Chemistry - Thermodynamics - II Question with Solution | TestHub

For the equilibrium,$\mathrm{A}\to \mathrm{B}$
${\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}-{\text{TΔS}}^{\text{o}}$
${\text{ΔG}}^{\text{o}}=-2.303\text{ RT}{\text{log}}_{\text{10}}\text{K}$($\text{K}$is equilibrium constant)
$-2.303\text{ RT}{\text{log}}_{\text{10}}\text{K}={\text{ΔH}}^o-{\text{TΔS}}^{\text{o}}$
$\text{2}\text{.303 RT}{\text{log}}_{\text{10}}\text{K}\text{=}{\text{TΔS}}^{\text{o}}-{\text{ΔH}}^{\text{o}}$
${\log }_{10}\mathrm{K}=\frac{\mathrm{T}∆{\mathrm{S}}^{\mathrm{o}}-∆{\mathrm{H}}^{\mathrm{o}}}{2.303\mathrm{ }\mathrm{R}\mathrm{T}}=\frac{298\times 10+54.07\times 1000}{2.303\times 8.314\times 298}=10$
Hence$\text{(B)}$is correct option.