A 150.0 mL sample of a 1.50 M solution of CuSO4

is 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 ΔHrxn for this reaction in units of kJ / mol of copper (II) hydroxide.Assume
the solutions is dilute enough that the specific heat and density of the solution is the same as that
of water

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To calculate ΔHrxn for this reaction, we need to apply the principle of constant heat capacity. The heat released by the reaction will be absorbed by the solution and the calorimeter, and the temperature change can be used to calculate the enthalpy change of the reaction.

First, we need to calculate the heat absorbed by the calorimeter and the solution. The formula for heat absorbed or released is given by:

q = mcΔT

Where:
q = heat absorbed or released
m = mass of the solution
c = specific heat capacity of water (since the solution is assumed to have the same properties as water)
ΔT = change in temperature

Since we have equal volumes of CuSO4 and KOH solutions, we can assume the masses of the solutions are the same. Given that the density of water is roughly 1 g/mL, the mass of the solutions will be approximately 150.0 g.

Now, let's calculate the heat absorbed by each solution and the calorimeter.

q(CuSO4) = mcΔT = (150.0 g)(4.18 J/g°C)(31.3°C - 25.2°C)
q(CuSO4) = 150 J

q(KOH) = mcΔT = (150.0 g)(4.18 J/g°C)(31.3°C - 25.2°C)
q(KOH) = 150 J

q(calorimeter) = mcΔT = (24.2 J/K)(31.3°C - 25.2°C)
q(calorimeter) = 147 J

The total heat absorbed or released by the system is the sum of the heat absorbed by the solutions and the calorimeter:

ΔHtotal = q(CuSO4) + q(KOH) + q(calorimeter)
ΔHtotal = 150 J + 150 J + 147 J
ΔHtotal = 447 J

Now, we can calculate ΔHrxn using the following equation:

ΔHrxn = -ΔHtotal

Since we want the answer in units of kJ/mol, we need to convert the heat change from J to kJ and divide it by the number of moles of copper (II) hydroxide produced in the reaction.

To find the moles of CuSO4 reacting, we need to use the molarity and volume of CuSO4:

moles of CuSO4 = molarity x volume
moles of CuSO4 = 1.50 M x 0.150 L
moles of CuSO4 = 0.225 mol

Since the reaction is a 1:1 ratio between CuSO4 and copper (II) hydroxide, the moles of CuSO4 will be the same as moles of copper (II) hydroxide.

ΔHrxn = -ΔHtotal / moles of Cu(OH)2
ΔHrxn = -447 J / 0.225 mol
ΔHrxn = -1987 J/mol

We can convert this answer to kJ/mol by dividing by 1000:

ΔHrxn = -1987 J/mol / 1000
ΔHrxn = -1.987 kJ/mol

Therefore, the ΔHrxn for this reaction is -1.987 kJ/mol of copper (II) hydroxide.