Question

Catalytic decomposition of nitrous oxide by gold at ${900}^{0}C$ at an initial pressure of 200 mm was 50 % in 53 min and 73 % in 100 min. The amount that will compose in 100 min at the same temperature but at an initial pressure of 600 mm is (multiply the answer by 100) :

Answer 1

We have, $K=\left(\frac{2.303}{t}\right)log10\left[\frac{a}{(a–x)}\right]$ --- (1)
Case I : $t_{1/2}=53minutes$
$\therefore K_1=\left(\frac{2.303}{53}\right)log10\left[\frac{200}{(200–100)}\right]=1.307\times {10}^{2}minut{e}^{-1}$

Case II : $t_{73}=100minutes$
$\therefore K_2=\left(\frac{2.303}{100}\right)log10\left[\frac{200}{(200–146)}\right]=1.309\times {10}^{2}minut{e}^{-1}$

Since the value of K is constant. Hence it is first order reaction, the time required to complete any fraction is independent of its initial concentration of reactant.

$\therefore$ 73% of $N_{2}O$ will decomposes when the initial concentration is 600 mm which corresponds to a pressure of
$⟹\frac{600mm}{100}=4.38mm$