Question

# Use the thermodynamic data provided below to determine ΔG (in kJ/mol) for the condensation of NaCl...

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

 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

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