Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYEasy

Hydrolysis of sucrose is given by the following reaction: $\text{Sucrose} + \text{H}_2\text{O} \rightleftharpoons \text{Glucose} + \text{Fructose}$ . If the equilibrium constant ( $K_c$ ) is $2 \times 10^{13}$ at $300 \text{ K}$ , the value of $\Delta_r G^\ominus$ at the same temperature will be:

$8.314 \text{ J mol}^{-1} \text{ K}^{-1} \times 300 \text{ K} \times \ln (2 \times 10^{13})$

$8.314 \text{ J mol}^{-1} \text{ K}^{-1} \times 300 \text{ K} \times \ln (3 \times 10^{13})$

$-8.314 \text{ J mol}^{-1} \text{ K}^{-1} \times 300 \text{ K} \times \ln (4 \times 10^{13})$

$-8.314 \text{ J mol}^{-1} \text{ K}^{-1} \times 300 \text{ K} \times \ln (2 \times 10^{13})$

Step-by-Step Solution

The relationship between standard Gibbs free energy change ( $\Delta_r G^\ominus$ ) and the equilibrium constant ( $K_c$ ) is given by the equation: $\Delta_r G^\ominus = -RT \ln K_c$

Given the values: Universal gas constant, $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$ Temperature, $T = 300 \text{ K}$ Equilibrium constant, $K_c = 2 \times 10^{13}$

Substituting these values into the equation, we get: $\Delta_r G^\ominus = -8.314 \text{ J mol}^{-1} \text{ K}^{-1} \times 300 \text{ K} \times \ln (2 \times 10^{13})$

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