Hess’s law can be used to determine enthalpy changes for reactions which cannot be obtained directly.
An example is the reaction of anhydrous copper(II) sulfate with water to form hydrated copper(II) sulfate, CuSO4.5H2O.
The following outline procedure was carried out.
Step 1 |
42.75 g of deionised water was weighed out in a polystyrene cup and the temperature measured. |
Step 2 | 0.0250 mol of hydrated copper(II) sulfate was added to the water in the polystyrene cup with stirring, making a total of 45.00 g of water. |
Step 3 | The temperature change was recorded. |
Step 4 |
Steps 1 to 3 were repeated using 45.00 g of deionised water and 0.0250 mol of anhydrous copper(II) sulfate. |
Calculate the mass of 0.0250 mol of hydrated copper(II) sulfate, CuSO4.5H2O.
The reaction of hydrated copper(II) sulfate with water is shown.
CuSO4.5H2O (s) + aq → CuSO4 (aq) ∆H1 = +18.2 kJ mol−1
Calculate the temperature change that would have given this enthalpy change for the stated experimental procedure.
Give your answer to a measurable number of significant figures and state whether the temperature increases or decreases.
[Specific heat capacity of the solution = 4.18 J g−1 oC−1]
The reaction of anhydrous copper(II) sulfate with water is shown.
CuSO4 (s) + aq → CuSO4 (aq) ∆H2 = −84.5 kJ mol−1
CuSO4 (s) + 5H2O (l) → CuSO4.5H2O (s)
(1)
State why the enthalpy change for the reaction of one mole of anhydrous copper(II) sulfate with five moles of water to form hydrated copper(II) sulfate, CuSO4.5H2O, cannot be measured directly.
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