Question 12.3: Determining a Rate Law from Initial Rates Initial rate data ......
Determining a Rate Law from Initial Rates
Initial rate data at 25 °C are listed in the table for the reaction
NH_{4}^{+}(aq) + NO_{2}^{-}(aq) → N_2(g) + 2 H_2O(l)(a) What is the rate law?
(b) What is the value of the rate constant including the proper units?
(c) What is the initial rate when the initial concentrations are [ NH_{4}^{+} ] = 0.39 M and [ NO_{2}^{-} ] = 0.052 M?
STRATEGY
(a) The rate law for the reaction can be written as
\text { Rate }=-\frac{\Delta\left[\mathrm{NH}_4^{+}\right]}{\Delta t}=k\left[\mathrm{NH}_4^{+}\right]^m\left[\mathrm{NO}_2^{-}\right]^nwhere m is the order of the reaction in NH_{4}^{+} and n is the order of the reaction in NO_{2}^{-} . To find the values of the exponents m and n, compare the change in the initial rate with the change in the initial concentration of one reactant in pairs of experiments in which the concentration of the other reactant is held constant.
(b) To find the value of the rate constant k, solve the rate law for k and then substitute in the data from any one of the three experiments.
(c) To calculate the initial rate, substitute into the rate law the rate constant found in part (b) and the given initial concentrations of NH_{4}^{+} (0.39 M) and NO_{2}^{-} (0.052 M).
| Experiment | Initial [ NH_{4}^{+} ] | Initial [ NO_{2}^{-} ] | Initial Rate of Consumption of NH_{4}^{+} (M/s) |
| 1 | 0.12 | 0.10 | 3.6 \times 10^{-6} |
| 2 | 0.24 | 0.10 | 7.2 \times 10^{-6} |
| 3 | 0.12 | 0.15 | 5.4 \times 10^{-6} |
(a) Comparing Experiments 1 and 2 shows that doubling the initial concentration of NH_{4}^{+} from 0.12 M to 0.24 M while holding the NO_{2}^{-} concentration constant increases the initial rate by a factor of 2, from 3.6 × 10^{-6} M/s to 7.2 × 10^{-6} M/s. Therefore, the initial rate is proportional to [ NH_{4}^{+} ] and m = 1.
Comparing Experiments 1 and 3 shows that increasing the initial concentration of NO_{2}^{-} by a factor of 1.5 from 0.10 M to 0.15 M while holding the NH_{4}^{+} concentration constant increases the initial rate by a factor of 1.5, from 3.6 × 10^{-6} M/s to 5.4× 10^{-6} M/s. And so the initial rate is proportional to \left[\mathrm{NO}_2^{-}\right] and n = 1. Because the reaction is first order in both NH_{4}^{+} and NO_{2}^{-}, the rate law is
\text { Rate }=-\frac{\Delta\left[\mathrm{NH}_4^{+}\right]}{\Delta t}=k\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{NO}_2^{-}\right]A more formal way to approach this problem begins with the rate law for two experiments in which the concentration of only one reactant varies. We’ll illustrate this approach for Experiments 1 and 2, which yields the value of m. (A similar procedure applied to Experiments 1 and 3 would give the value of n.) The rate laws for Experiments 1 and 2 are
\begin{aligned}& (\text { Rate })_1=k\left[\mathrm{NH}_4^{+}\right]_1^m\left[\mathrm{NO}_2^{-}\right]_1^n=k(0.12 \mathrm{M})^m(0.10 \mathrm{M})^n \\& (\text { Rate })_2=k\left[\mathrm{NH}_4^{+}\right]_2^m\left[\mathrm{NO}_2^{-}\right]_2^n=k(0.24 \mathrm{M})^m(0.10 \mathrm{M})^n\end{aligned}If we then divide the second equation by the first, we obtain
\frac{(\text { Rate })_2}{(\text { Rate })_1}=\frac{\cancel{k} (0.24 \mathrm{M})^m \cancel{(0.10)^n}}{\cancel{k} (0.12 \mathrm{M})^m \cancel{(0.10)^n}}=\left(\frac{0.24}{0.12}\right)^mThe right-hand side of this equation equals the ratio of the experimental rates:
\left(\frac{0.24}{0.12}\right)^m=\frac{7.2 \times 10^{-6} \mathrm{M} / \mathrm{s}}{3.6 \times 10^{-6} \mathrm{M} / \mathrm{s}}Finally, we can solve for m if we take the logarithm of both sides of the last equation and remember that log x^m = m log x (see Appendix A.2 for a review of logarithms):
\begin{aligned}\log \left(\frac{0.24}{0.12}\right)^m & =m \log \left(\frac{0.24}{0.12}\right)=\log \left(\frac{7.2 \times 10^{-6} \mathrm{M} / \mathrm{s}}{3.6 \times 10^{-6} \mathrm{M} / \mathrm{s}}\right) \\m & =\frac{\log \left(\frac{7.2 \times 10^{-6} \mathrm{M} / \mathrm{s}}{3.6 \times 10^{-6} \mathrm{M} / \mathrm{s}}\right)}{\log \left(\frac{0.24}{0.12}\right)}=\frac{\log 2}{\log 2}=1\end{aligned}A similar procedure applied to Experiments 1 and 3 yields n = 1, and therefore the rate law is
\text { Rate }=-\frac{\Delta\left[\mathrm{NH}_4^{+}\right]}{\Delta t}=k\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{NO}_2^{-}\right](b) Solving the rate law for k and substituting in the data from the frst experiment gives
k=\frac{\text { Rate }}{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{NO}_2^{-}\right]}=\frac{3.6 \times 10^{-6} \frac{\mathrm{M}}{\mathrm{s}}}{(0.12 \mathrm{M})(0.10 \mathrm{M})}=3.0 \times 10^{-4} \mathrm{M}^{-1} \mathrm{~s}^{-1}(c) Substituting the initial concentrations ([ NH_{4}^{+} ] = 0.39 M; [ NO_{2}^{-} ] = 0.052 M) and the rate constant from part (b) ( 3.0 \times 10^{-4} \ M^{-1} s^{-1} ) into the rate law gives
\begin{aligned}\text { Rate }=-\frac{\Delta\left[\mathrm{NH}_4^{+}\right]}{\Delta t}=k\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{NO}_2^{-}\right] & =\left(3.0 \times 10^{-4} \mathrm{M}^{-1} \mathrm{~s}^{-1}\right)(0.39 \mathrm{M})(0.052 \mathrm{M}) \\& =6.1 \times 10^{-6} \mathrm{M} / \mathrm{s}\end{aligned}BALLPARK CHECK
(b) It’s a good idea to check the units of the rate constant. The units of k, M^{-1} \ s^{-1} , are the expected units for a reaction that is second order overall.
(c) Because the initial [ NH_4^{+} ] is about 3 times that in Experiment 1 and the initial [ NO_{2}^{-} ] is about 1/2 times that in Experiment 1, the reaction rate is estimated to be about 3/2 times that in Experiment 1, or about 5.4 \times 10^{-6} M/s, which agrees with the solution.