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.

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