JEE MAIN - Chemistry (2022 - 29th June Morning Shift - No. 14)
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 : $$\ln 10 = 2.3$$, R = 8.3 J K$$-$$1 mol$$-$$1]
Answer
1
Explanation
$$
\begin{aligned}
& \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right) \\\\
& \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=\frac{532611}{8.3} \times\left(\frac{10}{310 \times 300}\right)
\end{aligned}
$$
where $\mathrm{K}_2$ is at $310 \mathrm{~K} \,\& \mathrm{~K}_1$ is at $300 \mathrm{~K}$
$$ \begin{aligned} & \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=6.9 \\\\ & =3 \times \ln 10 \\\\ & \ln \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\ln 10^3 \\\\ & \mathrm{~K}_2=\mathrm{K}_1 \times 10^3 \\\\ & \mathrm{~K}_1=\mathrm{K}_2 \times 10^3 \end{aligned} $$
So K = 1
where $\mathrm{K}_2$ is at $310 \mathrm{~K} \,\& \mathrm{~K}_1$ is at $300 \mathrm{~K}$
$$ \begin{aligned} & \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=6.9 \\\\ & =3 \times \ln 10 \\\\ & \ln \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\ln 10^3 \\\\ & \mathrm{~K}_2=\mathrm{K}_1 \times 10^3 \\\\ & \mathrm{~K}_1=\mathrm{K}_2 \times 10^3 \end{aligned} $$
So K = 1
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