Consider These Reactions

Question

Consider these reactions, where M represents a generic metal. 2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)Δ𝐻1=−626.0 kJ

HCl(g)⟶HCl(aq) Δ𝐻2=−74.8 kJ

H2(g)+Cl2(g)⟶2HCl(g) Δ𝐻3=−1845.0 kJ

MCl3(s)⟶MCl3(aq) Δ𝐻4=−225.0 kJ

Use the given information to determine the enthalpy of the reaction 2M(s)+3Cl2(g)⟶2MCl3(s)

Answer 1

Hess' Law:

2M(s) + 3Cl2(g) ==> 2MCl3(s) TARGET EQUATION

Given:

eq.1: 2M(s) + 6HCl(aq)⟶2MCl3(aq) + 3H2(g) ... Δ𝐻1=−626.0 kJ

eq.2: HCl(g)⟶HCl(aq) ... Δ𝐻2=−74.8 kJ

eq.3: H2(g) + Cl2(g)⟶2HCl(g) ... Δ𝐻3=−1845.0 kJ

eq.4: MCl3(s)⟶MCl3(aq) ... Δ𝐻4=−225.0 kJ

Procedure:

copy eq.1: 2M(s) + 6HCl(aq)⟶2MCl3(aq) + 3H2(g) ... Δ𝐻=−626.0 kJ

copy eq.3 & x3: 3H2(g) + 3Cl2(g)⟶6HCl(g) ... Δ𝐻=−5535 kJ

copy eq.2 &x6: 6HCl(g)⟶6HCl(aq) ... Δ𝐻 = −448.8 kJ

reverse eq.4 &x2: 2MCl3(aq) —> 2MCl3(s) … Δ𝐻 = +450.0 kJ

_______________________________________add all equations and cancel like terms to get...

2M(s) + 3Cl2(g) —> 2MCl3(s) TARGET EQUATION

∆H = -620 + (-5535) + (-448.8) + 450

∆H = -6154 kJ