Question

Ammonia (NH3(g), es001-1.jpgHf = –46.19 kJ/mol) reacts with hydrogen chloride (HCl(g), es001-2.jpgHf = –92.30 kJ/mol) to form ammonium chloride (NH4Cl(s), es001-3.jpgHf = –314.4 kJ/mol) according to this equation: NH3(g) + HCl(g) es001-4.jpg NH4Cl(s) What is es001-5.jpgHrxn for this reaction? kJ

Answer 1

Answer is: enthalpy for this reaction is -175.91 kJ.
Chemical reaction: NH₃ + HCl → NH₄Cl.
ΔrH = ∑ΔfH(products of reaction) - ∑ΔfH(reactants).
ΔrH = ΔfH(NH₄Cl) - (ΔfH(NH₃) + ΔfH(HCl)).
ΔrH = -314.4 kJ/mol · 1 mol - (-46.19 kJ/mol · 1 mol + (-92.30 kJ/mol) · 1 mol).
ΔrH = -314.4 kJ + 138.49 kJ.
ΔrH = -175.91 kJ.

Answer 2

The answer is -175.9 kJ and is an exothermic reaction