Question

a. Which ${\mathbf{K}}_{\mathbf{eq}}$corresponds to a negative value of localid="1648198317845" ${\mathbf{\Delta G}}^{\mathbf{\circ }}$, localid="1648198334107" ${\mathbf{K}}_{\mathbf{eq}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1000}$ or K_eq= 0.001 ?

b. Which ${\mathbf{K}}_{\mathbf{eq}}$corresponds to a lower value oflocalid="1648198364090" ${\mathbf{\Delta G}}^{\mathbf{\circ }}$, localid="1648198396853" ${\mathbf{K}}_{\mathbf{eq}}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}$or localid="1648198411007" ${\mathbf{K}}_{\mathbf{eq}}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$?

Short answer

Answer

a. ${\mathrm{K}}_{\mathrm{eq}}=1000$ corresponds to a negative value of ${\mathrm{\Delta G}}^{\circ }$.

b. ${\mathrm{K}}_{\mathrm{eq}}={10}^{-2}$ corresponds to a lower value of ${\mathrm{\Delta G}}^{\circ }$.

Step-by-step solution

Step-by-Step SolutionStep 1: Equilibrium constant

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

Relation between free energy and equilibrium constant

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

${\mathrm{\Delta G}}^{\circ }=-2.303\mathrm{RT}{\mathrm{logK}}_{\mathrm{eq}}$

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" $$\begin{gathered} {\mathrm{\Delta G}}^{°}=-\log 1 \\ =0 \end{gathered}$$

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

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

Determining the sign of ΔG°

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 ${\mathrm{\Delta G}}^{°}$.

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

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