When a biochemical reaction is carried out in a laboratory without the presence of enzymes,the rate of reaction is $10^{-6}$ times slower. What will be the activation energy $(E_a)$ in the presence of enzymes?

For the reaction $A \to B$,$K_1 = 10^8 \, e^{-6000/8.34T}$ and for the reaction $P \to Q$,$K_2 = 10^{10} \, e^{-8000/8.34T}$. At what temperature $T$ will $K_1 = K_2$ (in $K$)?

The activation energy of a reaction is zero. Its rate constant at $280 \ K$ is $1.6 \times 10^{-6} \ s^{-1}$,the rate constant at $300 \ K$ is

The increase in rate constant of a chemical reaction with increasing temperature is due to the fact$(s)$ that:

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 $R = 8 \, J \, mol^{-1} K^{-1}$)

Plots showing the variation of the rate constant $k$ with temperature $T$ are given below. The plot that follows Arrhenius equation is