Answer to Question #71209 in Organic Chemistry for Nancy

Question

A scientist measures the standard enthalpy change for the following reaction to be -79.6 kJ:

2HBr(g) + Cl2(g)2HCl(g) + Br2(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of Br2(g) is: kJ/mol.

Answer 1

ΔHf(HCl(g)) = -92.3 kJ/mol
ΔHf(HBr(g)) = -36.3 kJ/mol
ΔHr = 2ΔHf(HCl(g)) + ΔHf(Br2(g)) - 2ΔHf(HBr(g)) - ΔHf(Cl2(g))
ΔHf(Br2(g)) = ΔHr + 2ΔHf(HBr(g)) + ΔHf(Cl2(g)) - 2ΔHf(HCl(g)) = -79.6 – 2∙36.3 + 0 + 2∙92.3 = 32.4 (kJ/mol)
Answer: ΔHf(Br2(g)) = 32.4 (kJ/mol)