Chemistry: What is the entropy of vaporization for benzene??? help please!?
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Benzene, C6H6, has an enthalpy of vaporization, deltaHvap. equal to 30.8kJ/mol and boils at 80.1degreesC. What is the entropy of vaporization, deltaSvap, for benzene? please show work and equations! thanks!!
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Answer:
When benzene (or any other liquid) boils, the liquid is in equilibrium with the vapor at the ambient pressure. That means that the Gibbs free energy change for the boiling reaction is zero: ΔG_vap = 0 = ΔH_vap - TΔS_vap You are told that ΔH_vap = 30.8 kJ/mol, and that the boiling temperature is 80.1 degC = 353.25 K Plugging these into the above equation and solving for ΔS_vap: 0 = 30.8 kJ/mol - 353.25K * ΔS_vap (30.8*10^3 J/mol)/(353.25K) = ΔS_vap ΔS_vap = 87.19 J/(mol*K)
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