Answer to Question #167440 in Chemistry for Trexia

Question

2 PbCl2 → Pb + PbCl4

Pb + Cl2 → PbCl2 ΔH = −223 kJ

PbCl2 + Cl2 → PbCl4 ΔH = −87 kJ

Answer 1

Solution:

2PbCl₂ → Pb + PbCl₄

The following is known:

(1): Pb + Cl₂ → PbCl₂, ΔH = −223 kJ

(2): PbCl₂ + Cl₂ → PbCl₄, ΔH = −87 kJ

Reaction 1 has to be flipped so that PbCl₂ is a reactant. The value of ∆H is also changed from (-) to (+).

PbCl₂ → Pb + Cl₂, ΔH₁ = +223 kJ

Reaction 2 does not need to be changed because PbCl₄ is already a product.

PbCl₂ + Cl₂ → PbCl₄, ΔH₂ = −87 kJ

The two reactions are summed:

PbCl₂ + PbCl₂ + Cl₂ → Pb + Cl₂ + PbCl₄

2PbCl₂ → Pb + PbCl₄, ΔH_r

ΔH_r = ΔH₁ + ΔH₂ = +223 kJ + (−87 kJ) = +136 kJ

ΔH_r = 136 kJ.

Answer: ΔH_r = 136 kJ.