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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Chemistry
Chapter
Chemical Kinetics

BITSAT 2023 · Paper 2 · Chemistry · Chemical Kinetics

For the decomposition of azomethane, CH_3 N_2 CH_3( g) arrow CH_3...

For the decomposition of azomethane, CH3N2CH3(g)→CH3CH3(g)+N2(g)\mathrm{CH}_3 \mathrm{ N}_2 \mathrm{CH}_3(\mathrm{ g}) \rightarrow \mathrm{CH}_3 \mathrm{CH}_3(\mathrm{ g})+\mathrm{N}_2(\mathrm{ g}) a first order reaction, the variation in partial pressure with time at 600K600 \mathrm{ K} is given as
For the decomposition of azomethane, CH_3 N_2 CH_3( g) arrow CH_3 CH_3( g)+N_2( g) a first order reaction, the

The half life of the reaction is.

Options

  1. A2×10−5s2 \times 10^{-5} s
  2. B2×10−4s2 \times 10^{-4} \mathrm{ s}
  3. C4×10−4s4 \times 10^{-4} s
  4. D2×10−3s2 \times 10^{-3} \mathrm{ s}

Correct answer

A. 2×10−5s2 \times 10^{-5} s

Step-by-step solution

For first order reactionk=1tln⁡(P0P)⇒In⁡(P0P)=kt⇒ln⁡(PP0)=−kt\begin{aligned} \\ & k=\frac{1}{t} \ln \left(\frac{P_0}{P}\right) \Rightarrow \operatorname{In}\left(\frac{P_0}{P}\right)=k t \\ \\ & \Rightarrow \ln \left(\frac{P}{P_0}\right)=-k t \\ \end{aligned}On comaring with straight line equation y=mx\mathrm{y}=\mathrm{mx}k= slope =3.465×104t1/2=ln⁡2k=0.6933.465×104=2×10−5s\begin{aligned} \\ & \mathrm{k}=\text{ slope }=3.465 \times 10^4 \\ \\ & \mathrm{t}_{1/2}=\frac{\ln 2}{\mathrm{k}}=\frac{0.693}{3.465 \times 10^4}=2 \times 10^{-5} \mathrm{ s} \\ \end{aligned}$$
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