Question

The decomposition of $N_2{O}_5$ takes place according to I order as $2N_2{O}_5\to 4N{O}_2+O_2$
Calculate:
(a) The rate constant, if instantaneous rate is $1.4\times {10}^{-6}mollitr{e}^{-1}{\sec }^{-1}$ when concentration of $N_2{O}_5$ is $0.04M$.
(b) The rate of reaction when concentration of $N_2{O}_5$ is $1.20M$.
(c) The concentration of $N_2{O}_5$ when the rate of reaction will be $2.45\times {10}^{-5}mollitr{e}^{-1}{\sec }^{-1}$.

Answer 1

(a) $Rate=K\left[N_2{O}_5\right]$

$\therefore K=\frac{1.4\times {10}^{-6}}{0.04}=3.5\times {10}^{-5}{\sec }^{-1}$

(b) Also $rate=K\left[N_2{O}_5\right]$

$=3.5\times {10}^{-5}\times 1.20=4.2\times {10}^{-5}mollitr{e}^{-1}{\sec }^{-1}$

(c) Further $rate=K\left[N_2{O}_5\right]$

$\therefore \left[N_2{O}_5\right]=\frac{2.45\times {10}^{-5}}{3.5\times {10}^{-5}}=0.7mollitr{e}^{-1}$.