Standard Enthalpy Change

Question

Given the standard enthalpy changes for the following two reactions: 1. 2Pb (s) + O2 (g) yields 2PbO (s) delta H = -434.6 2. 2Zn (s) + O2 (g) yields 2ZnO (s) delta H = -696.6 What is the standard enthalpy change for the following reaction? 3. PbO (s) + Zn (s) yields Pb (s) + ZnO (s) delta H = ?

 Standard enthalpy change = kJ

Answer 1

Hess' Law

Target equation:

PbO (s) + Zn (s) ==> Pb (s) + ZnO (s)

Given:

eq.1: 2Pb (s) + O2 (g) ==> 2PbO (s) ∆H = -434.6

eq.2: 2Zn (s) + O2 (g) ==> 2ZnO (s) ∆H = -696.6

Procedure:

Reverse eq.1 and divide by 2: PbO(s) ==> Pb(s) + 1/2 O2(g) ∆H = +217.3

Copy eq.2 and divide by 2: Zn(s) + 1/2 O2(g) ==> ZnO(s) ∆H = -348.3

Add these together and combine/cancel like terms:

PbO(s) + Zn(s) ==> Pb(s) + ZnO(s) which is the target equation

∆H = +217.3 - 348.3

∆H = -131 kJ