Question

A 50.0-mL sample of a 1.00 M solution of is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both solutions was 20.3 ∘C before mixing and 26.2 ∘C after mixing. The heat capacity of the calorimeter is 12.1 J/K. From these data calculate ΔH (in kJ/mol) for the process:

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

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

Answer 1

Total volume = 50 + 50 = 100 ml

Mass = volume * density

= 100 * 1 = 100 grams

Heat gained by solution

Q = mass * specific heat * ∆T

= 100 * 4.184 * ( 26.2 -20.3) = 2468.56 J

Heat gained by calorimeter

Q2 = C(cal) * ∆T

= 12.1 * ( 26.2 - 20.3 )

= 71.39 J

Q (total) = 71.39 + 2468.56 = 2539.95 J

Moles of CuSO4 = molarity * volume ( in L )

= 1.00 * (50*10^-3)

= 0.05

0.05 moles of CuSO4 release = 2539.95 J

1 mole will release = 2539.95 / 0.05

= 50799 J or 50.799 KJ or 51 KJ / kol of CuSO4

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