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For the reaction \( A(aq) + B(aq) \rightleftharpoons C(aq) + D(aq) \), the equilibrium constant is 23.0 at 25°C and 37.6 at 50°C. What is the change in the Gibbs standard free enthalpy (in kJ) of this reaction at 75°C?

Answer :

The change in the Gibbs standard free enthalpy (ΔG°) of this reaction at 75°C is approximately 30.824 kJ/mol. This value represents the amount of free energy released or absorbed during the reaction under standard conditions at 75°C.

To determine the change in the Gibbs standard free energy (ΔG°) of the reaction at 75°C, we can use the Van 't Hoff equation, which relates the equilibrium constant (K) to temperature changes:

ln(K₂/K₁) = -ΔH°/R * (1/T₂ - 1/T₁)

Where:

K₁ and K₂ are the equilibrium constants at temperatures T₁ and T₂, respectively.

ΔH° is the standard enthalpy change of the reaction.

R is the gas constant (8.314 J/(mol*K)).

T₁ and T₂ are the temperatures in Kelvin.

We are given the equilibrium constants at 25°C and 50°C, so we can rewrite the equation as:

ln(37.6/23.0) = -ΔH°/R * (1/(50 + 273) - 1/(25 + 273))

Simplifying the equation:

ln(1.6348) = -ΔH°/R * (0.003874 - 0.007299)

ln(1.6348) = -ΔH°/R * (-0.003425)

We can rearrange the equation to solve for ΔH°:

ΔH° = -R * ln(1.6348) / (-0.003425)

Substituting the values:

ΔH° = -8.314 J/(mol*K) * ln(1.6348) / (-0.003425)

ΔH° ≈ 30,824 J/mol

To convert the value to kJ/mol:

ΔH° ≈ 30.824 kJ/mol

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