Final answer:
To calculate the boiling point of a compound given its heat of vaporization and entropy of vaporization, you use the Gibbs-Helmholtz equation, which simplifies to the boiling point being equal to the heat of vaporization divided by the entropy of vaporization. Upon converting units to J/mol for consistency, the boiling point of the compound is found to be 331.0 K.
Step-by-step explanation:
To determine the boiling point of the compound using its heat of vaporization (ΔHvap) and entropy of vaporization (ΔSvap), you can employ the Clausius-Clapeyron equation, which describes the phase transition between two states of matter, such as liquid and gas. However, there is a more straightforward method called the Gibbs-Helmholtz equation, which relates ΔHvap and ΔSvap directly to the boiling point at a given pressure. For a compound at its normal boiling point, the condition ΔG = 0 (Gibbs free energy change) applies, and therefore ΔHvap = TΔSvap. Rearranging this to solve for the boiling point (T), you get T = ΔHvap / ΔSvap.
Given ΔHvap = 31.31 kJ/mol and ΔSvap = 94.56 J/mol, you first need to ensure that both values are in the same units. Since there are 1,000 J in a kJ, ΔHvap in J/mol would be 31,310 J/mol. Now you can calculate the boiling point:
T = ΔHvap / ΔSvap = 31,310 J/mol / 94.56 J/mol·K = 331.0 K
Thereby, the boiling point of this compound is 331.0 K.