Question

The catalytic decomposition of $N_{2}O$(g) by gold at 900C and at an initial pressure of 200 mm is 50% complete in 53 minutes and 73% complete in 100 minutes.
(i) What is the order of the reaction?
(ii)Calculate the velocity constant.
(iii) How much of $N_{2}O$(g) will decompose in 100 min. at the same temperature but at initial pressure of 600 mm?

Answer 1

Let $\left[A_o\right]=100$

Then $\left[A_t\right]=100-50=50$ at $53$ min

$\left[A_t\right]=100-73=27$ at $100$ min

(i) & (ii)

$k=\frac{2.303}{t}\log \frac{\left[A_o\right]}{\left[A_t\right]}$

$\Rightarrow k_{53}=\frac{2.303}{53}\log \frac{100}{50}=0.013mi{n}^{-1}$

$\Rightarrow k_{100}=\frac{2.303}{100}\log \frac{100}{27}=0.013mi{n}^{-1}$

$\Rightarrow k$ values indicate the order of reaction is $1$ .

(iii) For $1^{st}$ order the time required to complete any fraction is independent of $\left[A_o\right]$

$\therefore 73$ % of $N_{2}O$ will decompose when $\left[A_o\right]=600mm$ which corresponds to a pressure of $\frac{600}{100}mm=438mm$