Hydrolysis of sucrose is given by the following reaction: Sucrose + H2O Glucose + Fructose. If the equilibrium constant (K_c) is 2 × 1013 at 300 K, the value of _r G^° at the same temperature will be:

NEET 2020 · ChemistryPrevious Year Question

Hydrolysis of sucrose is given by the following reaction: $Sucrose + H X_{2} O Glucose + Fructose$ . If the equilibrium constant ( $K_{c}$ ) is $2 \times 1 0^{13}$ at $300 K$ , the value of $Δ_{r} G^{\circ}$ at the same temperature will be:

Answer: (C) $- 8.314 J m o l^{- 1} K^{- 1} \times 300 K \times ln (2 \times 1 0^{13})$ . \textbf{Answer:} (C) $- 8.314 \times 300 \times ln (2 \times 1 0^{13})$ \textbf{Solution:} The relation between standard Gibbs energy and the equilibrium constant is $Δ_{r} G^{\circ} = - R T ln K_{c}$ .

A. $8.314 J m o l^{- 1} K^{- 1} \times 300 K \times ln (3 \times 1 0^{13})$
B. $- 8.314 J m o l^{- 1} K^{- 1} \times 300 K \times ln (4 \times 1 0^{13})$
C. $- 8.314 J m o l^{- 1} K^{- 1} \times 300 K \times ln (2 \times 1 0^{13})$ ✓
D. $8.314 J m o l^{- 1} K^{- 1} \times 300 K \times ln (2 \times 1 0^{13})$

Correct Answer

Solution & Explanation

\textbf{Answer:} (C) $- 8.314 \times 300 \times ln (2 \times 1 0^{13})$ \textbf{Solution:} The relation between standard Gibbs energy and the equilibrium constant is $Δ_{r} G^{\circ} = - R T ln K_{c}$ . Substituting $R = 8.314 J m o l^{- 1} K^{- 1}$ , $T = 300 K$ , and $K_{c} = 2 \times 1 0^{13}$ gives $Δ_{r} G^{\circ} = - 8.314 \times 300 \times ln (2 \times 1 0^{13}) J m o l^{- 1}$ .

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Hydrolysis of sucrose is given by the following reaction: Sucrose+HX2​O​Glucose+Fructose. If the equilibrium constant (Kc​) is 2×1013 at 300 K, the value of Δr​G∘ at the same temperature will be:

Solution & Explanation

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Hydrolysis of sucrose is given by the following reaction: $Sucrose + H X_{2} O Glucose + Fructose$ . If the equilibrium constant ( $K_{c}$ ) is $2 \times 1 0^{13}$ at $300 K$ , the value of $Δ_{r} G^{\circ}$ at the same temperature will be: