[kay2]

I did an experiment of application of Hess' law. Using 5g of CuSO4 and 10g of CUSO4.5H2O, here are the results obtained.

Temperature of water : 21°C
Temperature of anhydrous CuSO4 : 23°C
Temperature of hydrated CuSO4.5H20 : 19°C

Specific heat capacity of water : 4.184J/g*K
1. Calculate the ∆H solution for one mole anhydrous CUSO4 (159.62g/mol)

No. of mol of CuSO4 (159.62 g/mol):

(5 g)/(159.62 g/mol) = 0.031 mol

∆H solution for one mole of anhydrous CuSO4:
50/1000  ×4.184 ×(23°C-21°C)= 0.418 kJ
0.418 ×  1/0.031  = 13.484 kJ mol^-1

(Is it right?)

2. Calculate the ∆H solution for one mole hydrated CUSO4.5H20 (249.68g/mol)

No. of mol of CuSO4.5H2O (249.68 g/mol):

(10 g)/(249.68 g/mol)  =0.04 mol

∆H solution for one mole of hydrated CUSO4.5H2O
50/1000  ×4.184 ×(19℃-21℃)= -0.418 kJ
-0.418 ×  1/0.04  =-10.45 kJ mol^-1

3. Combine the two thermochemical equations above to find the enthalpy change for the hydration (∆H hydration) of anhydrous CuSO4.
4. Explain the differences between ∆H solution of anhydrous CuSO4 and hydrated CuSO4.5H2O.
5. Is it not possible to calculate the enthalpy of formation(∆H formation) of CO directly. Suggest a reason for this. In an experiment to determine ∆ formation of CO from the information below:

CO (g) + 1/2 O2 (g) -> CO2               ∆H = -286 kJ mol^-1
C (g) + o2 (g) -> CO2                      ∆H = -395 kJ mol^-1

Calculate the ∆H formation CO (g):

C (g) + 1/2 O2 (g) -> CO (g)             ∆H = ??

Need help for these questions. Much appreciated! Thank you!