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
The answer is in the bladder.
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