Chemistry - Thermodynamics - II Question with Solution | TestHub

Given,

 $$\begin{gathered} \mathrm{T}=400 \mathrm{K}, \\ {\mathrm{\Delta H}}^{\mathrm{o}}=77.2 \mathrm{kJ} {\mathrm{mol}}^{-1}, \\ {\mathrm{\Delta S}}^{\mathrm{o}}=122{\mathrm{JK}}^{-1} \end{gathered}$$

${\mathrm{\Delta G}}^{\mathrm{o}}={\mathrm{\Delta H}}^{\mathrm{o}}-{\mathrm{T\Delta S}}^{\mathrm{o}}$

$$\begin{gathered} =77.2\times {10}^3-400\times 122 \\ =28400 \mathrm{J} \end{gathered}$$

$\mathrm{\Delta G}°=-2.303 \mathrm{R} \mathrm{T} \log \mathrm{K}$

Where K is the equilibrium constant.

By putting the values in above equation,

$\Rightarrow 28400=-2.303\times 8.314\times 400\times \log \mathrm{K}$

$$\begin{gathered} \Rightarrow \log \mathrm{K}=-3.708 \\ =-37.08\times {10}^{-1} \end{gathered}$$