Calculate Delta H and Delta S (2 of 3)

Question

Calculate delta H and delta S for the following reaction and predict whether the reaction is either: always product favored, product favored at only at low temperatures, product favored only at high temperatures, or never product favored.

FeO3 (s) 2 Al (s) ===> 2 Fe (s) Al2O3 (s)

Answer 1

Fe2O3 (s)

dHf = - 822.16 kJ

S = 89.96J



2 Al (s)

dHf = 0

S = (2) (28.32 J) = 56.64J

2 Fe (s)

dHf = 0

S = (2) (27.15J) = 54.30J

Al2O3 (s)

dHf = - 1669.8 kJ

S = 51J

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Fe2O3 (s) &  2 Al (s) ===> 2 Fe (s) & Al2O3 (s)

dH = prod - reactants

dH = (- 1669.9) - ( - 822.16 kJ)

Your answer: dH = - 847.9 kJ

dS = prod - reactants

dS = (51 & 54.3) - (56.64  & 89.96J)

your answer is: - 41.3 J

with needing a dG = - to be spontaneous,

& dG = dH -TdS

your answer is: product favored at only at low temperatures,

& that is because the dH prodides the desired "-",  but @ high temps,

-TdS will provide a "+" that will eventually over power the "-" of the dH