Exercises
Exercise 1. Use the initial rate data below to determine the order with respect to [A] and [B] for the reaction 2A + B → C. What is the order of the overall reaction?
Rate = k[A]m[B]n
First, look at experiments 1 and 2 to determine the exponent m. It works best to have the larger initial rate in the numerator.
\(\displaystyle\require{cancel}\frac{rate_2}{rate_1}\;=\;\frac{\cancel{k}[0.040\;M]^m\;\cancel{[0.020\;M]^n}}{\cancel{k}[0.020\;M]^m\cancel{[0.020\;M]^n}}\;=\;\frac{2.1\times 10^{-3}\;M/s}{1.1\times 10^{-3}\;M/s}\)
2m = 1.9 ≅ 2, and m = 1. The reaction is 1st order with respect to [A].
Next, we find the exponent n. We can compare experiments 2 and 4.
\(\displaystyle\require{cancel}\frac{rate_4}{rate_2}\;=\;\frac{\cancel{k}\cancel{[0.040\;M]^m\;}{[0.040\;M]^n}}{\cancel{k}\cancel{[0.040\;M]^m}{[0.020\;M]^n}}\;=\;\frac{2.1\times 10^{-3}\;M/s}{2.1\times 10^{-3}\;M/s}\)
2n = 1, and n = 0. The reaction is zero order in [B]. The overall order for the reaction is 1st order.
The rate law is Rate = k[A]
Exercise 2. Use the initial rate data to determine
b) the rate law
c) the rate constant and the value of the rate constant, k
The reaction is 2 NO (g) + 2 H2 (g) → N2 (g) + 2 H2O (g)
Rate = k[NO]m[H2]n
First, look at experiments 4 and 5 to determine the exponent m. It works best to have the larger initial rate in the numerator.
\(\displaystyle\require{cancel}\frac{rate_2}{rate_1}\;=\;\frac{\cancel{k}[150]^m\;\cancel{[200]^n}}{\cancel{k}[145]^m\cancel{[200]^n}}\;=\;\frac{7.2\times 10^{-1}\;M/s}{7.0\times 10^{-1}\;M/s}\)
1m = 1 and m = 1. The reaction is 1st order with respect to [H2].
Next, we find the exponent n. We can compare experiments 1 and 2.
\(\displaystyle\require{cancel}\frac{rate_2}{rate_1}\;=\;\frac{\cancel{k}\cancel{[200\;mmHg]^m\;}{[150\;mmHg]^n}}{\cancel{k}\cancel{[200\;mmHg]^m}{[75\;mmHg]^n}}\;=\;\frac{5.4\times 10^{-1}\;mmHg/s}{1.4\times 10^{-1}\;mmHg/s}\)
2n = 3.9 ≅ 4, and n = 2. The reaction is second order in [NO]. The overall order for the reaction is 3rd order.
The rate law is Rate = k[NO]2[H2]
To determine the value of the rate constant, we can use any one of the experiments in the table. Here, we will use experiment 1. Solve the rate law for k.
\(\displaystyle k\;=\;\frac{Rate}{[NO]^2[H_2]}\)
Put the concentrations and the initial rate into the equation.
\(\displaystyle k\;=\;\frac{1.4\times 10^{-1}\;mmHg/s}{[75\;mmHg]^2[200\;mmHg]}\;=\;\mathrm{1.2\times 10^{-7}/mmHg⋅s}\)
Exercise 3. Consider the following reaction.
NO2 (g) + ÇO (g) → NO (g) + CO2 (g)
The initial rate data is in the table below.
Write the rate law.
What is the value of the rate constant?
First determine the order with respect to the reactants. Compare experiments 1 and 2.
\(\displaystyle\require{cancel}\frac{rate_2}{rate_1}\;=\;\frac{\cancel{k}[1.6\;M]^m\;\cancel{[0.4\;M]^n}}{\cancel{k}[0.40\;M]^m\cancel{[0.40\;M]^n}}\;=\;\frac{0.320\;M/s}{0.020\;M/s}\)
4m = 16 and m = 2. The reaction is 2nd order with respect to [NO2].
