Question

Chlorine trifluoride, CIF3, is a toxic, intensely reactive gas. It was used in World War II to make incendiary bombs. It reacts with ammonia and forms nitrogen, chlorine, and hydrogen fluoride gases. When two moles of chlorine trifluoride reacts, 1196 kj of heat is evolved. What is the standard enthalpy of formation of chlorine trifluoride?
2CIF3(g) + 2NH3 = N2 + 6HF + Cl2

Answer 1

$2ClF_3 + 2NH_3 \to N_2 + 6H_F + Cl_2 +1196kJ$

$ΔH_f°reaction=∑ΔH_f(products)−∑ΔH_f(Reactants)$

ΔHf°[$N_2$] = 0 kJ/mol(plain elements have 0kj/mol)

ΔHf°[$6HF$] = 6(-271.1) kj/mol = -1626.6 kj/mol

ΔHf°[$Cl_2$] = 0 kJ/mol(plain elements have 0kj/mol)

ΔHf°[$2NH_3$] = 2(-46) kJ/mol = -92 kJ/mol

∆Hf°[$2NH_3$]) = ?

Putting all the values into the equation above, we have;

-1196 = 0 - 1626.6 + 0 -(-92 + ∆Hf°[$2NH_3$])

∆Hf°[$2NH_3$] = 1196 - 1626.6 + 92

∆Hf°[$2NH_3$]) = -338.6

∆Hf°[$NH_3$]) = -338.6/2

∆Hf°[$NH_3$]) = -169.3 kJ/mol

$\therefore$ The standard enthalpy of formation of chlorine trifluoride when 1196 kJ of heat is evolved is -169.3kJ/mol.