Answer:
3H2(g)+Fe2O3(s)→2Fe(s)+3H2O N2(g)+3H2(g)→2NH3(g)
ΔH∘rxn = 98.8 kJ -91.8 kJ
ΔS∘rxn = 0.141.5 kJ/K 0.198 kJ/K
ΔG∘rxn = 56.78 kJ -150 .80 kJ
Explanation:
The ΔH∘rxn is given by Hess law utilizing the standard heat of formation of reactants and products. The calculation for ΔS∘rxn, proceeds in a similar manner using the change in the standard entropy of formation of reactants and products. Finally for ΔG∘rxn we use the relation:
ΔG∘rxn = ΔH∘rxn - T ΔS∘rxn where T is the standard temperature 298 K.
We need the values for the standard enthaply of formation of the reactants and products which can be consulted in an appropiate handbook such as Lange, CRC, etc or in your text. The same goes to the absolute entropies of formation.
With that in mind, lets solve the question.
3H2(g)+Fe2O3(s)→2Fe(s)+3H2O(g)
ΔH∘f (kJ/mol) 0 -824.2 0 -241.8
Sº J/K 130.6 87.4 27.3 188.7
ΔH∘rxn = 3(-241.8) + 0 - ( 0 +(-824.2)) = -725.4 + 822.2 = 98.8 kJ
ΔS∘rxn = 2(27.3) + 3(188.7) - ( 3(130.6) + 87.4) = 141.5 J/k = 0.141.5 kJ/mol (Note need to convert the joules to kilo joules)
ΔG∘rxn = 98.8 kJ - 298 K (0.141 kJ/K) = 56.78 kJ
N2(g)+3H2(g)→2NH3(g)
ΔH∘f (kJ/mol) 0 0 -45.94
Sº J/K 191.6 130.5 192.5
ΔH∘rxn = 2(-45.94) - (0+0) = -91.8 kJ/mol
ΔS∘rxn = 2(192.5) -( 191.6 + 3(130.5)) = -198.1 J = 0. 198
ΔG∘rxn = = -91.8 kJ -298 K(0.198 kJ/K) = -150.80 kJ
