Question

Calculate standard free energy change using AHp and So values. Consider the reaction 2NH3(8) + 3N20(g)—__4N2(g) + 3H2O(g) Calculate the standard free energy change for this reaction at 25°C from AH°rxn and AS°rxn- AGºrxn = kJ/mol

Answer 1

Given:

Hof(NH3(g)) = -46.11 KJ/mol

Hof(N2O(g)) = 82.05 KJ/mol

Hof(N2(g)) = 0.0 KJ/mol

Hof(H2O(g)) = -241.818 KJ/mol

Balanced chemical equation is:

2 NH3(g) + 3 N2O(g) ---> 4 N2(g) + 3 H2O(g)

ΔHo rxn = 4*Hof(N2(g)) + 3*Hof(H2O(g)) - 2*Hof( NH3(g)) - 3*Hof(N2O(g))

ΔHo rxn = 4*(0.0) + 3*(-241.818) - 2*(-46.11) - 3*(82.05)

ΔHo rxn = -879.384 KJ

Given:

Sof(NH3(g)) = 192.45 J/mol.K

Sof(N2O(g)) = 219.85 J/mol.K

Sof(N2(g)) = 191.61 J/mol.K

Sof(H2O(g)) = 188.825 J/mol.K

Balanced chemical equation is:

2 NH3(g) + 3 N2O(g) ---> 4 N2(g) + 3 H2O(g)

ΔSo rxn = 4*Sof(N2(g)) + 3*Sof(H2O(g)) - 2*Sof( NH3(g)) - 3*Sof(N2O(g))

ΔSo rxn = 4*(191.61) + 3*(188.825) - 2*(192.45) - 3*(219.85)

ΔSo rxn = 288.465 J/K

So we have:

ΔHo rxn = -879.384 KJ/mol

ΔSo rxn = 288.465 J/mol.K

= 0.28847 KJ/mol.K

T= 25.0 oC

= (25.0+273) K

= 298 K

Now use:

ΔGo rxn = ΔHo rxn - T*ΔSo rxn

= -879.384 - 298.0*0.28847

= -965.3466 KJ/mol

Answer: -965.3 KJ/mol