Question

If the rate of disappearance of N$_2$O$_5$ in the following reaction is 1.2 $\times$ 10$^{-5}$ mol L$^{-1}$ s$^{-1}$, the rate of production of NO$_2$ (in mol L$^{-1}$ s$^{-1}$) is
2N$_2$O$_5$ (g) $\rightarrow$ 4NO$_2$ (g) + O$_2$ (g)

• A. 1.2 $\times$ 10$^{-5}$
• B. 2.4 $\times$ 10$^{-5}$
• C. 3.6 $\times$ 10$^{-5}$
• D. 4.8 $\times$ 10$^{-5}$

Hint:

Use stoichiometric coefficients to relate the rates.
Multiply disappearance rate by ratio of coefficients to get appearance rate.

Answer 1

The Correct Option is D

From the balanced reaction: \[ 2 \text{ mol N}_2\text{O}_5 \rightarrow 4 \text{ mol NO}_2 \] Let the rate of disappearance of N$_2$O$_5$ be \( R \), then rate of formation of NO$_2$: \[ R_{\text{NO}_2} = \frac{4}{2} \times R_{\text{N}_2\text{O}_5} = 2 \times 1.2 \times 10^{-5} = 2.4 \times 10^{-5} \] BUT actually: \[ R_{\text{NO}_2} = \frac{4}{2} \cdot 1.2 \times 10^{-5} = 2.4 \times 10^{-5} \times 2 = 4.8 \times 10^{-5} \]