Question

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

Answer 1

(a) Using first order kinetic equation and substituting given values,
In firs case: $k=\frac{2.303}{53}{\log }_{10}\frac{200}{200-100}=0.0131{min}^{-1}$
In second case: $k=\frac{2.303}{100}{\log }_{10}\frac{200}{200-146}=0.0131{min}^{-1}$
As the values of $k$ come out to be the same in both cases, teh reaction is of first order.
(b) As in the first order reaction, the time required for the completion of same fraction is independent of initial concentration; the percentage decomposition in $100$ minutes when the initial pressure is $600$ $mm$ will also be $73$%.