Question 15.2: First-Order Rate Laws I The decomposition of N2O5 in the gas...
First-Order Rate Laws I
The decomposition of N_{2}O _{5} in the gas phase was studied at constant temperature:
The following results were collected:
| [N_{2}O _{5}] (mol / L) | Time (s) |
| 0.1000 | 0 |
| 0.0707 | 50 |
| 0.0500 | 100 |
| 0.0250 | 200 |
| 0.0125 | 300 |
| 0.00625 | 400 |
Using these data, verify that the rate law is first order in [N_{2}O _{5}] , and calculate the value of the rate constant, where the rate = – d [N_{2}O _{5}] /dt.
We can verify that the rate law is first order in [N_{2}O _{5}] by constructing a plot of \ln [N_{2}O _{5}] versus time. The values of \ln [N_{2}O _{5}] at various times are given below, and the plot of \ln [N_{2}O _{5}] versus time is shown in Fig. 15.3.
| \ln [N_{2}O _{5}] | Time (s) |
| – 2.303 | 0 |
| – 2.649 | 50 |
| – 2.996 | 100 |
| – 3.689 | 200 |
| – 4.382 | 300 |
| – 5.075 | 400 |
The plot is a straight line, confirming that the reaction is first order in N_{2}O _{5} , since it follows the equation \ln [N_{2}O _{5}] = – kt + \ln [N_{2}O _{5}]_{0} .
Since the reaction is first order, the slope of the line equals – k. In this case
k = – (slope) = 6.93 × 10 ^{-3} s ^{-1}
