General Chemistry

We mentioned in the previous post that the order of a reaction can be determined only by experiment. Most often, this experiment consists of measuring the initial rate of the reaction by changing the concentration of the reactant and monitoring how it affects the rate.

For example, the rate law for a hypothetical reaction where molecule A transforms into products can be written as:

A → Products

Rate = k[A]n

where k is the rate constant and n is the reaction order.

Our objective is to determine the reaction order by calculating the n from a set of experiments. Keep in mind that:

• If n = 0, the reaction is zero-order, and the rate is independent of the concentration of A.
• If n = 1, the reaction is first-order, and the rate is directly proportional to the concentration of A.
• If n = 2, the reaction is second-order, and the rate is proportional to the square of the concentration of A.

Now, suppose we run three experiments and the following data is obtained for the concentration-rate correlation:

In every experiment, the concentration of A is doubled, and what we see is that the rate of the reaction doubles as well. Therefore, the initial rate is directly proportional to the initial concentration, and thus, we have a firs-order reaction:

Rate = k[A]1

If it was a zero-order reaction, the following data for the concentration-rate relationship would have been obtained:

The data for a zero-order reaction indicates that the rate does not depend on the concentration of reactants.

For a second-order reaction, doubling the concertation quadrupoles the reaction rate, and therefore, we would expect the following data:

If the numbers are not obvious for determining how the rate changes with concentration, you can pick the data from any set of two experiments, write the rate law, and divide them to see how the rate changed.

For example, going back to the data for a first-order reaction, we can divide the rate of experiments 1 and 2:

$\frac{{Rate\;2}}{{Rate\;1}}\;{\rm{ = }}\;\frac{{k{{[{{\rm{A}}_2}]}^{\rm{n}}}}}{{k{{[{{\rm{A}}_1}]}^{\rm{n}}}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.050}}\;M{\rm{/s}}}}{{{\rm{0}}{\rm{.025}}\,M{\rm{/s}}}}$

$\frac{{Rate\;2}}{{Rate\;1}}\;{\rm{ = }}\;\frac{{\cancel{k}{{{\rm{(0}}{\rm{.20}}\,\cancel{M}{\rm{)}}}^{\rm{n}}}}}{{\cancel{k}{{{\rm{(0}}{\rm{.10}}\,\cancel{M}{\rm{)}}}^{\rm{n}}}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.050}}\;\cancel{{M{\rm{/s}}}}}}{{{\rm{0}}{\rm{.025}}\,\cancel{{M{\rm{/s}}}}}}$

$\frac{{{{{\rm{(0}}{\rm{.20}}\,{\rm{)}}}^{\rm{n}}}}}{{{{{\rm{(0}}{\rm{.10}}\,{\rm{)}}}^{\rm{n}}}}}\;{\rm{ = }}\;2$

2n = 2, therefore,

n = 1

Determining the Value of Rate Constant

To determine the value of the rate constant, write the rate law expression:

Rate = k[A]

Now, you can pick data from any experiment and plug the numbers into a rate law experiment. Let’s use the data from experiment 1.

Rate1 = k[A1]

$k\, = \,\frac{{{\rm{rat}}{{\rm{e}}_{\rm{1}}}}}{{\left[ {{{\rm{A}}_{\rm{1}}}} \right]}}\; = \;\frac{{0.025\,M/s}}{{0.10\,M}}\; = \;0.25\,{s^{ – 1}}$

Let’s now do another example with a real reaction between carbon dioxide and hydrogen and detained the reaction order with respect to each reactant, the overall order, and the value of the rate constant.

Example:

Carbon dioxide, CO2, reacts with hydrogen to give methanol (CH3OH), and water.

CO2(g) + 3H2(g) ⇆ CH3OH(g) + H2O(g)

In a series of experiments, the following initial rates of disappearance of CO2 were obtained:

 Exp. [CO2] [H2] Initial rate, M/s 1 0.640 0.220 2.7 x 10-3 2 1.28 0.220 1.08 x 10-2 3 0.640 0.440 5.4 x 10-3

Determine the rate law and calculate the value of the rate constant for this reaction.

Solution:

To determine the overall reaction order, we need to determine it with respect to both reactants. Let’s first determine the order in CO2. Find two experiments where the concentration of H2 is kept constant while the concentration of CO2 is changed. In experiments 1 and 2, the concentration of CO2 is doubled from 0.640 M to 1.28 M while the concentration of His kept at 0.220 M. We see from the table, that doubling the concentration of CO2 quadruples the rate of the reaction (1.08 x 10-2 ÷ 2.7 x 10-3 = 4). Therefore, the reaction is second-order in CO2.

Now, let’s find two experiments where the concentration of CO2 is kept constant while that of His changed. In experiments 1 and 3, the concentration of CO2 is kept at 0640 M while the concentration of His doubled from 0.220 M to 0.440 M. We see from the table, that doubling the concentration of H2 had doubled the reaction rate (5.4 x 10-3 ÷ 2.7 x 10-3 = 2 ). Therefore, the reaction is first-order in H2.

