In: Chemistry
Calculate the standard entropy of vaporization of ammonia at 210.0K, given that the molar heat capacities at constant pressure of liquid ammonia and ammonia vapor are 80.8 J.K-1.mol-1 and 35.1 J.K-1mol-1, respectively, in this range.
The temperature of standard condition is 273.15K.
The boiling point of ammonia is 239.8 K, where the heat of vaporization is 25.8 kJ/mol.
The entropy change for vaporization at this temperature is
ΔSv = ΔHv/T = 25 800 J/mol / 239.8 K = 107.59 J.mol-1K-1
The calculation temperature is 210.1 K.
hl = 80.8 J.K-1.mol-1
hg = 35.1 J.K-1mol-1
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Consider 1 mol liquid ammonia at 210.1 K (initial state):
First calculate the entropy change for heating at 239.8 K from 210.1 K :
ΔS1 = hl ln (Tfinal/Tinitial) = 80.8 J.K-1.mol-1 x ln (239.8/210.1) = 10.68 J.mol-1K-1
Then add the entropy change for vaporization ΔSv at 239.8K, calculated above, 107.59 J.mol-1K-1
Then add the entropy change for cooling vapor from 239.8K to 210.1K:
ΔS2 = hg ln (Tfinal/Tinitial) = 35.1 J.mol-1K-1 x ln(210.1/239.8) = -4.64 J.mol-1K-1
Ammonia vapor at 210.1K is the final state (conventional). For the entire process
ΔS = ΔS1 + ΔSv + ΔS2 = 10.68 + 107.59 - 4.64 = 113.6 J.mol-1K-1 = 114 J.mol-1K-1