Question

Use the thermodynamic data provided below to determine ΔG (in kJ/mol) for the condensation of NaCl at 138.08 °C if the initial partial pressure of NaCl is 1.62 atm. Report your answer to one decimal place in standard notation (i.e. 123.4 kJ/mol).

Substance		ΔH°f (kJ/mol)		S° (J mol-1K-1)
NaCl (l)		-385.9		95.1
NaCl (g)		-181.4		229.8

Answer 1

Substance		ΔH°f (kJ/mol)		S° (J mol-1K-1)
NaCl (l)		-385.9		95.1
NaCl (g)		-181.4		229.8

NaCl(g) --> NaCl (l)
S = 229.8J mol^-1K^-1 --> 95.1 J mol^-1K^-1

dS = prod - reactant
dS =95.1 J mol^-1K^-1-229.8 J mol^-1K^-1

= - 134.7 J mol^-1K^-1

=- 0.1347 kJ mol^-1K^-1

NaCl(g) --> NaCl (l)
H = -181.4KJ mol^-1 --> -385.9K J mol^-1

dH = prod - reactant
dH =-385.9 K J mol^-1-(-181.4) KJ mol^-1

dH = - 204.5 KJ mol^-1

138.08Celsius .... Kelvin temp = 411.23 K

dG = dH - TdS
dG = - 204.5 KJ mol^-1- ( 411.23K) (- 0.1347kJ/K mol)
dG = - 204.5 KJ mol^-1+ 55.39 kJ / mol
dG = - 259.89 kJ/mol or -259.9 kJ/mol