The rate law, therefore, is:

Rate = k[CO2]2[H2]

And the overall order of the reaction is 2+1 = 3 – it is a third-order reaction.

To calculate the value of the rate constant, use the numbers from any experiment for the following equation:

$k\; = \;\frac{{{\rm{rate}}}}{{{{\left[ {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right]}^2}\left[ {{{\rm{H}}_{\rm{2}}}} \right]}}$

$k\; = \;\frac{{{\rm{2}}{\rm{.7 \times 1}}{{\rm{0}}^{{\rm{ – 3}}}}\;\cancel{{\rm{M}}}{\rm{/s}}}}{{{{\left( {{\rm{0}}{\rm{.640}}\;{\rm{M}}} \right)}^{\rm{2}}}\left( {{\rm{0}}{\rm{.220}}\;\cancel{{\rm{M}}}} \right)}}\;{\rm{ = }}\;{\rm{3}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ – 2}}}}\;{{\rm{M}}^{{\rm{ – 2}}}}{{\rm{s}}^{{\rm{ – 1}}}}\;$

Check Also

Practice

1.

Iron(II) ion is oxidized by hydrogen peroxide in an acidic solution.

2Fe2+ (aq) + H2O2(aq) + 2H+(aq) → 2Fe3+(aq) + 2H2O(l)

The rate law for the reaction is determined to be rate = k[H2O2 ][Fe2+]. The rate constant, at certain temperature, is 2.56 x 1024/M · s. Calculate the rate of the reaction at this temperature if [H2O2 ] = 0.48 M and [H2O2] = 0.070 M.

8.60 x 102 M/s

Solution

rate = k[H2O2 ][Fe2+]

rate = 2.56 x 104/M · s (0.48 M) (0.070 M) = 8.60 x 102 M/s

2.

For the kinetics of the reaction

2NO(g) + Cl2(g) → 2NOCl(g)

The following data were obtained:

 Exp. [NOCl] [Cl2] Initial rate, M/s 1 0.25 0.35 0.68 2 0.25 0.70 1.36 3 0.50 0.70 2.72

a) What is reaction order in Cl2 and NO?

b) What is the rate law?

c) What is the value of the rate constant?

a) First-order in both NO and Cl2

b) Second-order

c) 7.77 s-1 M-1

Solution

Let’s put the table here and work on the questions:

 Exp. [NO] [Cl2] Initial rate, M/s 1 0.25 0.35 0.68 2 0.25 0.70 1.36 3 0.50 0.70 2.72

a) To determine the rate law, we need to know the order of the reaction in both reactants. To determine the reaction order in NO, find two experiments where the concentration of Cl2 is kept constant while the concentration of NO is changed. In experiments 2 and 3, the concentration of Cl2 is kept at 0.70 M, while that of NO is doubled from 0.25 M to 0.50 M.

Now, let’s see how the rate has changed in these two experiments:

$\frac{{{\rm{rat}}{{\rm{e}}_{\rm{3}}}}}{{{\rm{rat}}{{\rm{e}}_{\rm{2}}}}} = \frac{{2.72}}{{1.36}}\; = \;2$

The rate of the reaction has also doubled as a result of this change, and therefore, the reaction with respect to NO is first order.

To determine the reaction order in Cl2, find two experiments where the concentration of NO is kept constant while the concentration of Cl2 is changed. In experiments 1 and 2, the concentration of NO is kept at 0.25 M, while that of Cl2 is doubled from 0.35 to 0.70.

Now, let’s see how the rate has changed in these two experiments:

$\frac{{{\rm{rat}}{{\rm{e}}_{\rm{2}}}}}{{{\rm{rat}}{{\rm{e}}_{\rm{1}}}}} = \frac{{1.36}}{{0.68}}\; = \;2$

The rate of the reaction has also doubled as a result of this change, and therefore, the reaction with respect to Cl2 is first order.

b) Because the reaction is first-order in each reactant, the rate law of the reaction is rate = k[NO][Cl2]

c) To determine the value of the rate constant, we need to write the expression for the reaction rate:

rate = k[NO][Cl2]

Now, you can pick data from any experiment and plug the numbers into a rate law experiment. Let’s use the data from experiment 1.

rate = k[NO][Cl2]

$k = \frac{{{\rm{rat}}{{\rm{e}}_{\rm{1}}}}}{{\left[ {{\rm{NO}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}\;$

$k = \frac{{{\rm{0}}{\rm{.68}}\;\cancel{{\rm{M}}}{\rm{/s}}}}{{{\rm{0}}{\rm{.25}}\;\cancel{{\rm{M}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.35}}\;{\rm{M}}}}\; = \;7.77\;{s^{ – 1}}\;{M^{ – 1}}\;$

3.

The date for the initial rate of the following reaction is listed in the table below:

A + B → C + D

(a) What is the order of reaction with respect to A and to B?

(b) What is the overall reaction order?

