The activation energy for the reaction $X \rightarrow Y$ is $150 \text{ kJ mol}^{-1}$. The change in enthalpy for the above reaction is $-135 \text{ kJ mol}^{-1}$. What is the activation energy for the reverse reaction $Y \rightarrow X$?

The rate of gas phase chemical reactions generally increases rapidly with a rise in temperature. This is mainly because

The rate of a reaction doubles when its temperature changes from $300 \, K$ to $310 \, K.$ Activation energy of such a reaction will be .......... $kJ \, mol^{-1}$. $(R= 8.314 \, J \, K^{-1} \, mol^{-1}$ and $\log 2=0.301)$

For a chemical reaction $A \rightarrow B$,the activation energy in the forward direction is $E_a$. Then the activation energy in the reverse direction is.........

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 of reaction increases with a rise in temperature because of: