Given CH₃CO₂H(aq) H⁺(aq)+ CH₃CO₂^-(aq)at 25°C,K_a = 1 $\times$ 10^-⁵ $\Delta$ G° at 25°C?
A)-27

Question

Given CH₃CO₂H(aq) H⁺(aq)+ CH₃CO₂^-(aq)at 25°C,K_a = 1.84 $\times$ 10^-⁵.What is $\Delta$G° at 25°C?

Quizplus · Accepted Answer

The given equation can be simplified as:

HS1U1P1(aq) + CHS1U1B13S1U1B0COS1U1B12S1U1B0S1U1P1-S1S1P0(aq) + H(aq) ⇌ CHS1U1B13S1U1B0COS1U1B12S1U1B0H(aq) + HS1U1P1-S1S1P0(aq)

Using the balanced equation and the given equilibrium constant:

∆G° = -RT ln(K)

Where R is the gas constant (8.314 J/K*mol), T is temperature in Kelvin (25°C = 298 K), and ln is the natural logarithm.

Substituting the values, we get:

∆G° = -8.314 J/K*mol * 298 K * ln(1.84 * 10^10 / 55)

∆G° ≈ 27.0 kJ

Therefore, the correct answer is B.

Given CH3CO2H(aq) H+(aq)+ CH3CO2-(aq)at 25°C,Ka = 1 ×\times× 10-5 Δ\DeltaΔ G° at 25°C? A)-27