Answer:
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
