Question

Use the following information to find ΔH°_f of gaseous HCl:

N₂(g) + 3H₂(g) → 2NH₃(g)                     ΔH°_rxn = - 91.8 kJ

N₂(g) + 4H₂(g) + Cl₂(g) → 2NH₄Cl(s)   ΔH°_rxn = - 628.8 kJ

NH₃(g) + HCl(g) → NH₄Cl(s)               ΔH°_rxn = - 176.2 kJ

Answer 1

N2(g) + 3H2(g) → 2NH3(g) ΔH°rxn = - 91.8 kJ 1equation

N2(g) + 4H2(g) + Cl2(g) → 2NH4Cl(s) ΔH°rxn = - 628.8 kJ 2 equation

NH3(g) + HCl(g) → NH4Cl(s) ΔH°rxn = - 176.2 kJ 3 equation

Multiply the third equation by -2. The (-) will reverse the direction of the reaction.
Multiply the first reaction by -1.
Sum the modified first and third and add the second. You will be left with
H2(g) + Cl2(g) ---> 2HCl(g).

Now perform the same operations on the ΔH. e.g., third value times -2, first value times -1. Add them now and add the second value. That's it.

(-2)*(-176.2) + (-1)*(-91.8) + 1*(-628.8) = -184.6/2 =so, ∆HRx = −92.3 kJ mol−1 of HCl