Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\mathrm{\Delta }{\mathrm{G}}^{\mathrm{o}}=-\mathrm{R}\mathrm{T}\text{ ln K}$

$\mathrm{\Delta }{\mathrm{H}}^{\mathrm{o}}-\mathrm{T}\mathrm{\Delta }{\mathrm{S}}^{\mathrm{o}}=-\mathrm{R}\mathrm{T}\text{ ln K}$

$-\frac{{\text{ΔH}}^{\text{o}}}{\text{RT}}+\frac{{\text{ΔS}}^{\text{o}}}{\text{R}}=\ln \text{K}$

Therefore$\ln \mathrm{ }\text{K vs }\frac{1}{\mathrm{T}}$graph will be a straight line with slope equal to$-\frac{\mathrm{\Delta }{\mathrm{H}}^{\mathrm{o}}}{\mathrm{R}}$. Since reaction is exothermic, therefore$\mathrm{\Delta }{\mathrm{H}}^{\mathrm{o}}$itself will be negative resulting in positive slope.