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 $K_1 = K_2$?

The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \cdot e^{-2000/T}$ and $10^{15} \cdot e^{-1000/T},$ respectively. The temperature at which $k_1 = k_2$ is

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

$A$ reaction has an activation energy of $209 \, kJ \, mol^{-1}$. The rate increases $10$-fold when the temperature is increased from $27^{\circ} C$ to $X^{\circ} C$. The temperature $X$ is closest to
[Gas constant,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$ ]

Collision theory is applicable to

For the equilibrium $A_{(g)} \rightleftharpoons B_{(g)}$,$\Delta H$ is $-40 \ kJ/mol$. If the ratio of the activation energies of the forward $(E_f)$ and reverse $(E_b)$ reactions is $\frac{2}{3}$,then: