Answer to Question #167441 in Chemistry for Trexia

Question

Given the thermochemical equations of lead that react to chlorine to produced lead (II) chloride.

Determine the ΔH for 2 PbCl2 → Pb + PbCl4

Pb + Cl2 → PbCl2 ΔH = −223 kJ

PbCl2 + Cl2 → PbCl4 ΔH = −87 kJ

Answer 1

Solution:

2PbCl₂ → Pb + PbCl₄, ΔH_x = ???

(1^st equation): Pb + Cl₂ → PbCl₂, ΔH₁ = −223 kJ

(2^nd equation): PbCl₂ + Cl₂ → PbCl₄, ΔH₂ = −87 kJ

According to Hess's law, the heat of reaction depends upon Initial and final conditions of reactants and does not depend of the intermediate path of the reaction.

1) Modify the two given equations to get the target equation:

a) (1^st eq): flip it so as to put PbCl₂ on the reactant side.

b) (2^nd eq): do nothing. We need one PbCl₄ on the product side and that's what we have.

2) Rewrite the two equations with the changes made (including changes in their enthalpy):

(1*^st equation): PbCl₂ → Pb + Cl₂, ΔH*₁ = +223 kJ

(2^nd equation): PbCl₂ + Cl₂ → PbCl₄, ΔH₂ = −87 kJ

Thus, adding modified equations and canceling out the common species on both sides, we get:

2PbCl2 → Pb + PbCl₄

3) Add the ΔH values of (1*^st) and (2^nd) equations to get your answer:

∆H_x = (+223 kJ/mol) + (-87 kJ/mol) = +136 kJ/mol

Hence, ΔH_x is +136 kJ/mol.

Answer: ΔH_x is 136 kJ/mol.