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5 August, 06:22

Calculate ΔHrxn for the following reaction: Fe2O3 (s) + 3CO (g) →2Fe (s) + 3CO2 (g) Use the following reactions and given ΔH′s. 2Fe (s) + 3/2O2 (g) →Fe2O3 (s), ΔH = - 824.2 kJ CO (g) + 1/2O2 (g) →CO2 (g), ΔH = - 282.7 kJ

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  1. 5 August, 07:23
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    The answer to your question is: ΔHrxm = - 23.9 kJ

    Explanation:

    dа ta:

    2Fe (s) + 3/2O2 (g) →Fe2O3 (s), ΔH = - 824.2 kJ (1)

    CO (g) + 1/2O2 (g) →CO2 (g) ΔH = - 282.7 kJ (2)

    Reaction:

    Fe2O3 (s) + 3CO (g) →2Fe (s) + 3CO2 (g)

    We invert (1) and change the sign of ΔH

    Fe2O3 (s) → 2Fe (s) + 3/2O2 (g) ΔH = 824.2 kJ

    We multiply (2) by 3

    3 (CO (g) + 1/2O2 (g) →CO2 (g) ΔH = - 282.7 kJ) (2)

    3CO (g) + 3/2O2 (g) →3CO2 (g) ΔH = - 848.1 kJ

    We add (1) and (2)

    Fe2O3 (s) → 2Fe (s) + 3/2O2 (g) ΔH = 824.2 kJ

    3CO (g) + 3/2O2 (g) →3CO2 (g) ΔH = - 848.1 kJ

    Fe2O3 (s) + 3CO (g) + 3/2O2 (g) → 2Fe (s) + 3/2O2 + 3CO2 (g)

    Simplify

    Fe2O3 (s) + 3CO (g) →2Fe (s) + 3CO2 (g) and ΔHrxm = - 23.9 kJ
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