Given that $\Delta$H_vap is 60.3 kJ/mol,and the boiling point is 83.4°C,1 atm,if one mole of this substance is vaporized at 1 atm,calculate $\Delta$S.

Question

Quizplus · Accepted Answer

The correct answer can be found using the formula ΔS = ΔH/T, where ΔH is the enthalpy change (60.3 kJ/mol, which is 60300 J/mol for vaporization) and T is the temperature in Kelvin. The boiling point is given as 83.4°C, which is 356.5 K (83.4 + 273.15). Therefore, ΔS = 60300 J/mol / 356.5 K = 169 J/K mol.

Given That Δ\DeltaΔ Hvap Is 60 Δ\DeltaΔ S A)-169 J/K Mol B)169 J/K Mol C)723 J/K Mol

Given That $\Delta$ H_vap Is 60 $\Delta$ S
A)-169 J/K Mol
B)169 J/K Mol
C)723 J/K Mol