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A student wishes to determine all the standard the
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Answer #1

Apply gibb's free energy equation

\Delta G =\Delta H - T\Delta S

Ag+

\Delta H = \Delta G + T\Delta S

= 77.11 + (298x73.45)/1000 (Consider T = 298.15K/ 25 C)

= 98.99 kJ mol-1

Ag (s)

Enthaply (del H) for formation of an element is 0

\Delta G =\Delta H - T\Delta S

= 0- (298x 42.6)/1000

= -12.69 kJ mol-1

Fe3+

Change in Gibb's free energy for reaction is 0 for spontaneous reaction

Del G = -12.69 - (77.11 - 84.94)

Del G = -12.69 +7.83

Del G = -4.86 kJ/mol

TS = -47.69 -(-4.86) = -- 42.86 KJ

S = -143.7 J/K

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