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^ovap = nCp*ΔT

i.e. ΔH^ovap = 1 mol * 116.45 J/(mol·K) * (345.6-298.15) K

i.e. ΔH^ovap = 5525.55 J

Formula (iii): ΔG^ovap = ΔH^ovap - TΔSvap

Now, ΔG^ovap = 5525.55 J - 298.15 K * 17.198 J/K

i.e. ΔG^ovap = 397.97 J

(iii) Formula (iv): ΔG^ovap = -RT lnK

i.e. K = e^-ΔGovap/RT

Here, T = 191 ^oC = 464.15 K and R = universal gas constant

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

i.e. K = 0.902