Verified Solution • 2025–26 Session moderate

QJEEM22Ch1a95: JEE Main 2022 Chemistry - Class 12 JEEM22Ch1a95 Solution: The activation energy of one of the reactions in a biochemical…

The activation energy of one of the reactions in a biochemical process is 532611 J mol −- 1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300 = x ×\times 10 −- 3 k310. The value of x is ____________.

[Given : ln⁡10=2.3\ln 10 = 2.3 , R = 8.3 J K −- 1 mol −- 1]

Step-by-Step Verified Solution

amp;ln⁡(K2 K1)=EaR(1 T1−1 T2)amp;ln⁡(K2 K1)=5326118.3×(10310×300)\begin{aligned}& \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right) \\\\& \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=\frac{532611}{8.3} \times\left(\frac{10}{310 \times 300}\right)\end{aligned}

where K2\mathrm{K}_2 is at 310 K & K1310 \mathrm{~K} \,\& \mathrm{~K}_1 is at 300 K300 \mathrm{~K}

amp;ln⁡(K2 K1)=6.9amp;=3×ln⁡10amp;ln⁡K2 K1=ln⁡103amp; K2=K1×103amp; K1=K2×103\begin{aligned}& \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=6.9 \\\\& =3 \times \ln 10 \\\\& \ln \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\ln 10^3 \\\\& \mathrm{~K}_2=\mathrm{K}_1 \times 10^3 \\\\& \mathrm{~K}_1=\mathrm{K}_2 \times 10^3\end{aligned}

So K = 1

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