[pharm1986]

Hi,

Was just wondering if anyone could offer help on a practice exam question i'm stuck on?

"Anhydrous copper(II) sulfate is a white powder that reacts with water to give the familiar light blue crystals of copper(II) sulfate-5-water.
CuSO4(s) + 5H2O(l) --> CuSO4.5H2O

Calculate the standard enthalpy change for this reaction from the heats of solution.

Compound: CuSO4(s)
dH0solution/kj/mol: -66.5

Compound: CuSO4.5H2O(s)
dH0solution/kj/mol: +11.7"

So far, I've managed:
1) CuSO4(s) -->CuSO4 (aq) dH0=-66.5
2) CuSO4.5H2O(s) --> CuSO4(l) + 5H2O(l) dH0=+11.7

However, the next part of the solutions has 2 subtracted from 1 to give:
3) CuSO4.5H2O(s) --> CuSO4.5H2O(s)
which then allows:
(-66.5) - (+11.7) to give a total enthalpy of -78.2kJ/mol

I'm not sure why subtracting 2 from 1 gives 3 and there's no explanation in the solutions. i'm just looking at it as a mathematical subtraction and getting a weird answer, but there must be something more complex that i seem to have missed in class?

Also, does anyone know of a simpler/briefer approach to this kind of question?

Thanks for your time.

[Yggdrasil]

Instead of thinking about subtraction, think about switching reaction 2 around so that it reads:

4)  CuSO4(aq) + 5H2O(l) --> CuSO4.5H2O(s)   dH0=-11.7

Here, the change in enthalpy switches signs because the reaction is running backwards.  Now add equation 1 to equation 4 and cancel out molecules that appear on both the reactant and product side.

[pharm1986]

oh, makes sense now - much easier that way!
thanks very much