The rate constant of a reaction varies with temperature according to the equation: $\log K = \text{constant} - \frac{E_a}{2.303 RT}$. If a plot of $\log K$ versus $1/T$ yields a straight line with a slope of $-5632$,then the activation energy of the reaction is .......... $kJ \ mol^{-1}$.

The rate of a certain biochemical reaction at physiological temperature $T$ occurs $10^{6}$ times faster with an enzyme than without. The change in the activation energy upon adding the enzyme is

The activation energy of a chemical reaction can be determined by:

The activation energy of a reaction at a given temperature is found to be $2.303 \ RT \ J \ mol^{-1}$. The ratio of rate constant to the Arrhenius factor is

For a reaction,given below is the graph of $\ln k$ vs $\frac{1}{T}$. The activation energy for the reaction is equal to $...... \ cal \ mol^{-1}$. (Nearest integer). (Given : $R = 2 \ cal \ K^{-1} \ mol^{-1}$)

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 $k_{300} = x \times 10^{-3} \, k_{310}$. The value of $x$ is $.....$ [Given: $\ln 10 = 2.3$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$]