Calculate enthalpy change

Question

Given the following two reactions:

(1)Pb(s) + Cl₂(g)PbCl₂(s)     H(1) = -359.4 kJ

(2)Mg(s) + Cl₂(g)MgCl₂(s)     H(2) = -641.3 kJ

calculate the enthalpy change for the following reaction:

(3) Pb(s) + MgCl₂(s) PbCl₂(s) + Mg(s)     H(3) = ???? kJ

Answer 1

This is an example of using Hess' Law

Pb(s) + MgCl2(s) ==> PbCl2(s) + Mg(s) ... TARGET EQUATION

Given:

(1). Pb(s) + Cl2(g) ==> PbCl2(s) ... ∆H = -359.4 kJ

(2). Mg(s) + Cl2(g) ==> MgCl2(s) ... ∆H = -641.3 kJ

Procedure:

Copy (1): Pb(s) + Cl2(g) ==> PbCl2(s) ... ∆H = -359.4 kJ

Reverse (2): MgCl2(s) => Mg(s) + Cl2(g) ... ∆H = +641.3 kJ (note to change sign)

Add them and combine and/or cancel like terms to get...

Pb(s) + MgCl2(s) ==> PbCl2(s) + Mg(s) ... TARGET EQUATION

∆H = -359.4 kJ + 641.3 kJ

∆H = +281.9 kJ