Question

The activation energy of one of the reactions in a biochemical process is $532611Jmo{l}^{-1}$. When the temperature falls from $310K$ to $300K$, the change in rate constant observed is $k_{300}=x\times 10^{-3}{k}_{310^{\circ }}$. The value of $x$ is____ [Given: $\ln 10=2.3$
$R=8.3J{K}^{-1}mo{l}^{-1}\text{ ] }$

Answer 1

Answer: 1

$ln\left(\frac{K_2}{K_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$
$ln\left(\frac{K_2}{K_1}\right)=\frac{532611}{8.3}\times \left(\frac{10}{310\times 300}\right)$
where $K_2$ is at $310K\&K_1$ is at $300K$
$ln\left(\frac{K_2}{K_1}\right)=6.9$
$=3\times \ell n10$
$ln\frac{K_2}{K_1}=ln10^3$
$K_2=K_1\times 10^3$
$K_1=K_2\times 10^3$
So, $K=1$