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?

Consider the following graph of the kinetic energy distribution among molecules at temperature $T.$ If the temperature is increased,how would the resulting graph differ from the one above $:-$

$A$ chemical reaction was carried out at $300 \, K$ and $280 \, K$. The rate constants were found to be $K_1$ and $K_2$ respectively. The energy of activation is $1.157 \times 10^4 \, cal \, mol^{-1}$ and $R = 1.987 \, cal \, K^{-1} \, mol^{-1}$. Then:

$A$ rise in temperature increases the velocity of a reaction. It is because it results in

In most cases,for a rise of $10 \ K$ temperature,the rate constant is doubled or tripled. This is due to the reason that

The number of given statement/s which is/are correct is $.....$.
$(A)$ The stronger the temperature dependence of the rate constant,the higher is the activation energy.
$(B)$ If a reaction has zero activation energy,its rate is independent of temperature.
$(C)$ The stronger the temperature dependence of the rate constant,the smaller is the activation energy.
$(D)$ If there is no correlation between the temperature and the rate constant then it means that the reaction has negative activation energy.