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Dec2-04, 12:53 PM
I have to measure the enthalpy change for the reaction:

CuSO4(s) + 5H2O --> CuSO4 · 5H2O

I have to use the hess law since the above reaction is slow.

CuSO4 · 5H2O + H2O --> CuSO4 + H2O (1)
solid CuSO4 + H2O --> CuSO4 (2)

The techer said the entalphy change for this reaction should be -79kJ/mol and the heat capacity of the solution should be taken as 4.2 kJ/mol.

However when I react 50 cm3 water with 0.025 moles CuSO4 · 5H2O in equation 1, I get a temperature change of -0.8 C.
In the second reaction, 50cm3 of water reacted with 0.025mol of CuSO4. Here the temperature change was +6.5 C.

But when I plug it into the equation E = m c ∆t; I get an endothermic value. What is wrong? Should the reaction 2 be exothermic?

so-crates
Dec4-04, 03:21 AM
I have to measure the enthalpy change for the reaction:

CuSO4(s) + 5H2O --> CuSO4 · 5H2O

I have to use the hess law since the above reaction is slow.


Hess' Law has nothing to do with rates of reactions (kinetmatics)


CuSO4 · 5H2O + H2O --> CuSO4 + H2O (1)
solid CuSO4 + H2O --> CuSO4 (2)


Those reactions don't make any sense, you never have the same compound on both sides of the reaction for (1) and for (2) you don't have H20 anywhere on the right hand side. Are you sure you got these reactions right?