Question

$ {NO_2}$ required for a reaction is produced by the decomposition of $ {N2O5}$ in $ {CCl4}$ as per the equation $ { 2N2O5(g) -> 4NO2(g) + O2(g).}$ The initial concentration of $ {N2O5}$ is $3.00\, mol$ $ {L^{-1}}$ and it is $2.75\, mol$ $ {L^{-1}}$ after $30$ minutes. The rate of formation of $ {NO2}$ is :

• A. $\ce{2.083 \times 10^{-3} \; mol \; L^{-1} \; min^{-1}}$
• B. $\ce{4.167 \times 10^{-3} \; mol \; L^{-1} \; min^{-1}}$
• C. $\ce{8.333 \times 10^{-3} \; mol \; L^{-1} \; min^{-1}}$
• D. $\ce{1.667 \times 10^{-2} \; mol \; L^{-1} \; min^{-1}}$

Answer 1

The Correct Option is D

To determine the rate of formation of $NO_2$, we first need to understand the given chemical equation and the concept of reaction rates.

The chemical reaction is: $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$.

This indicates that for every 2 moles of $N_2O_5$ that decompose, 4 moles of $NO_2$ are formed.

First, we calculate the rate of decomposition of $N_2O_5$. The rate can be determined using the change in concentration over time:

The initial concentration of $N_2O_5$ is $3.00 \, \text{mol} \, \text{L}^{-1}$, and after 30 minutes, it is $2.75 \, \text{mol} \, \text{L}^{-1}$.

The change in concentration of $N_2O_5$ is: $(3.00 - 2.75) \, \text{mol} \, \text{L}^{-1} = 0.25 \, \text{mol} \, \text{L}^{-1}$.

Time duration = 30 minutes.

Rate of decomposition of $N_2O_5$ is: \[\frac{0.25 \, \text{mol} \, \text{L}^{-1}}{30 \, \text{min}} = 8.333 \times 10^{-3} \, \text{mol} \, \text{L}^{-1} \, \text{min}^{-1}\].

Since 2 moles of $N_2O_5$ produce 4 moles of $NO_2$, the rate of formation of $NO_2$ is double the rate of decomposition of $N_2O_5$:

\[\text{Rate of formation of } NO_2 = 2 \times 8.333 \times 10^{-3} = 1.667 \times 10^{-2} \, \text{mol} \, \text{L}^{-1} \, \text{min}^{-1}\].

Therefore, the rate of formation of $NO_2$ is 1.667 \times 10^{-2} \, \text{mol} \, \text{L}^{-1} \, \text{min}^{-1}.