Question

In a constant-pressure calorimeter of negligible heat capacity, 25 ml of 1. 00 m cacl2 is mixed with 25 ml of 2. 00 m kf, resulting in solid caf2 precipitating out of the solution. During this process, the temperature of the water rises from 25. 0°c to 26. 7°c. Assume the specific heat capacity of the solution is 4. 184 j/°c•g and the density of the solution is 1. 00 g/ml. Calculate the enthalpy of precipitation in kj per mole of caf2 precipitated.

Answer 1

The enthalpy of precipitation of calcium fluoride is -13.974 kJ/mol.

In order to solve this, first we need to write the balanced chemical reaction equation of the reaction in question:

CaCl₂(aq) + 2KF(aq) → CaF₂(s) + 2KCl(aq)

In order to calculate the enthalpy of precipitation (ΔH), we need the amount  of heat released (Q) and the number of moles of CaF₂ (n):

ΔH = Q/n

To calculate the number of moles, we can use the molarity (c = 1.00 M) of the calcium chloride solution and its volume (V = 25 mL = 0.025 L):

c = n/V ⇒ n = c*V

n = 1.00 M * 0.025 L = 0.025 mol

We calculate the amount of heat released (Q) using the following equation:

Q = (t₁ - t₂) * C * m

t₁ - initial temperature (25.0 ⁰C)

t₂ - final temperature (26.7 ⁰C)

C - specific heat capacity (4.184 J/⁰Cg)

m - the mass of the solution

We need the mass of the solution, which we can calculate using the density (d = 1.00 g/mL) and the volume (V = 25 mL + 25 mL = 50 mL) of the solution:

d = m/V ⇒ m = d*V

m = 1.00 g/mL * 50 mL = 50 g

Q = (25.0 ⁰C - 26.7 ⁰C) * 4.184 J/°Cg * 50 g

Q = -349.4 J

ΔH = -349.4 J / 0.025 mol

ΔH = -13974 J/mol = -13.974 kJ/mol

You can learn more about enthalpy here:
brainly.com/question/7827769

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