In the given graph,$E_{a}$ for the reverse reaction will be (in $kJ$)

For the equilibrium $A_{(g)} \rightleftharpoons B_{(g)}$,$\Delta H$ is $-40 \ kJ/mol$. If the ratio of the activation energies of the forward $(E_f)$ and reverse $(E_b)$ reactions is $\frac{2}{3}$,then:

The number of correct statement/s from the following is:
$A.$ Larger the activation energy,smaller is the value of the rate constant.
$B.$ The higher is the activation energy,higher is the value of the temperature coefficient.
$C.$ At lower temperatures,the increase in temperature causes a larger change in the value of $k$ than at higher temperatures.
$D.$ $A$ plot of $\ln k$ vs $\frac{1}{T}$ is a straight line with a slope equal to $-\frac{E_a}{R}$.

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

In a reaction,for every $10\,^{\circ}C$ rise of temperature,the rate is doubled. If the temperature is increased from $10\,^{\circ}C$ to $100\,^{\circ}C,$ the rate of the reaction will become $.......$ times.

The energies of activation for forward and reverse reactions for $A_2 + B_2 \rightleftharpoons 2AB$ are $180 \, kJ \, mol^{-1}$ and $200 \, kJ \, mol^{-1}$ respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by $100 \, kJ \, mol^{-1}.$ The enthalpy change of the reaction $(A_2 + B_2 \rightarrow 2AB)$ in the presence of a catalyst will be (in $kJ \, mol^{-1}$)