Question

You are given the following thermodynamic data. 2 Fe(s) + 3/2 O2(g) → Fe2O3(s) ΔH° = -823 kJ 3 Fe(s) + 2 O2(g) → Fe3O4(s) ΔH° = -1120. kJ Calculate the ΔH° for the following reaction. 3 Fe2O3(s) → 2 Fe3O4(s) + ½ O2(g)

Answer 1

2Fe + 3/2O2 --> Fe2O3 , dH = -823 KJ ..................(1)

3Fe + 2O2 ---> Fe3O4 , dH = -1120 KJ .........................(2)

Now 2 times eq 2 - 3 times eq 3 gives us              ( - 3 times eq 3 means we add reverse of eq 3 multiplied by factor 3)

6Fe + 4O2 + 3Fe2O3 ---> 2Fe3O4 + 6Fe + 9/2O2

net equation is 3F2O3 ( s) --> 2Fe3O4 (s) + 1/2O2 ( g) , dH = 2( -1120) -3( -823) = 229 KJ