To determine the rate law for the reaction [tex]\( NO_2(g) + CO(g) \longrightarrow NO(g) + CO_2(g) \)[/tex], we need to find the reaction order with respect to each reactant. This is done by using the data provided from the different experiments.
Let's identify the initial rates and concentrations from the table:
Experiment 1:
- [tex]\([NO_2]_0 = 0.263 \, M\)[/tex]
- [tex]\([CO]_0 = 0.826 \, M\)[/tex]
- Initial rate [tex]\( = 1.44 \times 10^{-5} \, M/s \)[/tex]
Experiment 2:
- [tex]\([NO_2]_0 = 0.263 \, M\)[/tex]
- [tex]\([CO]_0 = 0.413 \, M\)[/tex]
- Initial rate [tex]\( = 1.44 \times 10^{-5} \, M/s \)[/tex]
Experiment 3:
- [tex]\([NO_2]_0 = 0.526 \, M\)[/tex]
- [tex]\([CO]_0 = 0.413 \, M\)[/tex]
- Initial rate [tex]\( = 5.76 \times 10^{-5} \, M/s \)[/tex]
Next, we determine the order of the reaction with respect to each reactant by comparing the experiments:
### Step 1: Determine the order with respect to [tex]\( CO \)[/tex]
Compare experiments 1 and 2 where [tex]\([NO_2]\)[/tex] is kept constant:
[tex]\[
\frac{\text{Rate}_1}{\text{Rate}_2} = \frac{(0.263)^{m}(0.826)^n}{(0.263)^{m}(0.413)^n}
\][/tex]
Since [tex]\(\text{Rate}_1 = \text{Rate}_2\)[/tex]:
[tex]\[
\frac{1.44 \times 10^{-5}}{1.44 \times 10^{-5}} = \left(\frac{0.826}{0.413}\right)^n
\][/tex]
[tex]\[
1 = 2^n
\][/tex]
This implies that [tex]\( n = 0 \)[/tex]. Therefore, the reaction is zero order with respect to [tex]\( CO \)[/tex].
### Step 2: Determine the order with respect to [tex]\( NO_2 \)[/tex]
Now, compare experiments 2 and 3 where [tex]\([CO]\)[/tex] is kept constant:
[tex]\[
\frac{\text{Rate}_3}{\text{Rate}_2} = \frac{(0.526)^m (0.413)^n}{(0.263)^m (0.413)^n}
\][/tex]
Since [tex]\(\text{Rate}_3 = 5.76 \times 10^{-5}\)[/tex] and [tex]\(\text{Rate}_2 = 1.44 \times 10^{-5}\)[/tex]:
[tex]\[
\frac{5.76 \times 10^{-5}}{1.44 \times 10^{-5}} = \left(\frac{0.526}{0.263}\right)^m
\][/tex]
[tex]\[
4 = 2^m
\][/tex]
This implies that [tex]\( m = 2 \)[/tex]. Therefore, the reaction is second order with respect to [tex]\( NO_2 \)[/tex].
### Conclusion
Combining these results, the overall rate law for the reaction is:
[tex]\[
\text{Rate} = k [NO_2]^2
\][/tex]
where [tex]\( k \)[/tex] is the rate constant. The reaction is zero order with respect to [tex]\( CO \)[/tex] and second order with respect to [tex]\( NO_2 \)[/tex]. Thus, we have the final rate law:
[tex]\[
\text{Rate} = k [NO_2]^2
\][/tex]
