Chemistry Homework

Question

Consider these reactions, where M represents a generic metal.

2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)Δ𝐻1=−928.0 kJ

HCl(g)⟶HCl(aq) Δ𝐻2=−74.8 kJ

H2(g)+Cl2(g)⟶2HCl(g) Δ𝐻3=−1845.0 kJ

MCl3(s)⟶MCl3(aq) Δ𝐻4=−446.0 kJ

Use the given information to determine the enthalpy of the reaction

2M(s)+3Cl2(g)⟶2MCl3(s)

Answer 1

This is a problem that requires some understanding of Hess' Law.

Given:

eq.1: 2M(s) + 6HCl(aq) ⟶ 2MCl₃(aq) + 3H₂(g) ... Δ𝐻1=−928.0 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=−446.0 kJ

Asked for ∆H for the following target equation:

2M(s) + 3Cl₂(g) ⟶ 2MCl₃(s)

Procedure:

Copy eq.1: 2M(s) + 6HCl(aq) ⟶ 2MCl₃(aq) + 3H₂(g) ... Δ𝐻1=−928.0 kJ

Copy eq.3 times 3: 3H₂(g) + 3Cl₂(g)⟶ 6HCl(g) ... Δ𝐻3=−1845.0 kJ x 3 = -5535 kJ

Copy eq.2 times 6: 6HCl(g) ⟶ 6HCl(aq) ... Δ𝐻2=−74.8 kJ x 6 = -449 kJ

Reverse eq.4: MCl₃(aq) ⟶ MCl₃(s) ... Δ𝐻4= +446.0 kJ

Add all together and combine/cancel like terms to obtain the target equation:

2M(s) + 3Cl₂(g) ⟶ 2MCl₃(s)

∆H = (-928 kJ) + (-5535 kJ) + (-449 kJ) + 446 kJ

∆H = -6466 kJ