Question

Consider the following elementary reaction:

Answer 1

Answer:

$[NO]=\frac{k_{-1}}{k_1} [N_2O_2]$

Explanation:

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In this case, since the reaction may be assumed in chemical equilibrium, we can write up the rate law as shown below:

$r=-k_1[NO]+k_{-1}[N_2O_2]$

However, since the rate of reaction at equilibrium is zero, due to the fact that the concentrations remains the same, we can write:

$0=-k_1[NO]+k_{-1}[N_2O_2]$

Which can be also written as:

$k_1[NO]=k_{-1}[N_2O_2]$

Then, we solve for the concentration of NO to obtain:

$[NO]=\frac{k_{-1}}{k_1} [N_2O_2]$

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