Question

Hydrolysis of sucrose gives glucose and fructose. The reaction takes place as: Sucrose $+H_{2}O\rightleftharpoons$ Glucose $+$ Fructose. The equilibrium constant $K_c$ for this reaction is $2\times {10}^{13}$ at $300K$. The $\Delta G^{\circ }$ at $300K$ is:

• A. $7.64\times {10}^{4}Jmo{l}^{-1}$
• B. $7.64\times {10}^{-4}Jmo{l}^{-1}$
• C. $-7.64\times {10}^{-4}Jmo{l}^{-1}$
• D. $-7.64\times {10}^{4}Jmo{l}^{-1}$

Answer 1

The correct option is D $-7.64\times {10}^{4}Jmo{l}^{-1}$

The relationship between the standard free energy change $\left(\Delta G^o\right)$ and the equilibrium constant $\left(K_p\right)$ is $\left(\Delta G^o\right)=-RTln{K}_c$, where, $R$ is the ideal gas constant and $T$ is the temperature.

Given, $K_c=2\times {10}^{13},T=300K,R=8.314Jmo{l}^{-1}{K}^{-1}$

Substituting these values in the above expression, we get

$\left(\Delta G^o\right)=-RTln{K}_c=-8.314\times 300\times ln(2\times {10}^{13})=-7.64\times {10}^{4}Jmo{l}^{-1}$

Hence, the standard free energy change $\left(\Delta G^o\right)=-7.64\times {10}^{4}Jmo{l}^{-1}$