Question

Ammonia (NH3(g), Delta.Hf = –46.19 kJ/mol) reacts with hydrogen chloride (HCl(g), Delta.Hf = –92.30 kJ/mol) to form ammonium chloride (NH4Cl(s), Delta.Hf = –314.4 kJ/mol) according to this equation:NH3(g) + HCl(g) Right arrow. NH4Cl(s)What is Delta.Hrxn for this reaction?

Answer 1

The enthalpy change for the reaction NH3(g) + HCl(g) → NH4Cl(s) is -175.91 kJ/mol.

Step-by-step explanation:

The enthalpy change for a reaction is also known as the heat of reaction and is represented by the symbol ΔHrxn. In this case, the reaction is:

NH3(g) + HCl(g) → NH4Cl(s)

The enthalpy change for the reaction is calculated by subtracting the sum of the enthalpies of the reactants from the sum of the enthalpies of the products. So, ΔHrxn = ΔHf(NH4Cl) - (ΔHf(NH3) + ΔHf(HCl))

Substituting the given values, ΔHrxn = (-314.4 kJ/mol) - ((-46.19 kJ/mol) + (-92.30 kJ/mol)) = -314.4 kJ/mol + 138.49 kJ/mol = -175.91 kJ/mol