Question

If ∆G◦ = 27.1 kJ at 25◦C for the reaction CH3COOH (aq) + H2O (l) → CH3COO− (aq) + H3O+ (aq), calculate Ka for this reaction at 298 K.1. 1.012. 5.63 × 1043. 1.15 × 10−114. 9.89 × 10−15. 1.78 × 10−5

Answer 1

Answer:

Ka = 1.78 × 10⁻⁵

Explanation:

Let's consider the following thermochemical equation.

CH₃COOH(aq) + H₂O(l) → CH₃COO⁻(aq) + H₃O⁺(aq)   ∆G° = 27.1 kJ/mol

At 25°C (298 K), we can find the equilibrium constant (Ka) using the following expression.

∆G° = - R × T × lnKa

where,

R: ideal gas constant

T: absolute temperature

27.1 × 10³ J/mol = - (8.314 J/K.mol) × 298 K × lnKa

Ka = 1.78 × 10⁻⁵