Question

Calculate the entropy of vaporization, ΔSvap, for A(l) at 25.0 °C given that the boiling point of A is 72.45 °C, and the molar heat capacity of A(l) is 116.45 J/(mol·K). Assume that the molar heat capacity of A(g) is 56.5% of that of A(l).

Calculate the standard Gibbs free energy of vaporization, ΔG°vap, at 25.0 °C.

Determine the equilibrium constant, K, for the vaporization at 191.0 °C.

Determine the equilibrium constant, K, for the vaporization at 191.0 °C.

Answer #1

(i). Formula
(i): **ΔSvap = nCp ln(T2/T1)**

Here, Cp = 116.45 J/(mol·K), T2 = 72.45 °C = 72.45 + 273.15 = 345.6 K and T1 = 25 °C = 298.15 K

Let's consider n = 1 mol

Now, ΔSvap = 1 mol * 116.45 J/(mol·K) * ln(345.6/298.15)

i.e. **ΔSvap = 17.198 J/K**

(ii) Formula
(ii): **ΔH ^{o}vap = nCp*ΔT**

i.e. ΔH^{o}vap = 1 mol * 116.45 J/(mol·K) *
(345.6-298.15) K

i.e. ΔH^{o}vap = 5525.55 J

Formula (iii):
**ΔG ^{o}vap = ΔH^{o}vap - TΔSvap**

Now, ΔG^{o}vap = 5525.55 J - 298.15 K * 17.198 J/K

i.e. **ΔG ^{o}vap = 397.97 J**

(iii) Formula
(iv): **ΔG ^{o}vap**

i.e. **K = e ^{-ΔGovap/RT}**

Here, T = 191 ^{o}C = 464.15 K and R = universal gas
constant

i.e. K = e^{-397.97/8.314*464.15}

i.e. **K = 0.902**

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