Question

Consider these reactions, where M represents a generic metal.

2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)    Δ?1=−840.0 kJ

HCl(g)⟶HCl(aq)  Δ?2=−74.8 kJHCl(g)⟶HCl(aq)  ΔH2=−74.8 kJ

H2(g)+Cl2(g)⟶2HCl(g) Δ?3=−1845.0 kJH2(g)+Cl2(g)⟶2HCl(g) ΔH3=−1845.0 kJ

MCl3(s)⟶MCl3(aq)  Δ?4=−152.0 kJMCl3(s)⟶MCl3(aq)  ΔH4=−152.0 kJ

Use the given information to determine the enthalpy of the reaction

2M(s)+3Cl2(g)⟶2MCl3(s)2M(s)+3Cl2(g)⟶2MCl3(s)

Δ?=____kJ

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 * (-840.0) +6 * (-74.8) +3 * (-1845.0) -2 * (-152.0)

= -6519.8 KJ

Answer: -6519.8 KJ