Question 12.3: Initial rate data for the decomposition of gaseous N2O5 at 5......
Initial rate data for the decomposition of gaseous N_{2}O_{5} at 55°C are as follows:
(a) What is the rate law?
(b) What is the value of the rate constant?
STRATEGY
(a) The rate law for the decomposition of can be written as
Rate = – \frac{\Delta [N_{2}O_{5}]}{\Delta t} = k [N_{2}O_{5}]^{m}
here m is both the order of the reaction in N_{2}O_{5} and the overall reaction order. To find the value of the exponent m, compare the change in the initial concentration of N_{2}O_{5} for experiments 1 and 2 with the change in the initial rate.
(b) To find the value of the rate constant k, solve the rate law for k and then substitute in the data from either experiment.
| Experiment | Initial [N_{2}O_{5}] | Initial Rate of Decomposition of N_{2}O_{5} (M / s) |
| 1 | 0.020 | 3.4 \times 10^{-5} |
| 2 | 0.050 | 8.5 \times 10^{-5} |
(a) Comparing experiments 1 and 2 shows that an increase in the initial concentration of N_{2}O_{5} by a factor of 2.5 increases the initial rate by a factor of 2.5:
\frac{[N_{2}O_{5}]_{2}}{[N_{2}O_{5}]_{1}} = \frac{0.050 M}{0.020 M} = 2.5 \quad\quad \frac{(\text{Rate})_{2}}{(\text{Rate})_{1}} = \frac{8.5 \times 10^{-5} M/s}{3.4 \times 10^{-5} M/s} = 2.5The rate depends linearly on the concentration of N_{2}O_{5} , and therefore the rate law is
\text{Rate} = – \frac{\Delta [N_{2}O_{5}]}{\Delta t} = k [N_{2}O_{5}]The reaction is first order in N_{2}O_{5} . If the rate had increased by a factor of (2.5)² = 6.25 , the reaction would have been second order in N_{2}O_{5} . If the rate had increased by a factor of (2.5)³ = 15.6 , the reaction would have been third order in N_{2}O_{5} and so on.
A more formal way to approach this problem is to write the rate law for each experiment:
(\text{Rate})_{1} = k[N_{2}O_{5}]_{1} = k(0.020 M)^{m}\quad\quad (\text{Rate})_{2} = k[N_{2}O_{5}]_{2} = k(0.050 M)^{m}If we then divide the second equation by the first, we obtain
\frac{(\text{Rate})_{2}}{(\text{Rate})_{1}} = \frac{k (0.050 M)^{m}}{k(0.020 M)^{m}} = (2.5)^{m}Comparing this ratio to the ratio of the experimental rates,
\frac{(\text{Rate})_{2}}{(\text{Rate})_{1}} = \frac{8.5 \times 10^{-5} M/s}{3.4 \times 10^{-5} M/s} = 2.5shows that the exponent m must have a value of 1. Therefore, the rate law is
\text{Rate} = – \frac{\Delta [N_{2}O_{5}]}{\Delta t} = k [N_{2}O_{5}](b) Solving the rate law for k and substituting in the data from the first experiment gives
k = \frac{\text{Rate}}{[N_{2}O_{5}]} = \frac{3.4 \times 10^{-5} \frac{M}{s}}{0.020 M} = 1.7 \times 10^{-3} s^{-1}The units of k, 1/s or s^{-1} are given by the ratio of the units for the rate and the con-centration. These are the expected units for a first-order reaction.