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?!?!?
I don't see anything wrong with what you have.
its supposed to be C+D-B-C
309
To calculate the standard enthalpy change for the reaction, you need to use the standard enthalpy of formation values for each substance involved. The standard enthalpy of formation (ΔH∘f) is the enthalpy change that occurs when one mole of a compound is formed from its elements in their standard states.
In this case, you are given the ΔH∘f values for substances A, B, C, and D.
The balanced equation for the reaction is:
2A + B ⇌ 2C + 2D
To calculate the standard enthalpy change, you need to consider the reactants and products separately.
Reactants:
ΔH∘r = [(2)(ΔH∘f A) + (ΔH∘f B)]
Products:
ΔH∘p = [(2)(ΔH∘f C) + (2)(ΔH∘f D)]
Standard enthalpy change (ΔH∘) = ΔH∘p - ΔH∘r
Substituting the given values:
ΔH∘r = [(2)(-251) + (-403)]
ΔH∘p = [(2)(213) + (2)(-511)]
ΔH∘ = ΔH∘p - ΔH∘r
= [(2)(213) + (2)(-511)] - [(2)(-251) + (-403)]
= [426 - 1022] - [-502 - 403]
= [426 - 1022] + [502 + 403]
= -596 + 905
= 309 kJ
Your calculation of 309 kJ is correct. The standard enthalpy change for the reaction is indeed 309 kJ. It seems that there might be a mistake in the provided correct answer.