Question

The rate constant for the first-order decomposition of N2O5 in the reaction 2 N2O5(g)→4 NO2(g)+O2(g) is kr=3.38×10−5 s−1 at 25 °C. What is the half-life of N2O5? What will be the pressure, initially 500 Torr, after (i) 50 s, (ii) 20min after initiation of the reaction?

Answer 1

Explanation:

$2 N_2O_5(g)\rightarrow 4 NO_2(g)+O_2(g)$

Rate of the reaction ,k= $3.38\times 10^{-5} s^{-1}$

Half life of the $N_2O_5=t_{\frac{1}{2}}$

$t_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{3.38\times 10^{-5} s^{-1}}$(first order kinetics)

$t_{\frac{1}{2}}=20,502.958 seconds$

Half life of the $N_2O_5$ is 20,502.958 seconds.

Integrated rate equation for first order kinetics in gas phase is given as:

$k=\frac{2.303}{t}\log\frac{p_o}{2p_o-p}$

p= pressure of the gas at given time t.

$p_o$ = Initial pressure of the gas

(i) When, t = 50 sec

$p_o=500 torr$

$3.38\times 10^{-5} s^{-1}=\frac{2.303}{50 s}\log\frac{500 Torr}{2(500 Torr)-p}$

p = 500.49 Torr

(ii)When, t = 20 min = 1200 sec

$p_o=500 torr$

$3.38\times 10^{-5} s^{-1}=\frac{2.303}{1200 s}\log\frac{500 Torr}{2(500 Torr)-p}$

p = 519.83 Torr