For a reaction,the rate constants $K_1$ and $K_2$ are given by $10^{16} \cdot e^{-2000/T}$ and $10^{15} \cdot e^{-1000/T}$ respectively. At what temperature will $K_1 = K_2$?

An increase in the concentration of the reactants of a reaction leads to a change in:

For a chemical reaction at $27^{\circ} C$, the activation energy is $600 R$. The ratio of the rate constants at $327^{\circ} C$ to that of at $27^{\circ} C$ will be

It has been found that for a chemical reaction with a rise in temperature by $9 \, K$,the rate constant gets doubled. Assuming a reaction to be occurring at $300 \, K$,the value of activation energy is found to be $...... \, kJ \, mol^{-1}$. [nearest integer]
(Given $\ln 10 = 2.3, R = 8.3 \, J \, K^{-1} \, mol^{-1}, \log 2 = 0.30$ )

$A$ reaction having equal activation energies for forward and reverse reaction has:

For a first-order reaction $A \to P$,the rate constant equation is given by $\log K = -2000 \, (1/T) + 6.0$. The pre-exponential factor $A$ and the activation energy $E_a$ are,respectively: