Question

A scientist measures the standard enthalpy change for the following reaction to be -327.2 kJ : P4O10(s) 6 H2O(l)4H3PO4(aq) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of P4O10(s) is kJ/mol.

Answer 1

Answer:

ΔH°f P4O10(s) = - 3115.795 KJ/mol

Explanation:

• P4O10(s) + 6H2O(l) ↔ 4H3PO4(aq)
• ΔH°rxn = ∑νiΔH°fi

∴ ΔH°rxn = - 327.2 KJ

∴ ΔH°f H2O(l) = - 285.84 KJ/mol

∴ ΔH°F H3PO4(aq) = - 1289.5088 KJ/mol

⇒ ΔH°rxn = (4)(- 1289.5088) - (6)(- 285.84) - ΔH°f P4O10(s) = - 327.2 KJ

⇒ ΔH°f P4O10(s) = - 5158.035 + 1715.04 + 327.2

⇒ ΔH°f P4O10(s) = - 3115.795 KJ/mol

Answer 2

The standard enthalpy of formation of P₄O₁₀(s) = - 3115.795 KJ/mol

Given data :

Standard enthalpy change for the reaction ( ΔH°rxn ) = -327.2 kJ

The reaction equation is :

P₄O₁₀ (s) + 6H₂O (l)  ⇄  4H₃PO₄ (aq)

also : ΔH°rxn = ∑νiΔH°fi  ----- ( 1 )

Standard enthalpy change of H₂O(l) ( ΔH°f  ) = - 285.84 KJ/mol

ΔH°F of  H₃PO₄(aq) = - 1289.5088 KJ/mol

Back to equation ( 1 )

ΔH°rxn = (4) * (- 1289.5088) - (6) * (- 285.84) - ΔH°f P₄O₁₀(s)

- 285.84 KJ/mol =  (4) * (- 1289.5088) - (6) * (- 285.84) - ΔH°f P₄O₁₀(s)

∴ Standard enthalpy of formation ( ΔH°f ) of  P₄O₁₀(s) = - 5158.035 + 1715.04 + 327.2 = - 3115.795 KJ/mol

Hence we can conclude that The standard enthalpy of formation of P₄O₁₀(s) = - 3115.795 KJ/mol.

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