If we plot a graph between $\log \, K$ and $\frac{1}{T}$ by Arrhenius equation,the slope is

If the activation energy of a reaction is $80.9 \, kJ \, mol^{-1}$,the fraction of molecules at $700 \, K$,having enough energy to react to form products is $e^{-x}$. The value of $x$ is ....... . (Rounded off to the nearest integer) $[$Use $R = 8.31 \, J \, K^{-1} \, mol^{-1}]$

In most cases,for a rise of $10 \ K$ temperature,the rate constant is doubled or tripled. This is due to the reason that

The $\Delta H$ value of the reaction $H_2 + Cl_2 \rightleftharpoons 2HCl$ is $-44.12 \ kcal$. If $E_1$ is the activation energy of the backward reaction and $E_2$ is the activation energy of the forward reaction,then for the above reaction:

For a reaction,the activation energy $E_{a} = 0$ and the rate constant at $200 \ K$ is $1.6 \times 10^{6} \ s^{-1}$. The rate constant at $400 \ K$ will be (given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$):

For a first order reaction $A \rightarrow P$,the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation $\log_{10} k = -(2000) \frac{1}{T} + 6$. The activation energy $(E_a)$ of the reaction in $kJ \, mol^{-1}$ will be ......... (Given: $\ln x = 2.3 \times \log_{10} x$ and $R = 8 \, J \, mol^{-1} K^{-1}$)