Asked by Ellie
what is the standard enthalpy formation of liquid diethylamine (CH3CH2)2 NH?
N2O5(g) + 8CH4(g) ==> 2(CH3CH2)2NH(l) +5H2O(l), delta H= -1103kJ
N2O5(g) + 8CH4(g) ==> 2(CH3CH2)2NH(l) +5H2O(l), delta H= -1103kJ
Answers
Answered by
DrBob222
Standard delta Hs are listed as kJ/mol so divide -1103 by 2.
Answered by
Anonymous
that gives me -551.5. that is not 1 of the choices listed.
Answered by
DrBob222
You might have done well to list the choices. It could be units or something else.
Answered by
Anonymous
-131kj/mol
-1452kl/mol
-421 kj/mol
+131kj/mol
+421kj/mol
-1452kl/mol
-421 kj/mol
+131kj/mol
+421kj/mol
Answered by
DrBob222
OK. I didn't read the problem correctly. You look up in tables for delta H formation and
delta Hrxn = (delta H products)-(delta H reactants). Solve for delta H of the amine.
delta Hrxn = (delta H products)-(delta H reactants). Solve for delta H of the amine.
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