P4O10 (s) = - 3110 kJ/mol H2O (l) = - 286 kJ/mol H3PO4 (s) = - 1279 kJ/mol Calculate the change in enthalpy for the following process: P4O10 (s) + 6H2O (l) → 4H3PO4 (s)

Question

Chemistry

Yasmine

13 October, 17:51

P4O10 (s) = - 3110 kJ/mol H2O (l) = - 286 kJ/mol H3PO4 (s) = - 1279 kJ/mol Calculate the change in enthalpy for the following process: P4O10 (s) + 6H2O (l) → 4H3PO4 (s)

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Answers (1)

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Joaquin Heath · Answer 1

-290KJ/mol

Explanation:

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn = 4ΔHH3PO4 - {6ΔHH2O + ΔHP4O10}

ΔHrxn = 4 (-1279) - [6 (-286) - 3110]

= - 5116 - (-1716-3110)

= - 5116 - (-4826)

= - 5116 + 4826 = - 290KJ/mol