Question

The acid dissociation constant of the 1,10-phenanthrolium ion (phen) was meeasured at different temperatures using spectroscopic methods.

The values of pKa measured at 298K and 308K were 4.812 and 4.731, respectively.

Part A: Calculate  ΔrGo at 298K. Express your answer in kJ/mol to three significant figures.

Part B: Calculate  ΔrGo at 308K. Express your answer in kJ/mol to three significant figures.​

Part C: Calculate  ΔrHo. Express your answer in kJ/mol to three significant figures.​

Part D: Calculate  ΔrSo. Express your answer in J/(K.mol) to three significant figures.​

Part E: Calculate the pKa of the acid at 359K. Express your answer to three significant figures​

Answer 1

We know that pK_a = -log K_a

Given pK_a (298 K) = 4.812; K_a (298 K) = 10^-pKa = 10^-4.812 = 1.5417*10^-5;

pK_a (308 K) = 4.731; K_a (308 K) = 10^-4.731 = 1.8578*10^-5

A) We know that

ΔG⁰ = -RTlnK_eq

Therefore, Δ_rG⁰₂₉₈ = -(8.314 J/K.mol)*(298 K)*ln (1.5417*10^-5) = 27451.596 J/mol = 27.451 kJ/mol (ans)

B) Δ_rG⁰₃₀₈ = - (8.314 J/mol.K)*(308 K)*ln (1.8578*10^-5) = 27895.199 J/mol = 27.895 kJ/mol (ans).

C) We know, from Vant Hoff equation,

ln (K₂/K₁) = -ΔH⁰/R*(1/T₂ – 1/T₁)

Therefore, for the present problem,

ln (1.8578*10^-5/1.5417*10^-5) = -Δ_rH⁰/(8.314 J/mol.K)*(1/308 – 1/298) K^-1

===> 0.1865 = -Δ_rH⁰/(8.314 J/mol.K)*(-1.0895*10^-4) K^-1

===> Δ_rH⁰ = 0.1865*8.314/1.0895*10^-4 J/mol = 14231.859 J/mol = 14.2318 kJ/mol ≈ 14.232 kJ/mol (ans).

D) Take the values at 298 K and use the equation

ΔG⁰ = ΔH⁰ – T*ΔS⁰

Here, we must have,

Δ_rG⁰₂₉₈ = Δ_rH⁰ – (298 K)*Δ_rS⁰

====> 27.451 kJ/mol = 14.232 kJ/mol – (298 K)*Δ_rS⁰

====> Δ_rS⁰ = (27.451 – 14.232) kJ/mol/298 K = 0.044359 kJ/mol.K = 44.359 J/mol.K (ans).