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 k_(300) = x x 10^(-3) k_(310). The value of x is ____________. [Given : ln 10 = 2.3, R = 8.3 J K^(-1) mol^(-1)]

ln (frac K_2 K_1)=frac E_ a R(frac1 T_1-frac1 T_2) \ ln (frac K_2 K_1)=(532611 / 8.3) x ((10 / 310 x 300)) where K_2 is at 310 K \ K_1 is at 300 K ln (frac K_2 K_1)=6.9 \ =3 x ln 10 \ ln frac K_2 K_1=ln 10^3 \ K_2= K_1 x 10^3 \ K_1= K_2 x 10^3 So K = 1

The activation energy of one of the reactions in a bioc... | null (Chemistry)

\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

Explanation

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