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- May 24, 2017
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On GC test 2 of a BC full length I am confused with this answer explanation provided below:
The ∆Hf for Br(g) is: (1/2) Br2(g) -> Br(g) ; ∆Hf = +193 kJ / mol
If we simply double this reaction, we find: Br2(g) → 2Br(g) ; ∆Hf = (2)(+193) = +386 kJ/mol
The part in bold is the issue. Since this is a delta H of formation reaction, shouldn't Br2 be a liquid in its standard state? So technically this would not qualify as a formation reaction? Any clarification would be appreciated!
The ∆Hf for Br(g) is: (1/2) Br2(g) -> Br(g) ; ∆Hf = +193 kJ / mol
If we simply double this reaction, we find: Br2(g) → 2Br(g) ; ∆Hf = (2)(+193) = +386 kJ/mol
The part in bold is the issue. Since this is a delta H of formation reaction, shouldn't Br2 be a liquid in its standard state? So technically this would not qualify as a formation reaction? Any clarification would be appreciated!
