For an exothermic reaction $X \rightarrow Y$,the activation energy is $30 \ kJ \ mol^{-1}$. If the enthalpy change $(\Delta H)$ for the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction is . . . . . . $kJ \ mol^{-1}$.

If we plot a graph between $\log \, K$ and $\frac{1}{T}$ by Arrhenius equation,the slope is

For the reaction $A_2 + B_2 \rightleftharpoons 2 AB$,the activation energies $(E_a)$ for the forward and backward reactions are $180 \ kJ \ mol^{-1}$ and $200 \ kJ \ mol^{-1}$ respectively. If a catalyst lowers the $E_a$ for both reactions by $100 \ kJ \ mol^{-1}$,which of the following statements is correct?

Activation energy is defined as:

Slope of the straight line obtained by plotting $\log_{10} k$ against $\frac{1}{T}$ represents which term?

$A \rightarrow B$
The rate constants of the above reaction at $200 \, K$ and $300 \, K$ are $0.03 \, min^{-1}$ and $0.05 \, min^{-1}$ respectively. The activation energy for the reaction is $.... \, J$ (Nearest integer).
(Given: $\ln 10 = 2.3$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$,$\log 5 = 0.70$,$\log 3 = 0.48$,$\log 2 = 0.30$)