Question

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13. Find the Hºrn for the reaction 2NH3(g) + 3Cl2(g) → N2(g) + 6HCl(g). AH°F (NH3) = -46 kJ/mol, AH°F (HCI) = -92.3 kJ/mol Answer: -461.8 kJ 14. Use the standard enthalpies of formation to calculate AH°rxn: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) AH°F (Fe2O3) = -826 kJ/mol, AH°F (CO) = -110.5 kJ/mol, AH°F (CO2) = -393.5 kJ/mol Answer: -23 kJ 15. Given the following reaction: B2H6(g) + 6H2O(l) → 2H3BO3(g) + 6H2(g) AH°= -493.4 kJ AH°t (H3BO3) = -1088.7 kJ/mol AH° (H20) = -285.5 kJ/mol Answer: 30.8 kJ

Answer 1

13)

Given:

Hof(NH3(g)) = -46.0 KJ/mol

Hof(Cl2(g)) = 0.0 KJ/mol

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

Hof(HCl(g)) = -92.3 KJ/mol

Balanced chemical equation is:

2 NH3(g) + 3 Cl2(g) ---> N2(g) + 6 HCl(g)

ΔHo rxn = 1*Hof(N2(g)) + 6*Hof(HCl(g)) - 2*Hof( NH3(g)) - 3*Hof(Cl2(g))

ΔHo rxn = 1*(0.0) + 6*(-92.3) - 2*(-46.0) - 3*(0.0)

ΔHo rxn = -461.8 KJ

Answer: -461.8 KJ

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