Question

If the rate of decomposition of $N_2{O}_5$ during a certain time internal is $2.4\times {10}^{-4}mol{L}^{-1}mi{n}^{-1}$.

$N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$

What is the rate of formation of $N{O}_2$ and $O_{2}mol{L}^{-1}mi{n}^{-1}$?

• A. $2.3\times {10}^{-5}$ and $1.2\times {10}^{-5}$ respectively
• B. $3.8\times {10}^{-4}$ and $0.6\times {10}^{-4}$ respectively
• C. $2.4\times {10}^{-4}$ and $1.5\times {10}^{-4}$ respectively
• D. $4.8\times {10}^{-4}$ and $1.2\times {10}^{-4}$ respectively

Answer 1

The correct option is D $4.8\times {10}^{-4}$ and $1.2\times {10}^{-4}$ respectively

Rate of reaction$=-\frac{d\left[N_2{O}_5\right]}{dt}=+\frac{1}{2}\frac{d\left[N{O}_2\right]}{dt}=+2\frac{d\left[O_2\right]}{dt}$

Rate of formation of $N{O}_2$ $=\frac{d\left[N{O}_2\right]}{dt}=2\frac{-d\left[N_2{O}_5\right]}{dt}=4.8\times {10}^{-4}mol{L}^{-1}mi{n}^{-1}$

Rate of formation of $O_2$ $=\frac{d\left[O_2\right]}{dt}=\frac{1}{2}\frac{-d\left[N_2{O}_5\right]}{dt}=1.2\times {10}^{-4}mol{L}^{-1}mi{n}^{-1}$