Catalytic decomposition of nitrous oxide by gold at 900°C at an initial pressure of 200 mm was 50% in 53 minutes and 73% in 100 minutes.
(i) What is the order of reaction?
(ii) How much will it decompose in 100 minutes at the same temperature but at an initial pressure of 600 mm?

(i) Let [A]₀ = 100.
Then, [A]_t at 53 minutes = (100 – 50) = 50 and [A], at 100 minutes = (100 – 73) = 27.
Substituting t and concentration values in the integrated rate equation for first-order reaction.
                                               $\mathrm{k} = \frac{2.303}{\mathrm{t}}\log \frac{{\left[\mathrm{A}\right]}_0}{{\left[\mathrm{A}\right]}_{\mathrm{t}}}$

At t = 53 min,      

               $\mathrm{k}=\frac{2.303}{53 \min }\log \frac{100}{50}=\frac{2.303}{53 \min }\log 2$

                                                $= \frac{2.303}{53 \min }\times 0.3010 = 0.013 {\min }^{-1}$

At t = 100 min,
               
                                     $$\begin{gathered} \mathrm{k}=\frac{2.303}{100 \min }\log \frac{100}{27} \\ = \frac{2.303}{53 \min }\times 0.5686 = 0.013 {\min }^{-1} \end{gathered}$$

Since the value of k is constant, the order of reaction is 1.

(ii) For a first order reaction, the time required to complete any fraction is independent of the initial concentration of reactant.
∴ 73% of N₂O will decompose when the initial concentration is 600 mm which corresponds to a pressure of $\frac{600 \mathrm{mm}}{100} = 438 \mathrm{mm}.$