JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Chemical Kinetics

For a reaction of order $n$ the half-life depends on the initial pressure as $t_{1/2}\propto P_0^{1-n}$. Comparing the two experiments, $\frac{t_{1/2,1}}{t_{1/2,2}}=(\frac{P_{0,1}}{P_{0,2}}{)}^{1-n}$, and the data give $\frac{340}{170}=(\frac{55.5}{27.8}{)}^{1-n}$, that is $2=2^{1-n}$. Equating the exponents gives $1-n=1$, so $n=0$: halving the initial pressure halves the half-life, which is precisely the signature of a zero-order decomposition.