Given
CH 3 CO 2 H(aq) H + (aq) + CH 3 CO 2 - (aq)
at 25°C, K a = 1.8 × 10 -5
-What is ΔG at 25°C for a solution in which the initial concentrations are

Question

Given
CH 3 CO 2 H(aq)  H + (aq) + CH 3 CO 2 - (aq)
at 25°C, K a = 1.8 × 10 -5 
-What is ΔG at 25°C for a solution in which the initial concentrations are

Quizplus · Accepted Answer

To calculate ΔG, we need to use the equation: ΔG = ΔG° + RTlnQ where ΔG° = -RTlnK R = gas constant = 8.314 J/mol K T = temperature in Kelvin = 25°C = 298 K K = equilibrium constant = K_a = 1.8 × 10^-5 ΔG° = -RTlnK ΔG° = -(8.314 J/mol K)(298 K)ln(1.8 × 10^-5) ΔG° = -7889.9 J/mol Q = [H⁺][CH₃CO₂^-]/[CH₃CO₂H] Substituting the initial concentrations given in the problem, we get: Q = [(11eaaf91_9e15_34a5_892c_d1d2ed0719c1_TB6422_11)(11eaaf91_9e15_0d94_892c_1b191bbe001e_TB6422_11)]/[(11eaaf91_9e15_0d94_892c_1b191bbe001e_TB6422_00)] Q = 0.0036 ΔG = ΔG° + RTlnQ ΔG = (-7889.9 J/mol) + (8.314 J/mol K)(298 K)ln(0.0036) ΔG = -22897.3 J/mol ΔG = -22.9 kJ/mol Since we have one mole of the reaction, we can say that ΔG = -22.9 kJ. Therefore, the correct answer is D: -23 kJ.

Given CH3CO2H(aq) H+(aq) + CH3CO2-(aq)

Given CH₃CO₂H(aq) H⁺(aq) + CH₃CO₂^-(aq)