In a reversible reaction,the catalyst

In the Arrhenius equation,the pre-exponential factor represents ...

The rate constant of a reaction depends upon

The decomposition of a hydrocarbon follows the equation $K = (4.5 \times 10^{11} \, s^{-1}) e^{-28000 \, K/T}$. What will be the value of activation energy in $KJ \, mol^{-1}$?

Subtract $(i)$ $\ln \, k_1 = - \frac{E_a}{R T_1} + \ln A$ and $(ii)$ $\ln \, k_2 = - \frac{E_a}{R T_2} + \ln A$ and write the resulting equation.

For a first order reaction,a plot of $\ln k$ ($y$-axis) and $\frac{1}{T}$ ($x$-axis) gave a straight line with a slope equal to $-10^3 \ K$ and an intercept equal to $2.303$ (on $y$-axis). What is the activation energy ($E_a$ in $kJ \ mol^{-1}$) of the reaction? (Given $R = 8.314 \ J \ mol^{-1} \ K^{-1}$)