Question

Hydrolysis of sucrose is given by the following reaction: $\mathrm{Sucrose}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{aq}\right)\rightleftarrows \mathrm{Glucose}+\mathrm{Fructose}$ if the equilibrium constant (${\mathrm{K}}_{\mathrm{c}}$) is $2\times {10}^{13}$ at $300\mathrm{K}$, value of ${∆}_{\mathrm{r}}{\mathrm{G}}^{°}$ at the same temperature will be:

Answer 1

Step 1: Given Data

• Equilibrium constant (${\mathrm{K}}_{\mathrm{c}}$)=$2\times {10}^{13}$
• Temperature($\mathrm{T}$)= $300\mathrm{K}$
• Gas constant ($\mathrm{R}$)= $8.314{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$

Step 2: Calculation of Gibbs free energy (${\mathbf{∆}}_{\mathbf{r}}{\mathbf{G}}^{\mathbf{°}}$)

The relation between Gibbs energy and equilibrium constant is given as:

${∆}_{\mathrm{r}}{\mathrm{G}}^{°}=-\mathrm{RTln}\left({\mathrm{K}}_{\mathrm{c}}\right)$

Putting the values in the above equation we get:

$$\begin{gathered} {∆}_{\mathrm{r}}{\mathrm{G}}^{°}=-\mathrm{RTln}\left({\mathrm{K}}_{\mathrm{c}}\right) \\ {∆}_{\mathrm{r}}{\mathrm{G}}^{°}=-8.314\times 300\times \ln \left(2\times {10}^{13}\right) \\ {∆}_{\mathrm{r}}{\mathrm{G}}^{°}=-76402.84\mathrm{J}{\mathrm{mol}}^{-1} \\ {∆}_{\mathrm{r}}{\mathrm{G}}^{°}=-76.402\mathrm{kJ}{\mathrm{mol}}^{-1} \end{gathered}$$

Therefore, the value of Gibbs free energy ${∆}_{\mathrm{r}}{\mathrm{G}}^{°}=-76.389\mathrm{kJ}{\mathrm{mol}}^{-1}$