1. -8.314 J mol ⁻¹ K ⁻¹×300 K ×ln(2×10¹³) 2. 8.314 J mol ⁻¹ K ⁻¹×300 K ×ln(2×10¹³) 3. 8.314 J mol ⁻¹ K ⁻¹×300 K ×ln(3×10¹³) 4. -8.314 J mol ⁻¹ K ⁻¹×300 K ×ln(4×10¹³)

Answer: 1. -8.314 J mol ⁻¹ K ⁻¹×300 K ×ln(2×10¹³) Standard Gibbs energy and equilibrium constant are related by Δ_rG^ominus=-RTln K . Substituting: Δ_rG^ominus=-8.314\ mathrmJ mol⁻¹K⁻¹×300\ K×ln(2×10¹³) . Since K is very large, Δ_rG^ominus is large and negative (spontaneous). Options (3) and (4) use a wrong K value, and option (2) has the wrong sign.

Topic: Gibbs energy change and equilibrium constant· Chemical Thermodynamics

Hydrolysis of sucrose is given by the following reaction.Sucrose + H_2O ⇌ Glucose + FructoseIf the equilibrium constant (K_c) is 2×10^13 at 300 K, the value of Δ_rG^ at the same temperature will be :

1. \(-8.314\) J mol\(^{-1}\)K\(^{-1}\times300\) K \(\times\ln(2\times10^{13})\)

2. \(8.314\) J mol\(^{-1}\)K\(^{-1}\times300\) K \(\times\ln(2\times10^{13})\)

3. \(8.314\) J mol\(^{-1}\)K\(^{-1}\times300\) K \(\times\ln(3\times10^{13})\)

4. \(-8.314\) J mol\(^{-1}\)K\(^{-1}\times300\) K \(\times\ln(4\times10^{13})\)

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Answer: 1. \(-8.314\) J mol\(^{-1}\)K\(^{-1}\times300\) K \(\times\ln(2\times10^{13})\)

Standard Gibbs energy and equilibrium constant are related by \(\Delta_rG^\ominus=-RT\ln K\).

Substituting: \(\Delta_rG^\ominus=-8.314\ \mathrm{J\,mol^{-1}K^{-1}}\times300\ \mathrm{K}\times\ln(2\times10^{13})\).

Since K is very large, \(\Delta_rG^\ominus\) is large and negative (spontaneous). Options (3) and (4) use a wrong K value, and option (2) has the wrong sign.

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Hydrolysis of sucrose is given by the following reaction.Sucrose + H_2O ⇌ Glucose + FructoseIf the equilibrium constant (K_c) is 2×10^13 at 300 K, the value of Δ_rG^ at the same temperature will be :

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