The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \times e^{-2000/T}$ and $10^{15} \times e^{-1000/T}$ respectively. The temperature at which $k_1 = k_2$ is

Consider the following plots of rate constant versus $\frac{1}{T}$ for four different reactions. Which of the following orders is correct for the activation energies of these reactions?

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?

The rate constant for the first order decomposition of a certain reaction is described by the equation $\ln k (s^{-1}) = 14.34 - \frac{1.25 \times 10^{4} \ K}{T}$. The energy of activation for this reaction is

Which of the following statements is correct?

For the following reactions:
$A \xrightarrow{700 \ K}$ Product
$A \xrightarrow[\text{catalyst}]{500 \ K}$ Product
it was found that $E_{a}$ is decreased by $30 \ kJ/mol$ in the presence of a catalyst. If the rate remains unchanged,the activation energy for the catalysed reaction is (Assume pre-exponential factor is same):