In an exothermic reaction $A \to B$,the evolved heat is $280 \ kJ \ mol^{-1}$ and the activation energy is $200 \ kJ \ mol^{-1}$. The activation energy of the reverse reaction $B \to A$ is $.......... \ kJ \ mol^{-1}$.

Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.

For a first order reaction $A \rightarrow P$,the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation $\log_{10} k = -(2000) \frac{1}{T} + 6$. The activation energy $(E_a)$ of the reaction in $kJ \, mol^{-1}$ will be ......... (Given: $\ln x = 2.3 \times \log_{10} x$ and $R = 8 \, J \, mol^{-1} K^{-1}$)

Activation energy of any reaction depends on

The rate constant for the decomposition of $N_{2}O_{5}$ at various temperatures is given below:

$T / ^{\circ}C$	$0$	$20$	$40$	$60$	$80$
$10^{5} \times k / s^{-1}$	$0.0787$	$1.70$	$25.7$	$178$	$2140$

Draw a graph between $\ln k$ and $1 / T$ and calculate the values of $A$ and $E_{a}.$ Predict the rate constant at $30^{\circ}C$ and $50^{\circ}C$.

The reason for almost doubling the rate of reaction on increasing the temperature of the reaction system by $10 \, ^\circ C$ is