91Ó°ÊÓ

Equilibrium constant (K): In the equilibrium reaction, the ratio of the concentration is of the reactant and the concentration of the product. If K's value is smaller than 1, the reaction should be moved to the left; if K is more than 1, the reaction should be moved to the right.

Also,

$$\begin{gathered} ∆\mathrm{G}°=-\mathrm{RTlnK} \\ \mathrm{K}={\mathrm{e}}^{\left(-∆\mathrm{G}°/\mathrm{RT}\right)} \end{gathered}$$

$∆G°$is the change in free energy.

K is the equilibrium constant.

R is Gas constant 8.314

T is Temperature.

Let's take the equilibrium constant $K_1$ at temperature $T_1$ and the equilibrium constant $K_2$ at temperature $T_2$.

$$\begin{gathered} K_1=e^{-∆G°/R{T}_1} \\ =e^{-\left(∆H°-T_1∆S°\right)/R{T}_1} \\ =e^{-∆H°-R{T}_1.}{e}^{∆S°/R} \end{gathered}$$

Similarly,

$K_2=e^{-∆H°-R{T}_2.}{e}^{∆S°-R}$

Dividing $K_1$ by $K_2$:

$$\begin{gathered} \frac{K_1}{K_2}=e^{-\left(∆H°/R\right)\left(1/T_1-1/T_2\right)} \\ ∆H°={\left(\frac{1}{T_2}-\frac{1}{T_2}\right)}^{-1}R\ln \frac{K_1}{K_2} \end{gathered}$$

Substituting:

$$\begin{gathered} K_1=1.479×{10}^{-5}at{T}_1=278.15K \\ {K}_2=1.570×{10}^{-5}at{T}_2=278.15K \end{gathered}$$

It gives:

$∆\mathrm{H}°=+7.82\mathrm{kJ}/\mathrm{mol}$

The equilibrium constant equation is:

$K=e^{-∆H°-RT.}{e}^{-∆S°-R}$

$\ln K=-\frac{∆H°}{R}\left(\frac{1}{T}\right)+\frac{∆S°}{R}$

and,

Compare with y = mx + b

The graph is formed between role="math" localid="1663326807525" $lnKvs.\frac{1}{T}$, and it gives a slope of $\frac{∆H°}{R}$.