A schematic plot of In ${\text{k}}_{\text{eq}}$versus inverse of temperature for a reaction is shown below 

 

 

 

 

 

 

 

The reaction must be

The graphs show that reaction is exothermic. $\text{log\hspace{0.17em}k\hspace{0.17em}=\hspace{0.17em}}\frac{-\text{\hspace{0.17em}ΔH}}{\text{RT}}\text{\hspace{0.17em}+I}$ For exothermic  reaction $\text{ΔH\hspace{0.17em}<\hspace{0.17em}0}$or$\therefore \text{\hspace{0.17em}log\hspace{0.17em}k\hspace{0.17em}Vs\hspace{0.17em}}\frac{\text{1}}{\text{T}}$would be negative straight line with positive slope.