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

The slope of the graph drawn between $\ln k$ and $\frac{1}{T}$ as per Arrhenius equation gives the value ($R=$ gas constant,$E_a=$ Activation energy)

For a chemical reaction,the rate constant,activation energy,and Arrhenius factor at $25 \, ^\circ C$ are $3.0 \times 10^{-4} \, s^{-1}$,$104.4 \, kJ \, mol^{-1}$,and $6.0 \times 10^{14} \, s^{-1}$ respectively. Find the value of the rate constant as $T \rightarrow \infty$.

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

Decomposition of a hydrocarbon follows the equation $k = (5.5 \times 10^{11} \text{ s}^{-1}) e^{\frac{-28000 \text{ K}}{T}}$. The activation energy of the reaction is . . . . . . $\text{kJ mol}^{-1}$. (Nearest Integer) Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$

The activation energy of a reaction is $0$. The rate constant of the reaction: