JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Chemical Kinetics

When one reactant is consumed by two parallel first-order routes, the rates add, so the effective rate constant is $k=k_1+k_2$. Since a first-order rate constant is inversely proportional to its half-life, the reciprocals of the half-lives add: $\frac{1}{t_{1/2}}=\frac{1}{12}+\frac{1}{3}=\frac{1+4}{12}=\frac{5}{12}{\text{min}}^{-1}$, so the effective half-life is $t_{1/2}=\frac{12}{5}=2.4$ min. This is the time for half, that is fifty per cent, of the reactant to be used up, which to the nearest integer is two minutes.