Question

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 17.8 oC before mixing and 32.4 oC after mixing. The heat capacity of the calorimeter is 12.1 J/K. From these data, calculate ΔH for the process: CuSO4(1M)+ 2KOH(2M) →Cu(OH)2(s) + K2SO4(0.5M)

Answer 1

Answer:

see explanation

Explanation:

Step 1: Data given

Volume of 1M CusO4 = 50.0 mL = 0.05 L

Volume of 2M KOH = 50.0 mL = 0.05 L

Temperature before mixing= 17.8 °C

Temperature after mixing = 32.4 °C

The heat capacity of the calorimeter is 12.1 J/K

Step 2: The balanced equation

CuSO4(aq)+ 2KOH(aq) →Cu(OH)2(s) + K2SO4(aq)

Step 3: Calculate mass of the solution

Suppose the density of the solution is 1 g/mL

Total volume = 100 mL

Mass of the solution = density * volume

Mass of the solution = 1g/mL * 100 mL = 100 grams

Step 4:

Q = m*c*ΔT

with m = the mass of the solution = 100 grams

with c= the heat capacity of the solution = 4.184 J/g°C

with ΔT = 32.4 - 17. 8 = 14.6 °C

Q = 6108.64 J

Step 5: Calculate the energy of the calorimeter

Q = c*ΔT

Q = 12.1 J/K * 14.6

Q = 176.66 J

Step 6: Calculate total heat

Qtotal = 6108.64 + 176.66 = 6285.3 J = 6.29 kJ (negative because it's exothermic)

Step 7: Calculate moles

Moles CuSO4 = 0.05 L * 1M = 0.05 moles

Moles KOH = 0.05 L * 2M = 0.10 moles

ΔH = -6.29 kJ / 0.05 moles = -125.8 kJ/mol