The rate of reaction becomes double when the temperature of the reaction is increased from $27\,^{\circ}C$ to $57\,^{\circ}C$. The activation energy of the reaction will be ....... $k\,cal$.

Given the activation energy for the forward direction $(E_{af})$ is $50 \ kJ$ and for the backward direction $(E_{ab})$ is $30 \ kJ$. The reaction would be:

The plot of $\log k_f$ versus $1 / T$ for a reversible reaction $A_{(g)} \rightleftharpoons P_{(g)}$ is shown. Pre-exponential factors for the forward and backward reactions are $10^{15} \ s^{-1}$ and $10^{11} \ s^{-1}$,respectively. If the value of $\log K$ for the reaction at $500 \ K$ is $6$,the value of $|\log k_b|$ at $250 \ K$ is $\qquad$ $[K = \text{equilibrium constant of the reaction}, k_f = \text{rate constant of forward reaction}, k_b = \text{rate constant of backward reaction}]$

The rate constant of a first-order reaction is $1.5 \times 10^7 \text{ s}^{-1}$ at $300 \text{ K}$ and $3.0 \times 10^7 \text{ s}^{-1}$ at $330 \text{ K}$. What is the activation energy $(E_a)$ for the reaction? $[R \times 2.303 = 19.15 \text{ J K}^{-1} \text{mol}^{-1}]$

The energies of activation for forward and reverse reactions for $A_2 + B_2 \rightleftharpoons 2AB$ are $180 \ kJ \ mol^{-1}$ and $200 \ kJ \ mol^{-1}$ respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by $100 \ kJ \ mol^{-1}$. The enthalpy change of the reaction $(A_2 + B_2 \to 2AB)$ in the presence of catalyst will be (in $kJ \ mol^{-1}$)

The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)