Answer:
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔH°rxn 74,89 kJ/mol
Explanation:
The change in enthalpy of formation (ΔHf) is defined as the change in enthalpy in the formation of a substance from its constituent elements. For HCl(g):
<em>(1) </em>¹/₂H₂(g) + ¹/₂ Cl₂ → HCl(g) ΔH = -92,3 kJ/mol
It is possible to sum ΔH of different reactions to obtain ΔH of a global reaction (Hess's law).
For the reactions:
<em>(2) </em>N₂(g) + 4H₂(g) + Cl₂(g) → 2NH₄Cl(s) ΔH°rxn = −630.78 kJ/mol
<em>(3)</em> N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol
The sum of -(2) + (3) gives:
<em>-(2) </em>2NH₄Cl(s) → N₂(g) + 4H₂(g) + Cl₂(g) ΔH°rxn = +630.78 kJ/mol
<em>(3)</em> N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol
<em>-(2) + (3) </em>2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g)
ΔH°rxn = +630.78 kJ/mol −296.4 kJ/mol = +334,38 kJ/mol
Now, the sum of -(2) + (3) + 2×(1)
<em>-(2) + (3) </em>2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g) ΔH°rxn = +334,38 kJ/mol
<em>2×(1) </em>H₂(g) + Cl₂(g)→ 2HCl(g) ΔH = 2×-92,3 kJ/mol
<em>-(2) + (3) + 2×(1) </em>2NH₄Cl(s) → 2NH₃(g) + 2HCl(g)
ΔH°rxn = +334,38 kJ/mol + 2×-92,3 kJ/mol = 149,78 kJ/mol
The reaction of:
<em>NH₄Cl(s) → NH₃(g) + HCl(g)</em>
<em>Has ΔH°rxn = 149,78kJ/mol / 2 = 74,89 kJ/mol</em>
I hope it helps!
