For the decomposition of azomethane. CH_3N_2CH_3 (g) \to CH_3CH_3 (g) + N_2 (g), a first order reaction, the variation in partial pressure with time at 600 K is given as The half life of the reaction is \underline\hspace2cm \times 10 - 5 s. [Nearest integer]

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2022 · 25 Jul · Shift 2 · Q17 with Solution

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Correct answer: 2

$A(g) \to B(g) + C(g)$

if initially only $A$ is present with pressure $P_0$ , and total pressure at time $t$ is $P_t$ , then:

This is because if decomposition amount is $x$ , then

$P_t = (P_0-x)+x+x = P_0+x \Rightarrow x=P_t-P_0$

Hence

$P_A = P_0-x = P_0-(P_t-P_0)=2P_0-P_t$

= \frac{2.303}{t} \log \frac{P_0}{2P_0-P_t}$$ 3. The given pressure-time data/plot at $600\,\text{K}$ gives the rate constant approximately as $$k \approx 3.47 \times 10^4\ \text{s}^{-1}$$ 4. For a first-order reaction, half-life is $$t_{1/2} = \frac{0.693}{k}$$ So, $$t_{1/2} = \frac{0.693}{3.47\times10^4} \approx 2.0\times10^{-5}\ \text{s}$$ 5. Therefore, in the form asked: $$t_{1/2} = 2 \times 10^{-5}\ \text{s}$$ So the required nearest integer is $2$.