In the decomposition reaction of a gas, reaction -rates are 7.25 and 5.14 `"mol .L"^(-1).s^(-1)` , respectively , for 5% and 20% decomposition of the gas . Determine order of the reaction.

Explanation

If the order of the reaction be n , then
rate= `k["reactant" ]^(n)` (k=rate constant).
Let the initial concentration of the reactant `=a "mol.L"^(-1)` .
At 5% decomposition , the concentration of the reactant
= `a-0.05a=0.95a" mol.L"^(-1)`
and rate `=k(0.95a)^(n)=7.25" "...[1]`
At 20% decomposition , the concentration of the reactant
= `a-0.02a=0.8a" mol.L"^(-1)`
and rate `=k(0.8a)^(n)=5.14" "...[2]`
Dividing equation [1] by [2] , we get
`((0.95)/(0.8))^(n)=((7.25)/(5.14))`
or, `(1.1875)^(n)=1.41=(1.1875)^(2)`
or , n = 2
Thus , order of the reaction = 2.