Question

A scientist measures the standard enthalpy change for the following reaction to be -613.2 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 H₃PO₄(aq) is kJ/mol.

Answer 1

_fH⁰ P₄O_10(s) = -3012.5 KJ/mol

_fH⁰ H₂O_(l) = -285.8 KJ/mol

_fH⁰ H₃PO_4(aq) = -1277 KJ/mol

H⁰ = _fH⁰ (Product) - _fH⁰ (Reactant)

H⁰ = [(4_fH⁰ H₃PO_4(aq)) ] - [(_fH⁰ P₄O_10(s) ) + (6_fH⁰ H₂O_(l) )]

= [(4 (-1277))] - [(-3012.5) + (6 (-285.8)]

= [-5108] - [-4727.3]

= [-5108] + [4727.3]

H⁰ = -380.7 KJ/mol

H⁰ for reaction = -380.7 KJ/mol