Question

Item 19

Given the data, calculate ΔSvap for each liquids. (ΔSvap=ΔHvap/T, where T is in K)

Compound	Name	BP(∘C)	ΔHvap(kJ/mol)atBP
C4H10O	Diethyl ether	34.6	26.5
C3H6O	Acetone	56.1	29.1
C6H6	Benzene	79.8	30.8
CHCl3	Chloroform	60.8	29.4
C2H5OH	Ethanol	77.8	38.6
H2O	Water	100	40.7

Part A

Diethyl ether

Express your answer with the appropriate units.

ΔSvap =

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Part B

Acetone

Express your answer with the appropriate units.

ΔSvap =

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Part C

Benzene

Express your answer with the appropriate units.

ΔSvap =

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Part D

Chloroform

Express your answer with the appropriate units.

ΔSvap =

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Part E

Ethanol

Express your answer with the appropriate units.

ΔSvap =

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Part F

Water

Express your answer with the appropriate units.

ΔSvap =

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Answer 1

At boling point of a substance, a state of equilibrium between the liquid and gas phases exitsts, ΔG = 0.

ΔG = ΔH – TΔS

at boiling point , ΔH – TΔS = 0

ΔS =  ΔH / T

i) Diethyl ether, given that ΔH = 26.5 kJ/mol and T = 34.6 ^oC = 307.6 K

ΔS =  ΔH / T = (26.5 kJ/mol) / 307.6 K = 0.08615 kJ mol^-1 K^-1 = 0.086 * 1000 J mol^-1 K^-1= 86.15 J mol^-1 K^-1

ii) Acetone, given that ΔH = 29.1 kJ/mol and T = 56.1 ^oC = 329.1 K

ΔS =  ΔH / T = 29.1 / 329.1 = 88.42 J mol^-1 K^-1

iii) Benzene, given that ΔH = 30.8 kJ/mol and T = 79.8 ^oC = 352.8 K

ΔS =  ΔH / T = 30.8 / 352.8 = 87.30 J mol^-1 K^-1

iv) Chloroform, given that ΔH = 29.4 kJ/mol and T = 60.8 ^oC = 333.8 K

ΔS =  ΔH / T = 29.4 / 333.8 = 88.07 J mol^-1 K^-1

v) Ethanol, given that ΔH = 38.6 kJ/mol and T = 77.8 ^oC = 350.8 K

ΔS =  ΔH / T = 38.6 / 350.8 = 88.07 J mol^-1 K^-1 = 110.03 J mol^-1 K^-1