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29 November, 04:08

P4 (s) + 5O2 (g) ⟶ P4 O10 (s) ΔG° = - 2697.0 kJ/mol 2H2 (g) + O2 (g) ⟶ 2H2 O (g) ΔG° = - 457.18 kJ/mol 6H2 O (g) + P4 O10 (s) ⟶ 4H3 PO4 (l) ΔG° = - 428.66 kJ/mol (a) Determine the standard free energy of formation, ΔGf °, for phosphoric acid.

(b) How does your calculated result compare to the value in Appendix G? Explain.

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  1. 29 November, 07:36
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    the standard free energy of formation of phosphoric acid H3 PO4 is - 1010 kJ/mol

    Explanation:

    Knowing that

    1) P4 (s) + 5O2 (g) ⟶ P4 O10 (s) ΔG° = - 2697.0 kJ/mol

    2) 2H2 (g) + O2 (g) ⟶ 2H2 O (g) ΔG° = - 457.18 kJ/mol

    3) 6H2 O (g) + P4 O10 (s) ⟶ 4H3 PO4 (l) ΔG° = - 428.66 kJ/mol

    since

    ΔG° reaction = ν * ΔGf ° products - v * ΔGf ° reactives

    for reaction 1

    ΔG° = ∑ν * ΔGf ° products - ∑v * ΔGf ° reactives

    ΔG° = 1 * ΔGf ° P4 O10 - (5 * ΔGf ° O2 + 1 * ΔGf ° P4)

    ΔG° = ΔGf ° P4 O10 - (5*0 + 1*0)

    ΔGf ° P4 O10 = ΔG° = - 2697.0 kJ/mol

    for reaction 2

    ΔG° = ∑ν * ΔGf ° products - ∑v * ΔGf ° reactives

    ΔG° = 2 * ΔGf ° H20 - (1*ΔGf ° O2 + 2 * ΔGf ° H2)

    ΔG° = 2 * ΔGf ° H20 - (1*0 + 2*0)

    ΔGf ° H20 = ΔG° / 2 = - 457.18 kJ/mol/2 = - 228.59 kJ/mol

    for reaction 3

    ΔG° = ∑ν * ΔGf ° products - ∑v * ΔGf ° reactives

    ΔG° = 4 * ΔGf ° H3 PO4 - (6*ΔGf ° H2O + 1*ΔGf ° P4O10)

    -428.66 kJ/mol = 4 * ΔGf ° H3 PO4 - [ 6 * (-228.59 kJ/mol) + 1 * (-2697.0 kJ/mol) ]

    -428.66 kJ/mol = 4 * ΔGf ° H3 PO4 + 3611.36 kJ/mol

    ΔGf ° H3 PO4 = (-428.66 kJ/mol - 3611.36 kJ/mol) / 4 = - 1010 kJ/mol
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