Verified Solution • 2025–26 Session moderate

QJEEM23Ch82ed: JEE Main 2023 Chemistry - Class 12 JEEM23Ch82ed Solution: A B The rate constants of the above reaction at 200 K and 300 K …

A →\to B

The rate constants of the above reaction at 200 K and 300 K are 0.03 min −1^{-1} and 0.05 min −1^{-1} respectively. The activation energy for the reaction is ___________ J (Nearest integer)

(Given : ln10=2.3\mathrm{ln10=2.3}

R=8.3 J K−1 mol−1\mathrm{R=8.3~J~K^{-1}~mol^{-1}}

log⁡5=0.70\mathrm{\log5=0.70}

log⁡3=0.48\mathrm{\log3=0.48}

log⁡2=0.30\mathrm{\log2=0.30} )

Step-by-Step Verified Solution

amp;log⁡k2k1=Ea2.3×8.3(1200−1300)amp;log⁡0.050.03=Ea2.3×8.3(1600)amp;(0.70−0.48)=Ea2.3×8.3×1600amp;amp;⇒0.22=Ea2.3×8.3×1600amp;⇒Ea=2.3×8.3×600×0.22amp;=2519.88amp;≈2520 J\begin{aligned}& \log \frac{\mathrm{k}_2}{\mathrm{k}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.3 \times 8.3}\left(\frac{1}{200}-\frac{1}{300}\right) \\\\& \log \frac{0.05}{0.03}=\frac{E_a}{2.3 \times 8.3}\left(\frac{1}{600}\right) \\\\& \begin{aligned}(0.70-0.48)=\frac{E_a}{2.3 \times 8.3} \times \frac{1}{600}\end{aligned} \\\\& \begin{aligned}& \Rightarrow 0.22 =\frac{E_a}{2.3 \times 8.3} \times \frac{1}{600} \\\\& \Rightarrow {E_a} = 2.3 \times 8.3 \times 600 \times 0.22 \\\\& =2519.88 \\\\& \approx 2520 \mathrm{~J}\end{aligned}\end{aligned}

Related Learning Resources & Exam Preparation