Question

In: Chemistry

Calculate the standard entropy of vaporization of ammonia at 210.0K, given that the molar heat capacities...

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.

Solutions

Expert Solution

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

...............................................

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


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