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]

$$ \begin{aligned} & \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right) \\\\ & \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=\frac{532611}{8.3} \times\left(\frac{10}{310 \times 300}\right) \end{aligned} $$<br/><br/> where $\mathrm{K}_2$ is at $310 \mathrm{~K} \,\& \mathrm{~K}_1$ is at $300 \mathrm{~K}$<br/><br/> $$ \begin{aligned} & \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=6.9 \\\\ & =3 \times \ln 10 \\\\ & \ln \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\ln 10^3 \\\\ & \mathrm{~K}_2=\mathrm{K}_1 \times 10^3 \\\\ & \mathrm{~K}_1=\mathrm{K}_2 \times 10^3 \end{aligned} $$<br/><br/> So K = 1