Question

The rate constant for the reaction,

$2N_2{O}_5⟶4{NO}_2+O_2$ is $3.0\times {10}^{-4}{s}^{-1}$.

If start made with $1.0$ $mol$ $L^{-1}$ of $N_2{O}_5$, calculate the rate of formation of ${NO}_2$ at the moment of the reaction when concentration of $O_2$ is $0.1mol$ $L^{-1}$ :

• A. $2.7\times {10}^{-4}mol$ $L^{-1}{s}^{-1}$
• B. $2.4\times {10}^{-4}mol$ $L^{-1}{s}^{-1}$
• C. $4.8\times {10}^{-4}mol$ $L^{-1}{s}^{-1}$
• D. $9.6\times {10}^{-4}mol$ $L^{-1}{s}^{-1}$

Answer 1

The correct option is D $9.6\times {10}^{-4}mol$ $L^{-1}{s}^{-1}$

$Mol$ $L^{-1}$ of $N_2{O}_5$ reacted $=2\times 0.1=0.2$

$\left[N_2{O}_5\right]$ left $=1.0-0.2=0.8mol$ $L^{-1}$

Rate of reaction $=k\times \left[N_2{O}_5\right]$

$=3.0\times {10}^{-4}\times 0.8$

$=2.4\times {10}^{-4}mol$ $L^{-1}{s}^{-1}$

Rate of formation of ${NO}_2$

$=4\times 2.4\times {10}^{-4}=9.6\times {10}^{-4}mol$ $L^{-1}{s}^{-1}$