91Ó°ÊÓ

The constant (mathematical expression) that describes the relationship between the concentration of reactants and products of a reaction at equilibrium is termed equilibrium constant.

The formula relating free energy $\left({\mathrm{Δ³Ò}}^{∘}\right)$and the equilibrium constant $\left({\mathrm{K}}_{\mathrm{eq}}\right)$is as follows:

${\mathrm{Δ³Ò}}^{∘}=-2.303\mathrm{RT}{\mathrm{logK}}_{\mathrm{eq}}$

From the above formula, it is observed that free energy is directly related to the negative logarithm of the equilibrium constant.

For${\mathrm{K}}_{\mathrm{eq}}=1,$

localid="1648200130333" $$\begin{gathered} {\mathrm{Δ³Ò}}^{°}=-\log 1 \\ =0 \end{gathered}$$

Therefore, for ${\mathrm{K}}_{\mathrm{eq}}>1$, the value of ${\mathrm{logK}}_{\mathrm{eq}}$is positive and hence ${\mathrm{Δ³Ò}}^{°}$is negative.

For ${\mathrm{K}}_{\mathrm{eq}}<1$, the value of ${\mathrm{logK}}_{\mathrm{eq}}$is negative and hence ${\mathrm{Δ³Ò}}^{°}$is positive.

a. In the given case, the ${\mathrm{K}}_{\mathrm{eq}}$that has a value greater than one $\left({\mathrm{K}}_{\mathrm{eq}}>1\right)$corresponds to a negative value of ${\mathrm{Δ³Ò}}^{°}$.

b. In the given case, the that has a larger value corresponds to a lower value of ${\mathrm{Δ³Ò}}^{°}$as $\mathrm{Δ³Ò}≈-{\mathrm{logK}}_{\mathrm{eq}}$

Therefore, ${\mathrm{K}}_{\mathrm{eq}}={10}^{-2}$corresponds to a lower valueof${\mathrm{Δ³Ò}}^{°}$.