Calculate the rate constant of a first-order reaction having a pre-exponential factor $A = 1.6 \times 10^{13} \ s^{-1}$ and $\frac{E_a}{2.303 RT} = 21$.

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

How does the graph of the fraction of molecules versus kinetic energy change at different temperatures?

Consider the following graph of the kinetic energy distribution among molecules at temperature $T.$ If the temperature is increased,how would the resulting graph differ from the one above $:-$

Choose the incorrect statement -