Next, we find the exponent n. We can compare experiments 1 and 3.
\(\displaystyle\require{cancel}\frac{rate_2}{rate_1}\;=\;\frac{\cancel{k}\cancel{[0.40\;M]^m\;}{[0.80\;M]^n}}{\cancel{k}\cancel{[0.40\;M]^m}{[0.40\;M]^n}}\;=\;\frac{0.02\;M/s}{0.02\;M/s}\)
2n = 1 and n = 0. The reaction is 0th order with respect to [CO].
Rate = k[NO2]2
To determine the value of the rate constant, we can use any one of the experiments in the table. Here, we will use experiment 1. Solve the rate law for k.
\(\displaystyle k\;=\;\frac{Rate}{[NO_2]^2}\)
Put the concentrations and the initial rate into the equation.
\(\displaystyle k\;=\;\frac{0.020\;M/s}{[0.40\;M]^2}\;=\;0.13\;M^{-1}s^{-1}\)
Exercise 4. Consider the following reaction and the initial rate data below.
NO2 (g) + ÇO (g) → NO (g) + CO2 (g)
What is the initial rate of reaction with the following concentrations; [NO2] = 0.465 M and [CO] = 0.32 M?
In problem 3, we found the rate law for this reaction to be Rate = k[NO2]2. We also determined the value of k to be 0.13 M-1s-1. The reaction is second order in NO2 and zero order in CO.
Rate = 0.13 M-1s-1 x [0.465 M]2 x [0.32 M]0 = 0.028 M/s
Exercise 5. Consider the reaction and and the initial rate data below to answer the following questions.
C2H4Br2 (aq) + 3 I– (aq) → C2H4 (g) + 2 Br– (aq) + I3– (aq)
b) What is the value of k with units?
c) What is the initial rate of formation of I3– if the concentrations of both reactants are 0.165 M?
First, determine the exponents m and n.
Rate = k[C2H4Br2]m[I–]n
We can compare experiments 1 and 2 to solve for m.
\(\displaystyle \frac{Rate_2}{Rate_1}\;=\;\require{cancel}\frac{\cancel{k}[0.686\;M]^m[\cancel{0.204\;M]^n}}{\cancel{k}[0.254\;M]^m[\cancel{[0.204\;M]^n}}\;=\;\frac{3.48\times 10^{-4}M/s}{1.29\times 10^{-4}M/s}\)2.7m = 2.7
m = 1
Next, we determine the exponent n. Note, the concentration of C2H4Br2 was not held constant in any of the experiments. We know the value of m, so we can put the concentrations in the rate experiments. We can compare experiments 2 and 3.
\(\displaystyle \frac{Rate_3}{Rate_2}\;=\;\frac{k[0.406\;M]^1[0.250\;M]^n}{k[0.686\;M]^1[0.204\;M]^n}\;=\;\frac{2.52\times 10^{-4}M/s}{3.48\times 10^{-4}\;M/s}\)
0.592 x 1.2n = 0.724
0.71n = 0.72 and n = 1
The rate law is Rate = k[C2H4Br2][I–]
To find the value of k, we can use any of the experiments. I will use experiment 1.
\(\displaystyle k\;=\;\frac{Rate}{[C_2H_4Br_2][I^-]}\;=\;\frac{1.29\times 10{-4}\;M/s}{[0.254\;M][0.204\;M]}\;=\;2.5\times 10^{-3}M^{-1}s^{-1}\)
For part c, we find the initial rate using the concentrations given and the value of k.
Rate = (2.5 x 10-3/M⋅s) 0.165 M x 0.165 M = 6.8 x 10-5 M/s
\(\displaystyle\frac{-\Delta{[I^-]}}{3\Delta{t}}\;=\;\frac{\Delta{[I_3]}}{\Delta{t}}\;=\;6.8\times 10^{-5}\;M/s\)
\(\displaystyle\frac{\Delta{[I_3]}}{\Delta{t}}\;=\;-3\times\;6.8\times 10^{-5}\;M/s\;=\;2.0\times 10^{-4}\;M/s\)