Consider a hypothetical reaction at 38 ºC, X₂ (aq) + 2 Y (s) ⇌ 3 Z (aq), with a ∆G of −75.8 kJ.
Concentrations of reactants and products: [X₂] = 1.4 M, [Y] = 0.34 M, [Z] = 2.6 M. Calculate K_eq of this given reaction.

• A. 4.13 × 10¹⁵
• B. 2.67 × 10¹¹
• C. 9.60 × 10¹²
• D. 6.67 × 10¹³

Answer: (D)

Verified step by step guidance

1. Start by understanding the relationship between Gibbs free energy change (∆G) and the equilibrium constant (Keq). The equation that connects these two is: ∆G = -RT ln(Keq), where R is the universal gas constant (8.314 J/mol·K) and T is the temperature in Kelvin.
2. Convert the temperature from Celsius to Kelvin. Since the temperature is given as 38 ºC, add 273.15 to convert it to Kelvin: T = 38 + 273.15 = 311.15 K.
3. Substitute the values into the equation ∆G = -RT ln(Keq). Remember to convert ∆G from kJ to J by multiplying by 1000: ∆G = -75.8 kJ = -75800 J.
4. Rearrange the equation to solve for Keq: ln(Keq) = -∆G / (RT). Substitute the values: ln(Keq) = -(-75800 J) / (8.314 J/mol·K * 311.15 K).
5. Calculate Keq by taking the exponential of both sides: Keq = e^(ln(Keq)). This will give you the equilibrium constant for the reaction.