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20 August, 18:32

Calculate ΔrG∘ at 298 K for the following reactions. CO (g) + H2O (g) →H2 (g) + CO2 (g) 2-Predict the effect on ΔrG∘ of lowering the temperature for the reaction above.ΔrG∘ will decrease with decreasing temperature.ΔrG∘ will increase with decreasing temperature.ΔrG∘ will change slightly with decreasing temperature.

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  1. 20 August, 18:46
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    1) ΔG°r (298 K) = - 28.619 KJ/mol

    2) ΔG°r will decrease with decreasing temperature

    Explanation:

    CO (g) + H2O (g) → H2 (g) + CO2 (g)

    1) ΔG°r = ∑νiΔG°f, i

    ⇒ ΔG°r (298 K) = ΔG°CO2 (g) + ΔG°H2 (g) - ΔG°H2O (g) - ΔG°CO (g)

    from literature, T = 298 K:

    ∴ ΔG°CO2 (g) = - 394.359 KJ/mol

    ∴ ΔG°CO (g) = - 137.152 KJ/mol

    ∴ ΔG°H2 (g) = 0 KJ/mol ... pure substance

    ∴ ΔG°H2O (g) = - 228.588 KJ/mol

    ⇒ ΔG°r (298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol) - ( - 137.152 KJ7mol)

    ⇒ ΔG°r (298 K) = - 28.619 KJ/mol

    2) K = e∧ (-ΔG°/RT)

    ∴ R = 8.314 E-3 KJ/K. mol

    ∴ T = 298 K

    ⇒ K = e∧ (-28.619 / (8.314 E-3) (298) = 9.624 E-6

    ⇒ ΔG°r = - RTLnK

    If T (↓) ⇒ ΔG°r (↓)

    assuming T = 200 K

    ⇒ ΔG°r (200 K) = - (8.314 E-3) (200) Ln (9.624E-3)

    ⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r (298 K) = - 28.619 KJ/mol
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