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.
At t = 53 min,
At t = 100 min,
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
This reaction follows second order kinetics.
So that, the rate equation for this reaction will
Rate, R = k[X]2 .............(1)
Let initial concentration is x mol L−1,
Plug the value in equation (1)
Rate, R1 = k .(a)2
Given that concentration is increasing by 3 times so new concentration will 3a mol L−1
Plug the value in equation (1) we get
Rate, R2 = k (3a)2
We have already get that R1 = ka2 plus this value we get
R2 = 9 R1
So that, the rate of formation will increase by 9 times.
Rate = k[A]2
If concentration of X is increased to three times,
Rate = k[3A]2
or Rate = 9 k A2
Thus, rate will increase 9 times.