For each of the following reactions, calculate δh∘rxn, δs∘rxn, and δg∘rxn at 25 ∘c. 3h2(g)+fe2o3(s)→2fe(s)+3h2o(g)n2(g) +3h2(g)→2nh3(g)

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

the answer is barium and copper bromide

i belive it is water