Use the following information to answer questions 5-10. A student determined the heat of neutralization of copper (II) sulfate (CuSO4) mixed with potassium hydroxide (KOH) using the procedure described part 2 of this experiment. A 150.0 mL sample of a 1.50 M solution of CuSO4 was mixed with a 150.0 mL sample of 3.00 M KOH in a coffee cup calorimeter. The temperature of both solutions and the calorimeter was 25.2 °C before mixing and 31.3 °C after mixing. The heat capacity of the calorimeter is 24.2 J/K. Calculate the heat of reaction (ΔHrxn) for this reaction in units of J/mole of copper (II) hydroxide (Cu(OH)2). Assume the solution is dilute enough that the specific heat capacity and density of the solution is the same as that of water, 4.18 J/g∙K and 1.00 g/mL respectively. Is the overall reaction endothermic or exothermic? Cannot be determined Endothermic Exothermic Next

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.

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-11-17T07:38:31+01:00

The heat of reaction is -34.65 kJ/mol.

The equation of the reaction is;

CuSO4(aq) + 2KOH(aq) ------> Cu(OH)2(s) + K2SO4(aq)

Number of moles of CuSO4 = 150/1000 × 1.50 M = 0.225 moles

Number of moles of KOH = 150/1000 × 3.00 M = 0.45 moles

Since 2 moles of KOH reacts with 1 mole of CuSO4

0.45 moles of KOH reacts with 0.45 moles × 1 mole/ 2 moles = 0.225 moles

We can clearly see that CuSO4 is the limiting reactant.

Total volume of solution = 150 mL + 150mL = 300mL

Mass of solution = 300 g

Temperature change = 31.3 °C - 25.2 °C = 6.1°C

Heat gained by calorimeter and solution= cpθ + mcθ

cp = Heat capacity of the calorimeter

θ = temperature rise

c = Heat capacity of solution

m = mass of solution

H = cpθ + mcθ

H = (24.2 J/K × 6.1°C) + (300 g × 4.18 J/g∙K × 6.1°C)

H = 147.62 + 7649.4 = 7.797 kJ

The heat of reaction is obtained from;

ΔH = -7.797 kJ/ 0.225

ΔH = -34.65 kJ/mol

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