(1) P4(s) + 3 O2(g) → P4O6(s) ΔH = –1640.1 kJ
(2) P4(s) + 5O2(g) → P4O10(s) ΔH = –2940.1 kJ
We shall modify the given reactions to arrive at the overall reaction. Any modification to the reaction is reflected in the value of ΔH for the reaction.
Calculate the enthalpy change for the reaction: P4O6(s) + 2 O2(g) → P4O10(s), given the following enthalpies of reaction:
P4(s) + 3 O2(g) → P4O6(s), ΔH = –1640.1 kJ
P4(s) + 5O2(g) → P4O10(s), ΔH = –2940.1 kJ
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