4.A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M

Question

4.A 50.0 mL sample of a 1.00  M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both solutions was 20.2 degree celsious before mixing and 26.3 degree celsious after mixing . The heat capacity of the calorimeter is 12.1 J/K. From the data, calculate delta Hfor the process

CuSO4(1 M)+2KOH(2 M)----------------------Cu(OH)2(s)+K2SO4(0.5 M)

Assume that the specific heat and density of the solution after mixing are the same as those of pure water and that the volumes are additive.

Explanation / Answer

For a process carried out at constant pressure, the change in energy is equal to the change in enthalpy, so delta-E = delta-H

The specific heat of water is 4.1855 J/(gm * K) (see source), and its density is ~1 gm/cm^3. The combined volume of solution in the calorimeter is 100 mL, and we are told to assume that the density and specific heat of this solution is the same as pure water. The heat capacity of the solution is therefore:

100 cm^3 * 1 gm/cm^3 * 4.1855 J/(gm* K)= 418.55 J/K

The total heat capacity of the system is:

(418.55 + 12.1) J/K = 430.65 J/K

The temperature change was 6.1 degrees C = 6.1 K, so

delta-E = delta-H = 430.65 J/K * 6.1 K = 2627 J

Now, this enthalpy change was for the reaction of 50 mL of a 1MCUS04 solution with 50 mL of 2M KOH solution. 50 mL of 1M CuSO4 solution only contains 0.05 liters * 1 mole/liter = 0.05 moles of CuSO4. If we want the enthalpy change PER MOLE of CuSO4 reacted, we have to divide the enthalpy we just calculated by 0.05:

delta-H (J/mole CuSO4) = (2627 J)/(0.05 mol CuSO4) = 5.254*10^4 J/(mole CuSO4)

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