Question

Given the following information, answer these questions.

P₄(g) + 5 O₂(g) → P₄O₁₀(s)

ΔG⁰298  = −2748 kJ/mol2 H₂(g) + O₂(g) → 2 H₂O(l)

ΔG⁰298  = −474.2 kJ/molP₄O₁₀(s) + 6 H₂O(l) → 4 H₃PO₄(l)

ΔG⁰298  = −349 kJ/mol

(a)

Determine the standard free energy of formation, ΔG⁰f (in kJ/mol), for phosphoric acid from this information.(Assume that ΔG⁰298 for all elemental substances is negligible.)

_______ kJ/mol

(b)

How does your calculated result compare to the value in the Standard State Thermodynamic Data table? Explain.

The listed value of ΔG⁰f (in kJ/mol) = _______ kJ/mol.

Answer 1

a) Given reactions are

P4(g) + 5 O2(g) → P4O10(s) ΔG⁰298  = −2748 kJ/mol2 --------------> [1]

2 H2(g) + O2(g) → 2 H2O(l) ΔG⁰298  = −474.2 kJ/mol --------------> [2]

P4O10(s) + 6 H2O(l) → 4 H3PO4(l) ΔG⁰298  = −349 kJ/mol --------------> [3]

the heat of formation means gibbs free energy chagnge for reactions where compound is made from its constituent elements for one mole of a compound

the required equation is

1/4 P4(g) + 2 O2(g) + 3/2 H2(g) → H3PO4(l)   ΔG⁰298 = ?

we can get the above reaction from 1,2 and 3 reactions by

[1] x1/4 +  [2] x 3/4 +  [3] x1/4

1/4 P4(g) + 5/4 O2(g) → 1/4 P4O10(s)

3/2 H2(g) + 3/4 O2(g) → 3/2 H2O(l)

1/4 P4O10(s) + 3/2 H2O(l) → H3PO4(l)

------------------------------------------------------------- by adding above reactions

1/4 P4(g) + 2 O2(g) + 3/2 H2(g) → H3PO4(l)

ΔG_rxn⁰298 = [−2748] x1/4 +  [ −474.2] x 3/4 +  [ −349 ] x1/4 = -687-355.65-87.25 = -1129.9 kJ/mol

b) the value in the Standard State Thermodynamic Data table for the formation for H3PO4 is -1123.6  kJ/mol and calculated value is -1129.9 kJ/mol which is almost equal. this will indicate the calculaion of heat of formation from defferent methods will give the same value irrespective of the reaction steps. ths is the Hess law of heat of summation.