Hydrolysis of sucrose is given by the following reaction: . If the equilibrium constant is at 300 K, the value of at the same temperature will be [2020]

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Q.
Hydrolysis of sucrose is given by the following reaction: $Sucrose + H_{2} O ⇌ Glucose + Fructose$ . If the equilibrium constant $(K_{C})$ is $2 \times 10^{13}$ at 300 K, the value of $Δ_{r} G °$ at the same temperature will be [2020]

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

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

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

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

Ans.
(1)

$Δ G = Δ G ° + R T \ln Q$

At equilibrium, $Δ G = 0$ and $Q = K_{C}$

$∴$ $0 = Δ G ° + R T \ln K_{C}$

$\Rightarrow Δ G ° = - R T \ln K_{C}$

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

Solution

Q.
Hydrolysis of sucrose is given by the following reaction: $Sucrose + H_{2} O ⇌ Glucose + Fructose$ . If the equilibrium constant $(K_{C})$ is $2 \times 10^{13}$ at 300 K, the value of $Δ_{r} G °$ at the same temperature will be [2020]

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

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

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

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

Ans.
(1)

$Δ G = Δ G ° + R T \ln Q$

At equilibrium, $Δ G = 0$ and $Q = K_{C}$

$∴$ $0 = Δ G ° + R T \ln K_{C}$

$\Rightarrow Δ G ° = - R T \ln K_{C}$

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

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Solution

Q. Hydrolysis of sucrose is given by the following reaction: Sucrose+H2O⇌Glucose+Fructose. If the equilibrium constant (KC) is 2×1013 at 300 K, the value of ΔrG° at the same temperature will be [2020]

1 -8.314 J mol-1K-1×300K×ln(2×1013)

2 8.314 J mol-1K-1×300K×ln(2×1013)

3 8.314 J mol-1K-1×300K×ln(3×1013)

4 -8.314 J mol-1K-1×300K×ln(4×1013)

Ans. (1) ΔG=ΔG°+RTlnQ At equilibrium, ΔG=0 and Q=KC ∴ 0=ΔG°+RTlnKC ⇒ ΔG°=-RTlnKC =-8.314 J K-1mol-1×300 K×ln(2×1013)

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Q.
Hydrolysis of sucrose is given by the following reaction: $Sucrose + H_{2} O ⇌ Glucose + Fructose$ . If the equilibrium constant $(K_{C})$ is $2 \times 10^{13}$ at 300 K, the value of $Δ_{r} G °$ at the same temperature will be [2020]

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

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

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

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

Ans.
(1)

$Δ G = Δ G ° + R T \ln Q$

At equilibrium, $Δ G = 0$ and $Q = K_{C}$

$∴$ $0 = Δ G ° + R T \ln K_{C}$

$\Rightarrow Δ G ° = - R T \ln K_{C}$

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