Calculate the standard enthalpy change for the reaction
2A+B⇌2C+2D
Use the following data:
Substance ΔH∘f (kJ/mol)
A -251
B -403
C 213
D -511
My thoughts:
[(2)(-511)+(213)(2)]-[(2)(-251)+(-403)]=309kJ
(but apparently that is not the correct answer)
What did I do wrong?!?!?
3 answers
I don't see anything wrong with what you have.
its supposed to be C+D-B-C
309