Question

Catalytic decomposition of nitrous oxide by gold at ${900}^{\circ }C$ at an initial pressure of $200mm$ was $50$% in $53$ minutes and $73$% in $100$ minutes. Find the order of the reaction. How much will it decompose in $100$ minutes at the same temperature but at an initial pressure of $600mm$?

Answer 1

We have $K=\frac{2.303}{t}\log \left(\right(1\left)cfracaa-x\right)$

Case (1) $a=100,a-x$(at $t=53min$)=$100-50=50$ & at $100min$ it is $(100-73)=27$

$\therefore K=\frac{2.303}{53}\log \frac{100}{50}=\frac{2.303}{53}\log 2=0.013mi{n}^{-1}$

Case (2) At $t=100min$

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

Since the value of $K$ is constant, the order of reaction is $1$.

For a first order reaction, the time required to complete any fraction is independent of initial concentration.

$\therefore 73$% of $N_{2}O$ will decompose when initial concentration is $600mm$, corresponds to a pressure of $\frac{600}{100}=438mm$