The nature of the reaction represented in the following graph is:

1.	Endothermic reaction
2.	Exothermic reaction
3.	Both endothermic and exothermic reactions are represented by the same graph.
4.	None of the above

The graph between lnK and 1/T is given below:

The value of activation energy would be:

The correct statement about X in the below mentioned graph:

For a general reaction A → B, the plot of concentration of A vs time is given below:

The order of the reaction would be- 

1. Zero 

2. First

3. Half

4. Second

For a general reaction A → B, the plot of concentration of A vs time is given below:

The unit of the rate constant would be:

For a reaction A$\to$B, enthalpy of reaction is $-4.2$ $\mathrm{kJ}\hspace{0.17em}{\mathrm{mol}}^{-1}$ and enthalpy of activation is $9.6$ $\mathrm{kJ}\hspace{0.17em}{\mathrm{mol}}^{-1}$. The correct potential energy profile for the reaction is:

1.		2.
3.		4.

The slope of Arrhenius Plot (ln k v/s $\frac{1}{\mathrm{T}}$) of the first-order reaction is $-5\times {10}^3$ $\mathrm{K}$. The value of E_a of the reaction is:

[Given R = 8.314 JK^–1 mol^–1]

The slope of the given below graph is -

 $1.$ $\frac{\mathrm{k}}{2.303}$
$$
$2.$ $-\frac{\mathrm{k}}{2.303}$
$$
$3.$ $-\frac{2.303}{\mathrm{k}}$
$$
$4.$ $\frac{2.303}{\mathrm{k}}$

Consider the following graph:
      

The instantaneous rate of reaction at t = 600 sec will be:

$1. - 4.75$ $\times {10}^{-4}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$2.$ $5.75\times {10}^{-5}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$3.$ $$ $6.75\times {10}^{-6}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$4.$ $-6.75\times {10}^{-6}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$

Based on the graph below, the average rate of reaction will be: 

1.  \(\frac{[R_{2}]-[R_{1}]}{t_{2}-t_{1}}\)
2.  \(-(\frac{[R_{2}]-[R_{1}]}{t_{2}-t_{1}})\)
3.  \(\frac{[R_{2}]}{t_{2}}\)
4.  \(-(\frac{[R_{1}]-[R_{2}]}{t_{2}-t_{1}})\)