Question

The initial concentration of $N _{2} O _{5}$ in the following first order reaction, $N _{2} O _{5(g)} \longrightarrow 2 NO _{2(g)}+1 / 2 O _{2(g)}$ was $1.24 \times 10^{-2} mol\, L ^{-1}$ at $318 K$. The concentration of $N _{2} O _{5}$ after 60 minutes was $0.20 \times 10^{-2} mol\, L ^{-1}$. The rate constant of the reaction at $318\, K$ is

• A. $k = 0.004 \,min^{-1} $
• B. $k = 0.0304 \,min^{-1} $
• C. $ k = 2.3 \,min^{-1} $
• D. $ k = 1.5 \,min^{-1} $

Answer 1

Answer: (B)

For a first order reaction
$\log \frac{[R]_{1}}{[R]_{2}}=\frac{k\left(t_{2}-t_{1}\right)}{2.303}$

$k=\frac{2.303}{\left(t_{2}-t_{1}\right)} \log \frac{[R]_{1}}{[R]_{2}}$

$=\frac{2.303}{(60 min -0 min )} \log \frac{1.24 \times 10^{-2} mol L ^{-1}}{0.20 \times 10^{-2}\, mol\, L ^{-1}}$

$=\frac{2.303}{60} \log 6.2 \,min ^{-1}$

$k=0.0304 \,min ^{-1}$