Indirect determination of ∆H via Hess's law
Indirect determination of ∆H via Hess's law
Conclusion & Evaluation
Research question:
What is the ∆H/mol of hydration of CuSO4 (anh)?
The ∆H/mol of hydration of CuSO4 (anh) is -70.2, according to the experiment done in class. It was determined by applying the Hess's law, using two reactions
1) CuSO4-5H2O ----> CuSO4 (aq) + 5H2O
2) CuSO4 (anh) -----> CuSO4 (aq) According to the Hess's Law, which states that the total enthalpy change for a chemical reaction is the same disregarding the route taken for the reaction; the reactions could be combined together to make the initial reaction shown below:
3) CuSO4 (anh)+5H2O ---> CuSO4-5H2O (aq)
As the equation of the first reaction should be flipped, the ∆H /mol of the reaction would change the sign; and then both ∆H/mol of reactions should be added together in order to calculate the ∆H/mol of initial reaction. Therefore, I carried out two reactions. The firsts reaction involved adding hydrated copper (II) sulfate in water. The initial and final temperatures were recorded, and the ∆T of the trials were calculated. As the reaction was endothermic, which means it absorbs the heat, the temperature goes down and therefore the ∆T of the reaction is positive. The table below represents the information about the experiment:
Data collection and processing of the dissolution of hydrated copper (II) sulfate
ΔH/mol of the solution was calculated by the formula ΔH = (cgΔT/1000)/mol, where c - specific heat (4.18 J for water), g - grams of water and ΔT - change of the temperature; division by 1000 shows that the data is calculated in kJ. Also the number was divided by the amount of moles used in each trial (which is shown in the table above). The final amount of ∆H sol/mol of CuSO4 was calculated by finding the average of all 5 trials, which is 11.4 kJ/mol (±7,91%). Experiment was done according to the equation below:
1) CuSO4-5H2O ----> CuSO4 (aq) + 5H2O
The equation represents the
Conclusion & Evaluation
Research question:
What is the ∆H/mol of hydration of CuSO4 (anh)?
The ∆H/mol of hydration of CuSO4 (anh) is -70.2, according to the experiment done in class. It was determined by applying the Hess's law, using two reactions
1) CuSO4-5H2O ----> CuSO4 (aq) + 5H2O
2) CuSO4 (anh) -----> CuSO4 (aq) According to the Hess's Law, which states that the total enthalpy change for a chemical reaction is the same disregarding the route taken for the reaction; the reactions could be combined together to make the initial reaction shown below:
3) CuSO4 (anh)+5H2O ---> CuSO4-5H2O (aq)
As the equation of the first reaction should be flipped, the ∆H /mol of the reaction would change the sign; and then both ∆H/mol of reactions should be added together in order to calculate the ∆H/mol of initial reaction. Therefore, I carried out two reactions. The firsts reaction involved adding hydrated copper (II) sulfate in water. The initial and final temperatures were recorded, and the ∆T of the trials were calculated. As the reaction was endothermic, which means it absorbs the heat, the temperature goes down and therefore the ∆T of the reaction is positive. The table below represents the information about the experiment:
Data collection and processing of the dissolution of hydrated copper (II) sulfate
ΔH/mol of the solution was calculated by the formula ΔH = (cgΔT/1000)/mol, where c - specific heat (4.18 J for water), g - grams of water and ΔT - change of the temperature; division by 1000 shows that the data is calculated in kJ. Also the number was divided by the amount of moles used in each trial (which is shown in the table above). The final amount of ∆H sol/mol of CuSO4 was calculated by finding the average of all 5 trials, which is 11.4 kJ/mol (±7,91%). Experiment was done according to the equation below:
1) CuSO4-5H2O ----> CuSO4 (aq) + 5H2O
The equation represents the