Answer:
ΔHrxn = -34,653.33 J /mol
Explanation:
In this question we have to calculate the heat of reaction for
CuSO₄ + 2 KOH ⇒ Cu(OH)₂ + K₂SO₄
in units of J/mole Cu(OH)₂
Now this heat by the law of conservation of energy is equal to the heat absorbed by the water plus the heat absorbed by the calorimeter. Thus what we need to do first is to calculate those numbers.
Heat absorbed by water
q H₂O = m x c x ΔT
where m is the mass of water mixed, c is the specific heat of water and ΔT is the change in temperature.
Total water volume = 150 mL + 150 mL
m water = 300 mL x 1 g/mL = 300 g
ΔT = T₂ - T₁ = 31.3 ºC - 25.2 º C = 6.1 ºC = 6.1 K
( the change is temperature in K and ºC is the same )
Note that this reaction is exothermic since heat is released which is absorbed by the water and calorimeter because ΔT is positive.
Now
q H₂O = 300 g x 4.18 J/g·K x 6.1 K = 7649 J
Heat absorbed by the calorimeter
q cal = Ccal x ΔT = 24.2 J/K x 6.1 K = 1476 J
Total Heat = q total = 7649 J + 1476 J = 7797 J
We need to calculate the # moles of Cu(OH)₂ to determine ΔHrxn
M = mol/V ⇒ mol = M x V
mol Cu(OH)₂ = 0.150 L x 1.5 mol/L = 0.225 mol
ΔHrxn = -q total / mol Cu(OH)₂ = 7797 J / 0.225 mol = -34,653.33 J /mol
Notice ΔHrxn is negative since it is an exothermic reaction.
