 × 10 −6 . If you begin the reaction with 1.0 mol of N 2 , 2.0 mol of H 2 , and sufficient C(s) in a 2.00 L container, what are the concentrations of N 2 and CH 3 NH 2 at equilibrium? What happens to K if the concentration of H 2 is doubled?

## Contributors

Modified by Joshua B. Halpern

## IMAGES

## VIDEO

## COMMENTS

1.E: Kinetics (Practice Problems with Answers) Page ID. A general chemistry Libretexts Textmap organized around the textbook. Chemistry: The Central Science. by Brown, LeMay, Bursten, Murphy, and Woodward. These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General ...

Solution: Evaluate yourself 1 1) Write the rate expression for the following reactions, assuming them as elementary reactions. i) 3A + 5B2 →4CD ii) X2 + Y2 →2XY 2). Consider the decomposition of N2O5 (g) to form NO2 (g) and O2 (g) . At a particular instant N2O5 disappears at a rate of 2.5x10-2 mol dm-3s-1 . At what rates are NO2 and O2 formed?

Practice Problems Chemical Kinetics: Rates and Mechanisms of Chemical Reactions 1. State two quantities that must be measured to establish the rate of a chemical reaction and cite several factors that affect the rate of a chemical reaction. Answer The rate of a reaction is defined as the change in concentration as a function of time.

Where you yourself need to first deduce the rate law, then plug in the values to solve for "k". 31. The initial rate data for the reaction 2N2O5(g) → 4NO2(g) + O2(g) is shown in the following table. Determine the value of the rate constant for this reaction. Experiment [N2O5](M) Rate (M/s) 1.28 × 102 22.5.

Course: MCAT > Unit 9. Lesson 18: Kinetics. Kinetics questions. Introduction to reaction rates. Rate law and reaction order. Worked example: Determining a rate law using initial rates data. First-order reaction (with calculus) Plotting data for a first-order reaction. Half-life of a first-order reaction.

Practice Problem 9: Acetaldehyde, CH 3 CHO, decomposes by second-order kinetics with a rate constant of 0.334 M-1 s-1 at 500C. Calculate the amount of time it would take for 80% of the acetaldehyde to decompose in a sample that has an initial concentration of 0.00750 M. Click here to check your answer to Practice Problem 9.

Introduction to reaction rates Rate law and reaction order Units of the rate constant Worked example: Determining a rate law using initial rates data Relationship between concentration and time Learn First-order reaction (with calculus) Plotting data for a first-order reaction Half-life of a first-order reaction Half-life and carbon dating

About this unit. This unit focuses on rates of change in chemical reactions and the factors that influence them. Learn about rate laws, reaction mechanisms, collision theory, catalysis, and more. Practice what you've learned and study for the AP Chemistry exam with 60 AP-aligned questions.

This chemistry video tutorial provides a basic introduction into chemical kinetics. It explains how to use the integrated rate laws for a zero order, first ...

NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics are the study materials that will help students in getting tuned in with the concepts involved in chemical kinetics. The NCERT Solutions for Class 12 Chemistry PDF for chemical kinetics are helpful for the students of CBSE Class 12.

Write the rate law for this reaction. rate = k[H2] Determine the value and units of the rate constant, k. plug and chug using the rate law & data from exp't 1 and solving for k, we get k = 0.0427 s-1. 7. Consider the reaction: SO2 + O3 → SO3 + O2. data that were obtained are shown in the table.

NCERT Solutions For Class 12 Chemistry Chapter 4 Chemical Kinetics Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics: NCERT TEXTBOOK QUESTIONS SOLVED 4.1.For the reaction R—>P, the concentration of reactant changes from 0.03 M to 0.02 M in 25 minutes.

Solved Examples - Chemical Kinetics | askIITians Solved Examples Example : 1 The experimental data for the reaction 2A + B2 → 2AB is as follows: Write the most probable equation for the rate of reaction giving reason for your answer. Solution :

The rate of a reaction measures the speed at which a chemical reaction takes place. This is done by dividing the change in concentration of reactants by the change in time. So it measures either the rate of disappearance of the reactants or the rate of appearance of the products. To calculate the rate let's first look at the general formula ...

Figure 13.2 provides one useful scheme for classifying chemical kinetic methods of analysis. Methods are divided into two main categories: direct-computation methods and curve-fitting methods. In a direct-computation method we calculate the analyte's initial concentration, [A] 0, using the appropriate rate law.

Answer: (c.) Explanation: The given graph represents that with the decrease in the concentration of A, the concentration of B increases. This implies that A is the reactant and B is the product of the reaction. The reaction can be represented as: A → B.

This video deals with derivation and solved problems on Parallel/Competitive Reaction.follow me on Unacademy: http://unacademy.com/user/N_Huda

Unit 1 Atoms, compounds, and ions. Unit 2 Mass spectrometry. Unit 3 Chemical reactions and stoichiometry. Unit 4 More about chemical reactions. Unit 5 Electronic structure of atoms. Unit 6 Periodic table. Unit 7 Chemical bonds. Unit 8 Gases and kinetic molecular theory. Unit 9 States of matter and intermolecular forces.

Solved Example on Chemical Kinetics. Let's solve some example problems on Chemical Kinetics. Example 1 : Identify the reaction order from each of the following rate constants. k = 2.3 × 10-5 L mol-1 s-1; k = 3 × 10-4 s-1; Solution: (1) The unit of second order rate constant is L mol-1 s-1,therefore k = 2.3 × 10-5 L mol-1 s-1 represents a ...

The "Solved Objective Problems: Chemical Kinetics NEET Questions" guide is a valuable resource for all aspiring students preparing for the NEET exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.

Metal diselenides-based anode materials for sodium ion batteries have attracted great attention owing to their high theoretical capacity and excellent conductivity. However, some critical issues such as notorious soluble polyselenides shuttle and volumetric expansion extremely limit their application in SIBs. Herein, the high-entropy strategy is applied to this material through embedding Mn ...

Solution: A The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows: K = [NO2]2[Cl2] [NOCl]2.