Question

The kinetics of the reaction, $2N_2{O}_5\to 4N{O}_2+O_2$, in liquid bromine medium was measured independently for three different initial concentrations of $N_2{O}_5$, which are $0.11,0.07\text{and}0.05mol{L}^{-1}$ respectively. The half life of the reaction was found to be $4.5hours$ for all these concentrations. The order of the reaction is:

• A. $0$
• B. $1$
• C. $2$
• D. Can't be predicted

Answer 1

The correct option is B $1$

Given,
$t_{\frac{1}{2}}=4.5h,\text{Initial Concentration of}{N}_2{O}_5{a}_1=0.11,a_2=0.07,a_3=0.05$
We know that,
The general formula for half life of $n^{th}$ order reaction is:
$t_{\frac{1}{2}}=\frac{2^{n-1}-1}{k(n-1)a_0^{n-1}}(\text{Where n is order of reaction and}n\ne 1)$
Half life of zero order reaction,
$t_{\frac{1}{2}}=\frac{\left[a_0\right]}{2k}$
$t_{\frac{1}{2}}\propto a_0(a_0=\text{Initial Concentration})$
Half life of first order reaction,
$t_{\frac{1}{2}}=\frac{0.693}{k}$
$t_{\frac{1}{2}}\text{is independent of initial concentration}$
Half life of second order reaction,
$t_{\frac{1}{2}}=\frac{1}{k\left[a_0\right]}$
$t_{\frac{1}{2}}\text{is inversely proportional to}{a}_0(a_0=\text{Initial Concentration})$
As it can be seen from the given information, three different initial concentration of $N_2{O}_5$ was taken and the half life was the same in each case. As half life is independent of the initial concentration, the reaction is a first order reaction.