Question

Nitrogen dioxide reacts with carbon monoxide to produce nitrogen monoxide and carbon dioxide. NO2(g) + CO(g) ⟶ NO(g) + CO2(g) A proposed mechanism for this reaction is 2NO2(g) ⟶ NO3(g) + NO(g) (fast, equilibrium) NO3(g) + CO(g) ⟶ NO2(g) + CO2(g) (slow) What is a rate law that is consistent with the proposed mechanism?

Answer 1

Answer: The rate law for the reaction is $\text{Rate}=k[NO_3][CO]$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

For the given chemical reaction:

$NO_2(g)+CO(g)\rightarrow NO(g)+CO_2(g)$

The intermediate reaction of the mechanism follows:

Step 1:  $2NO_2(g)\rightleftharpoons NO_3(g)+NO(g);\text{ (fast)}$

Step 2:  $NO_3(g)+CO(g)\rightarrow NO_2(g)+CO_2(g);\text{(slow)}$

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

$\text{Rate}=k[NO_3][CO]$

Hence, the rate law for the reaction is written above.