At $300 \ K$,the activation energy and rate constant for the given reaction are $10 \ kJ \ mol^{-1}$ and $2.4 \times 10^{-5} \ s^{-1}$ respectively. At what temperature (in $K$) will the value of $t_{1/2}$ be $2 \ hr$?
$2N_2O_5 \to 4NO_2 + O_2$

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 $\qquad$ $[K = \text{equilibrium constant of the reaction}, k_f = \text{rate constant of forward reaction}, k_b = \text{rate constant of backward reaction}]$

If the rate constants of a reaction at $500 \ K$ and $700 \ K$ are $0.002 \ s^{-1}$ and $0.06 \ s^{-1}$,respectively,the value of activation energy is $(R=8.314 \ J \ mol^{-1} \ K^{-1}, \log 3=0.477)$.

The activation energy of a forward reaction is $50 \, kcal$. What will be the activation energy of its reverse reaction?

The rate constants of a reaction at $500 \, K$ and $700 \, K$ are $0.02 \, s^{-1}$ and $0.07 \, s^{-1}$ respectively. Calculate the values of $E_{a}$ and $A$.

For $N_2 + 3H_2 \rightarrow 2NH_3$,$\Delta H = -22 \ kcal$,and $E_a = 70 \ kcal$. Hence $E_a$ for $2NH_3 \rightarrow N_2 + 3H_2$ is $.....$ $kcal$.