Question

For the reaction, $2N_{2}O_{5} \to 4NO_{2} + O_{2}$ rate and rate constant are $1.02 \times 10^{-4}\,mol\, L^{-1} s^{-1}$ and $3.4 \times 10^{-5} s^{-1}$ respectively. The concentration of $N_{2}O_{5}$ in mol $L^{-1}$ will be

• A. $3.4 \times 10^{-4}$
• B. $3.0$
• C. $5.2$
• D. $3.2 \times 10^{-5}$

Answer 1

Answer: (B)

$\left[N_{2}O_{5}\right]=\frac{Rate}{k}=\frac{1.02\times10^{-4}mol \,L^{-1}\,s^{-1}}{3.4\times10^{-5} s^{-1}}$
$=3\,mol\,L^{-1}$