Q24 / 90 · JEE Main 2022 (25 Jun Shift 2) · Chemistry

At $345 K$ , the half life for the decomposition of a sample of a gaseous compound initially at $55.5 kPa$ was $340 s$ . When the pressure was $27.8 kPa$ , the half life was found to be $170 s$ . The order of the reaction is - [integer answer]

Correct answer: 0

Solution

$\frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = {(\frac{P_{1}}{P_{2}})}^{1 - n}$

$\Rightarrow \frac{340}{170} = {(\frac{55.5}{27.8})}^{1 - n}$

${(2)}^{1} = {(2)}^{1 - n}$

$1 = 1 - n \Rightarrow n = 0$

For zero order reaction half life is directly proportional to the initial concentration.

Hence, the correct answer is 0.