Q24 / 160 · AP EAMCET 2024 (22 May Shift 1) · Chemistry

Isomerisation of gaseous cyclobutene to butadiene is a first order reaction. At $\mathrm{T}(\mathrm{K})$, the rate constant of the reaction is $3.3 \times 10^{-4} \mathrm{~s}^{-1}$. What is the time required (in min ) to complete $90 \%$ of this reaction at the same temperature? $(\log 2=0.3)$

  • Correct answer
    A
    116.67
  • B
    233.34
  • C
    58.34
  • D
    350.0

Solution

$\begin{aligned} & \text { } \mathrm{K}=3.3 \times 10^{-4} \mathrm{~s}^{-1} \\ & \mathrm{~K}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \\ & \therefore \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{100}{10} \\ & \mathrm{t}=\frac{2.303}{3.3 \times 10^{-4}} \times 1 \quad[\log 10=1] \\ & \mathrm{t}=\frac{2.303}{3.3} \times 10^4 \\ & \mathrm{t}=0.70 \times 10^4 \text { second }\end{aligned}$ $\begin{aligned} & t=7000 \mathrm{~s} \\ & \mathrm{t}=\frac{7000}{60}=116.6 \mathrm{~min}\end{aligned}$

Hence, the correct answer is option: 'A'.

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