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

the role of the microscope world in creating new materials .