Question

The decomposition of dinitrogen pentoxide $\left(N_{2}O_{5}\right)$ follows first order rate law. What will be the rate constant from the given data?
At $t = 800\, s$, $\left[N_{2}O_{5}\right]$ $= 1.45\, mol$ $L^{-1}$
At $t = 1600 \,s$, $\left[N_{2}O_{5}\right]$ $= 0.88 \,mol$ $L^{-1}$

• A. $3.12 \times 10^{-4}$ $s ^{-1}$
• B. $6.24 \times 10^{-4}$ $s ^{-1}$
• C. $2.84 \times 10^{-4}$ $s ^{-1}$
• D. $8.14 \times 10^{-4}$ $s ^{-1}$

Answer 1

Answer: (B)

$k=\frac{2.303}{\left(t_{2}-t_{1}\right)} \,log$ $\frac{\left[A_{1}\right]}{\left[A_{2}\right]}$
$k=\frac{2.303}{\left(1600-800\right)}\,log$ $\frac{1.45}{0.88}$ $=\frac{2.303}{800}\times0.2169$
$=6.24\times10^{-4}$ $s^{-1}$