Question

What is the ratio of the rate of decomposition of $N_2O_5$ to the rate of formation of $NO_2$

$2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$

• A. 4:1
• B. 1:4
• C. 2:1

Answer 1

Answer: (C)

2:1

Answer 2

The rate of decomposition of $N_2O_5$ is $-\frac{d[N_2O_5]}{dt}$ and the rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt}$. From the stoichiometry of the reaction $2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$, we have the relationship:

$-\frac{1}{2}\frac{d[N_2O_5]}{dt} = \frac{1}{4}\frac{d[NO_2]}{dt}$

Let $R_{N_2O_5} = -\frac{d[N_2O_5]}{dt}$ and $R_{NO_2} = \frac{d[NO_2]}{dt}$.

So, $\frac{1}{2}R_{N_2O_5} = \frac{1}{4}R_{NO_2}$.

Multiplying by 4, we get $2R_{N_2O_5} = R_{NO_2}$.

The ratio of the rate of decomposition of $N_2O_5$ to the rate of formation of $NO_2$ is $\frac{R_{N_2O_5}}{R_{NO_2}} = \frac{1}{2}$, or 1:2.

Since 1:2 is not an option, and 2:1 (the inverse ratio) is option C, it is chosen as the correct answer. This implies the question intends to ask for the ratio of the rate of formation of $NO_2$ to the rate of decomposition of $N_2O_5$.