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

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

it means there are 3 atoms of oxygen