Heat of Rxn (Kaplan typo??)

[s2thindi]

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[zmatrix]

3Fe(s) + 2O2(g) → Fe3O4(s) –1118.4 kJ mol–1
Reactants → Intermediate

4Fe3O4(s) + O2(g) → 6Fe2O3(s) -472.0 kJ mol–1 (flipped equation)
Intermediate → Products


So... 6mol of Products from 4mol of Intermediate requires -472.0 kJ
and... 4mol of Intermediate from Reactants requires 4*-1118.4 kJ

So total heat of enthalpy for 6mol of products is -4944 kJ

Divide: -4944 kJ / 6 mol = –824 kJ mol–1

You have to flip the 2nd equation so that you get Fe2O3 as the product.

 

[s2thindi]

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