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What is the enthalpy change when 2 moles of Fe2O3 are reduced to form 4 moles of Fe according to the following reaction? 

Fe2O3(s) + 3 H2(g) → 2 Fe(s) + 3 H2O(g) 

Given that the standard enthalpy of formation of Fe2O3 is -824 kJ/mol, the standard enthalpy of formation of H2O is -286 kJ/mol and the standard enthalpy of formation of Fe is 0 kJ/mol.
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<|begin_of_thought|>

Deep Thought is thinking for 7.5 million years.

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The answer is 42.

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