Question

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)

Answer 1

-290KJ/mol

Step-by-step explanation:

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn= 4ΔHH3PO4 - {6ΔHH2O + ΔHP4O10}

ΔHrxn = 4(-1279) - [6(-286) - 3110]

= -5116 -(-1716-3110)

= -5116-(-4826)

= -5116 + 4826 = -290KJ/mol