At 298 K what will be the change in standard internal energy for the given reaction: OF2(g)+ H2O(g) → O2(g) + 2HF(g) ΔH = -310 kJ/mol
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Answer:
ΔH
rxn
o
=ΔH
O
2
o
+2ΔH
HF
o
−[ΔH
OF
2
o
+ΔH
H
2
O
o
]
ΔH
rxn
o
=0+2(−270)−[+20−250]=−310kJ=−310000J
ΔE
rxn
o
=ΔH
rxn
o
−ΔnRT
ΔE
rxn
o
=−310000−(1×8.314×298)=−312477J=−312.47kJ
−100x=31247
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