Consider the following plots of $\log$ of rate constant $\mathrm{k}(\log \mathrm{k})$ vs $\frac{1}{\mathrm{~T}}$ for three different reactions. The correct order of activation energies of these reactions is :

Half life of zero order reaction $\mathrm{A} \rightarrow$ product is 1 hour, when initial concentration of reactant is $2.0 \mathrm{~mol} \mathrm{~L}{ }^{-1}$. The time required to decrease concentration of A from 0.50 to $0.25 \mathrm{~mol} \mathrm{~L}^{-1}$ is :

For $\mathrm{A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{AB}$

$\mathrm{E}_{\mathrm{a}}$ for forward and backward reaction are 180 and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively

If catalyst lowers $\mathrm{E}_{\mathrm{a}}$ for both reaction by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

Which of the following statement is correct?

Rate law for a reaction between $A$ and $B$ is given by

$$\mathrm{r}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}$$

If concentration of $A$ is doubled and concentration of $B$ is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{r_2}{r_1}\right)$ is