Question

Nitrogen dioxide reacts differently with carbon monoxide at different temperatures. The equation below shows the original reaction and reaction mechanism at two different temperatures. Original reaction: NO2 + CO > NO + CO2 Reaction mechanism at T > 770 K: 2 57 Competency Focused Practice Questions | Chemistry | Grade 12 NO2 + CO ----> NO + CO2 (slow) Reaction mechanism at T < 770 K: 2NO2---- > NO + NO3 (slow) NO3 + CO ----> NO2 + CO2 (fast) Write down the equation for the rate of the reaction for: (a) T > 770 K (b) T < 770 K

Answer 1

(a) Rate = k[NO2][CO] (b) Rate = k[NO2]^2

Answer 2

The rate of a reaction is determined by its slowest step, also known as the rate-determining step.

(a) For T > 770 K: The given reaction mechanism is:

$\text{NO}_2 + \text{CO} \xrightarrow{\text{slow}} \text{NO} + \text{CO}_2$

Since this is the only step and it is designated as slow, it is the rate-determining step. The rate law for an elementary step is derived directly from its stoichiometry.

Rate $= k[\text{NO}_2][\text{CO}]$

(b) For T < 770 K: The given reaction mechanism is:

1. $2\text{NO}_2 \xrightarrow{\text{slow}} \text{NO} + \text{NO}_3$
2. $\text{NO}_3 + \text{CO} \xrightarrow{\text{fast}} \text{NO}_2 + \text{CO}_2$

The first step is the slow step, so it is the rate-determining step. The rate law for this elementary step is:

Rate $= k[\text{NO}_2]^2$

Note that $\text{NO}_3$ is an intermediate, but it does not appear in the rate law because the rate-determining step only involves $\text{NO}_2$.

Explanation:

The rate law for a reaction mechanism is determined by the slowest step.

(a) For T > 770 K, the given mechanism has only one step which is slow. The rate law is derived directly from the stoichiometry of this slow step. (b) For T < 770 K, the first step is the slow step. The rate law is derived directly from the stoichiometry of this slow step. Intermediates (like $\text{NO}_3$) are not included in the final rate law expression.