Question

Consider these reactions, where M represents a generic metal.

1. 2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)ΔH1=−819.0 kJ2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)ΔH1=−819.0 kJ
2. HCl(g)⟶HCl(aq)  ΔH2=−74.8 kJHCl(g)⟶HCl(aq)  ΔH2=−74.8 kJ
3. H2(g)+Cl2(g)⟶2HCl(g) ΔH3=−1845.0 kJH2(g)+Cl2(g)⟶2HCl(g) ΔH3=−1845.0 kJ
4. MCl3(s)⟶MCl3(aq)  ΔH4=−258.0 kJMCl3(s)⟶MCl3(aq)  ΔH4=−258.0 kJ

Use the given information to determine the enthalpy of the reaction

2M(s)+3Cl2(g)⟶2MCl3(s)

Answer 1

Lets number the reaction as 1, 2, 3, 4, 5 from top to bottom
required reaction should be written in terms of other reaction
This is Hess Law
required reaction can be written as:
reaction 5 = +1 * (reaction 1) +6 * (reaction 2) +3 * (reaction 3) -2 * (reaction 4)

So, ΔHo rxn for required reaction will be:
ΔHo rxn = +1 * ΔHo rxn(reaction 1) +6 * ΔHo rxn(reaction 2) +3 * ΔHo rxn(reaction 3) -2 * ΔHo rxn(reaction 4)
= +1 * (-819.0) +6 * (-74.8) +3 * (-1845.0) -2 * (-258.0)
= -6286.8 KJ
Answer: -6286.8 KJ