Effect of temperature on the equilibrium process is analyzed by using the thermodynamics…

Effect of temperature on the equilibrium process is analyzed by using the thermodynamics. From the thermodynamics relation.
∆Gº = – 2.303 RT logK ...(i)
∆Gº : Standard free energy change
∆Gº = ∆Hº – T∆Sº ...(ii)
∆Hº : Standard heat of the reaction
From (i) and (ii)
–2.303 RT log K = ∆Hº – Τ∆Sº
∆Sº : Standard entropy change
\(\text { ⇒ }\) \(\log \mathrm{K}=-\frac{\Delta \mathrm{H}^{\circ}}{2.303 \mathrm{RT}}+\frac{\Delta \mathrm{S}^{\circ}}{2.303 \mathrm{R}}\) ...(iii)
If a plot of log K vs 1/T is made then it is a straight line having slope \(=\frac{-\Delta \mathrm{H}^{\circ}}{2.303 \mathrm{R}}\) and Y intercept \(=\frac{\Delta \mathrm{S}^{\circ}}{2.303 \mathrm{R}}\) 
If at temperature T₁, equilibrium constant be K₁ and at temperature T₂, equilibrium constant be K₂ then the above equation reduces to :
\(\text { ⇒ }\) \(\log \mathrm{K}_{1}=-\frac{\Delta \mathrm{H}^{\circ}}{2.303 \mathrm{RT}_{1}}+\frac{\Delta \mathrm{S}^{\circ}}{2.303 \mathrm{R}}\) 
\(\text { ⇒ }\) \(\log \mathrm{K}_{2}=-\frac{\Delta \mathrm{H}^{\circ}}{2.303 \mathrm{RT}_{2}}+\frac{\Delta \mathrm{S}^{\circ}}{2.303 \mathrm{R}}\)
Substracting (iv) from (v) we get.
\(\text { ⇒ }\) \[\log \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\Delta \mathrm{H}^{\circ}}{2.303 \mathrm{R}}\left(\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right)\] 
From the above relation we can conclude that the value of equilibrium constant increases with increase in temperature for an endothermic reaction and the same decreases with the increase in temperature for an exothermic reaction. Answer the following three questions based on the above information.
For exothermic reaction if ∆Sº < 0, then the sketch of log K vs 1/T may be