The velocity constant of a reaction at $290 \ K$ was found to be $3.2 \times 10^{-3}$. At $300 \ K$ it 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 $\ln 10 = 2.3, R = 8.3 \, J \, K^{-1} \, mol^{-1}, \log 2 = 0.30$ )

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 $R = k[NO]^2[O_2]$,the rate constant can be increased by:

At temperature $T \ K$,the rate constant of a reaction is $1/10$th of the rate constant at temperature $2T \ K$. What will be the activation energy of this reaction (in $RT$)?

For the reaction $A_2 + B_2 \rightleftarrows 2AB$,the activation energy for the forward and backward reactions are $180 \ kJ \ mol^{-1}$ and $200 \ kJ \ mol^{-1}$,respectively. If a catalyst lowers the activation energy of both forward and backward reactions by $100 \ kJ \ mol^{-1}$,what will be the enthalpy change $(\Delta H)$ for the reaction $A_2 + B_2 \rightarrow 2AB$ in the presence of the catalyst?