(c) What is the value of the rate constant, k?

a) First order in A, and second-order in B

b) Third-order overall

c) 0.108 M2s1

Solution

Let’s put the table here and work on the questions:

a) To determine the reaction order in B, find two experiments where the concentration of A is kept constant while the concentration of B is changed. In experiments 1 and 2, the concentration of A is kept at 0.370 M, while that of B is doubled from 0.248 M to 0.496 M.

Now, let’s see how the rate has changed in these two experiments:

$\frac{{{\rm{rat}}{{\rm{e}}_{\rm{3}}}}}{{{\rm{rat}}{{\rm{e}}_{\rm{1}}}}} = \frac{{4.9{\rm{ }} \times {\rm{ }}{{10}^{ – 3}}}}{{2.45{\rm{ }} \times {\rm{ }}{{10}^{ – 3}}}}\; = \;2$

The rate of the reaction has also doubled as a result of this change, and therefore, the reaction with respect to A is first order.

To determine the reaction order in A, find two experiments where the concentration of B is kept constant while the concentration of A is changed. In experiments 1 and 3, the concentration of B is kept at 0.248 M, while that of A is doubled from 0.370 to 0.740.

Now, let’s see how the rate has changed in these two experiments:

$\frac{{{\rm{rat}}{{\rm{e}}_{\rm{2}}}}}{{{\rm{rat}}{{\rm{e}}_{\rm{1}}}}} = \frac{{9.8{\rm{ }} \times {\rm{ }}{{10}^{ – 3}}}}{{2.45{\rm{ }} \times {\rm{ }}{{10}^{ – 3}}}}\; = \;4$

The rate of the reaction has quadrupled as a result of this change, and therefore, the reaction with respect to B is second order.

b) Because the reaction is first order in one reactant, and second-order in the other reactant, overall it is a third-order reaction.

c) To determine the value of the rate constant, you can pick data from any experiment and plug the numbers into a rate law experiment. Let’s use the data from experiment 1. Remember, the reaction is first order in A, and second order in B:

rate1 = k[A][B]2

$k = \frac{{{\rm{rat}}{{\rm{e}}_{\rm{1}}}}}{{\left[ {\rm{A}} \right]{{\left[ {\rm{B}} \right]}^{\rm{2}}}}}\;$

$k = \frac{{{\rm{2}}{\rm{.45 \times 1}}{{\rm{0}}^{{\rm{ – 3}}}}\;\cancel{{\rm{M}}}{\rm{/s}}}}{{{\rm{0}}{\rm{.370}}\;\cancel{{\rm{M}}}\;{{\left( {{\rm{0}}{\rm{.248}}\;{\rm{M}}} \right)}^{\rm{2}}}}}\; = \;0.108\;{{\rm{M}}^{{\rm{ – 2}}}}{{\rm{s}}^{{\rm{ – 1}}}}\;$

4.

Consider the reaction

A(g) + B(g) ⇌ C(g)

The following data were obtained at a certain temperature:

 Exp. [A] [B] Initial rate, M/s 1 2.40 3.60 4.8 x 10-2 2 2.40 7.20 4.8 x 10-2 3 4.80 3.60 9.6 x 10-2

Using the data, determine the order of the reaction and calculate the rate constant:

rate = k [A]

k = 0.020 s1

Solution

a) To determine the overall reaction order, we need to determine it with respect to both reactants. Let’s first determine the order in A. Find two experiments where the concentration of B is kept constant while the concentration of A is changed. In experiments 1 and 3, the concentration of A is doubled from 2.40 M to 4.80 M while the concentration of B is kept at 3.60 M. We see from the table, that doubling the concentration of A had also doubled the rate of the reaction. Therefore, the reaction is first order in A.

Now, let’s find two experiments where the concentration of A is kept constant while that of B is changed. In experiments 1 and 2, the concentration of A is kept at 2.40 M while the concentration of B is doubled from 3.60 M to 7.20 M. We see from the table, that doubling the concentration of B had no effect on the rate of the reaction as it remained at 4.8 x 10-2 M/s. Therefore, the reaction is zero order in B.

The overall order of the reaction is, therefore, first order:

rate = k [A]

b) To determine the value of the rate constant, we can pick data from any experiment and plug the numbers into a rate law experiment. Let’s use the data from experiment 1.

rate = k [A]

k = rate/ [A]

$k = \frac{{4.8{\rm{ }} \times {\rm{ }}{{10}^{ – 2}}\;\cancel{{\rm{M}}}{\rm{/s}}}}{{{\rm{2}}{\rm{.40}}\;\cancel{{\rm{M}}}\;}}\; = \;0.020\;{s^{ – 1}}$

$k = \frac{{4.8{\rm{ }} \times {\rm{ }}{{10}^{ – 2}}\;\cancel{{\rm{M}}}{\rm{/s}}}}{{{\rm{2}}{\rm{.40}}\;\cancel{{\rm{M}}}\;}}\; = \;0.020\;{s^{ – 1}}$