Calculate $\Delta$G for the vaporization of Br₂(l) at 25°C and 1 atm.

Question

Quizplus · Accepted Answer

Using the formula ΔG = ΔH - TΔS, where ΔG is the Gibbs free energy change, ΔH is the enthalpy change, T is the temperature in Kelvin, and ΔS is the entropy change.

For vaporization, ΔS is usually positive, as the number of particles increases in the gas phase. The enthalpy change can be found using the Clausius-Clapeyron equation:

ln(P₂/P₁) = -ΔH/R (1/T₂ - 1/T₁)

where P₁ and P₂ are the vapor pressures at two temperatures T₁ and T₂ respectively, ΔH is the enthalpy of vaporization, R is the gas constant. We can assume P₁ is negligible, so ln(P₂/P₁) ≈ ln(P₂).

Using data from NIST, we can find the vapor pressure of Br2(l) at 25°C to be 95.95 torr, or 0.126 atm.

ln(0.126/1) = -ΔH/(8.314 J/K/mol)(1/298 - 1/373)

ΔH = -8965 J/mol = -8.965 kJ/mol

ΔS can be found using the molar entropy values from NIST:

ΔS = (2 mol)(198.76 J/mol K) - (1 mol)(152.23 J/mol K) = 244.29 J/K/mol

ΔG = ΔH - TΔS = -8.965 kJ/mol - (298 K)(244.29 J/K/mol) = 3.15 kJ/mol.

Therefore, the answer is A) 3.15 kJ/mol.

Calculate Δ\DeltaΔ G for the Vaporization of Br2(l) at 25°C and 1

Calculate $\Delta$ G for the Vaporization of Br₂(l) at 25°C and 1