What is the value for delta G for the following reaction?

Question

Using the data:

Ag2O(s), = –31.1 kJ mol-1, S° = 121.3 J mol-1 K-1

Ag(s), = 0.00 kJ mol-1, S° = 42.55 J mol-1 K-1

O2(g), = 0.00 kJ mol-1, S° = 205.0 J mol-1 K-1

if we assume that, since the physical states do not change, enthalpy and entropy are independent of temperature between –50.0 °C and 950.0 °C, calculate the value of delta G for the reaction at 650 ºC:

Ag2O(s) yields 2 Ag(s) ½ O2(g)

Answer 1

Ag2O(s), dHf = –31.1 kJ mol-1,........ S° = 121.3 J mol-1 K-1

Ag(s), dHf = 0.00 kJ mol-1,..........     S° = 42.55 J mol-1 K-1

O2(g), dHf = 0.00 kJ mol-1, ......   .....S° = 205.0 J mol-1 K-1

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Ag2O(s)  -->   2 Ag(s)  &   ½ O2

dH reaction = prod - reactants]

dH rxn = [ zero & zero ] - ( -31.1kJ)

d H reaction = + 31.1 kJ

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Ag2O(s)  -->   2 Ag(s)  &   ½ O2

dS reaction = prod - reactants]

dS rxn = [(2)(42.55) & (1/2) (205.0)]  - ( 121.3J)

dS reaction = 85.1 & 102.5 -121.3

dS = + 66.3 Joules

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now, calculate the value of delta G for the reaction@ 650 C (aka 923Kelvin)

dG = dH - TdS

dG =  + 31.1 kJ - (923K) (0.0663 kJ)

dG = 31.1 - 61.2

your answer: dG = -30.1 kJ