The boiling point of a liquid is 64°C and the enthalpy change for the conversion of this liquid to the gas is 32.21 kJ/mole. What is the entropy change for vaporization, ΔSvap, in J/(mol·K)?
A
503.3 J/(mol·K)
B
97.3 J/(mol·K)
C
64.0 J/(mol·K)
D
322.1 J/(mol·K)
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1
Identify the given values: the boiling point of the liquid is 64°C, which needs to be converted to Kelvin by adding 273.15, resulting in 337.15 K. The enthalpy change for vaporization, ΔHvap, is given as 32.21 kJ/mol.
Recall the formula for calculating the entropy change for vaporization, ΔSvap, which is ΔSvap = ΔHvap / T, where T is the temperature in Kelvin.
Convert the enthalpy change from kJ/mol to J/mol to match the units of entropy change. Since 1 kJ = 1000 J, multiply 32.21 kJ/mol by 1000 to get 32210 J/mol.
Substitute the values into the formula: ΔSvap = 32210 J/mol / 337.15 K.
Perform the division to find the entropy change for vaporization, ΔSvap, in J/(mol·K).