Which of the following plots gives the value of activation energy?

For a first order decomposition of a certain reaction,rate constant is given by the equation $\log k \left( s^{-1} \right) = 7.14 - \frac{1 \times 10^4 \ K}{T}$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$ (in $.1$)

When the temperature of a reaction is increased from $300 \ K$ to $310 \ K$,the rate of the reaction increases by $2.5$ times. If the rate constant of the reaction at $300 \ K$ is $K$,what will be the rate constant at $310 \ K$?

At $27^{\circ}C$ in the presence of a catalyst, the activation energy of a reaction is lowered by $10 \ kJ \ mol^{-1}$. The logarithm ratio of $\frac{k(\text{catalysed})}{k(\text{uncatalysed})}$ is .... (Consider that the frequency factor for both the reactions is the same)

The temperature dependence of the rate constant $k$ is expressed as $k = A e^{-E_a / RT}$. When a plot between $\log k$ and $1/T$ is plotted,we get the graph as shown. What is the value of the slope in the graph?

The reaction rate at a given temperature becomes slower,then