Question

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.

Answer 1

The standard free energy of formation (ΔGf°) for H3PO4) is calculated to be -1293.66 kJ/mol by summing the given reactions and dividing by 4. This result should be compared with Appendix G's values to assess its accuracy.

Step-by-step explanation:

To determine the standard free energy of formation (ΔGf°) for phosphoric acid (H3PO4), we will use the given standard free energy changes for a series of reactions and manipulate them to represent the formation of one mole of phosphoric acid from its elements.

The first step is to use the values given for the overall reactions:

• P4(s) + 5O2(g) → P4O10(s) ΔG° = -2697.0 kJ/mol
• 2H2(g) + O2(g) → 2H2O(g) ΔG° = -457.18 kJ/mol
• 6H2O(g) + P4O10(s) → 4H3PO4(l) ΔG° = -428.66 kJ/mol

Using Hess's Law, we can sum these equations to cancel out intermediates (H2O and P4O10), so we are left with the formation of phosphoric acid from its elements:

P4(s) + 6H2(g) + 5O2(g) → 4H3PO4(l)

The sum of the ΔG° values for these reactions will give the ΔG for the formation of 4 moles of H3PO4. To find the ΔGf° per mole of H3PO4, we divide by 4:

ΔG° = -2697.0 kJ/mol (for P4O10 formation) + (2 x -457.18 kJ/mol for H2O formation) + (-428.66 kJ/mol for H3PO4 formation)

ΔGf°(H3PO4) = (-2697.0 + (2 x -457.18) - 428.66) / 4 = -1293.66 kJ/mol

To compare this with the value in Appendix G, one would need to look up the standard free energy of formation for H3PO4 in Appendix G and see if it agrees with the calculated value. If there is a discrepancy, it may be due to experimental error, approximations, or differences in the data sources used for Appendix G.

Answer 2

the standard free energy of formation of phosphoric acid H3 PO4 is -1010 kJ/mol

Step-by-step 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