The enthalpy changes for two reactions are given by the equations2Cr(s)+32O2(g)→Cr2O3(g),ΔH=−1130kJ ; C(s)+12O2(g)→CO(g),ΔH=−110kJ ; What is the enthalpy change (in kJ) for the reaction ? 3C(s) + Cr2O3(s) → 2Cr(s) + 3CO(g)

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-1460kJ

800kJ

-700kJ

650kJ

answer is B.

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Detailed Solution

Heat change of a reaction = ∆H = (Sum of the Heats of formation of products)- (Sum of the Heats of formation of reactants) ∆H = [2(0) + 3(-110)] – [ 3(0) + (-1130)] = 800kJ

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