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Heat of Solution

>> Parts of this equation/concept include:

A -	Finding the Heat of Solution from Calorimetry
B -	Finding the Heat of Solution from Lattice Energy and Heat of Hydration, Equation 13.2

Review calorimetry from Chapter 11. This chapter tends to use specific heat instead of heat capacity. Using specific heat, Equation 11.8 becomes

heat energy (q) = (mass)(specific heat)(change of temperature)

Because there is normally much more solvent than solute, it is usually assumed that the specific heat is that of the solvent. For water (the most common solvent),the specific heat is 4.184 J/g•°C. Remember that the mass will refer to the mass of the entire solution, since it is the entire solution undergoing the temperature change.

The heat of solution refers to the dissolution of a solute. For an ionic compound, he reaction is the salt forming its composite ions. Although water is required for the reaction, it is not written as part of the reaction. For example,

Na₂SO₄ 2 Na⁺ + SO₄^2–

So heat of solution is calculated from

Hsoln

=

q mol solute

Since the heat (q) comes from the chemical reaction of the solute, the H value is per moles solute, not per mol solution.

The value of H is positive for endothermic dissolutions(temperature decreases) and negative for exothermic dissolutions (temperature increase).

>> Example 1

Write the equation for the dissolution of copper(II) bromide. If 0.860 g copper(II) bromide is dissolved in 100.0 mL of water, the temperature changes from 23.10°C to 23.41°C. What is the heat of solution for copper(II) bromide? Assume the specific heat of water, 4.184 J/g•°C.

Solution:

Copper(II) bromide is made of a Cu²⁺ ion and Br^– ion, so the reaction is

CuBr₂ Cu²⁺ + 2 Br^–

To determine the amount of heat absorbed (q), use specific heat version of Equation 11.8. Remember that the heat energy is absorbed by the entire solution that includes the solvent as well as the solute. Recall that the density of water is 1.00 g/mL, so 100.0 mL of water is also 100.0 g water.

q = (mass)(specific heat)(T)

q = (100.860 g)(4.184 J/g•°C)(23.41 °C – 23.10 °C)

q = (100.860)(4.184)(0.31)

q = 130.8 J

Since the temperature increases, the H value will be negative.

H

=

–q mol

and

mol CuBr2

=

0.860 g

1 mol 223.35 g

=

0.00385 mol

so

H

=

–(130.8 J) 0.00385 mol

=

33974 J/mol

1 kJ 1000 J

=

–33.974 kJ/mol

However, in the calculation of q, T only has two significant figures. Therefore q only has two significant figures, so the answer is

H_soln = –3.4 x 10⁴ J/mol or –34kJ/mol

>> Example 2

What is the heat of solution and the dissolution reaction if when 1.893 g lithium fluoride dissolves in 250.0 mL water, the temperature changes from 23.69 °C to 19.59 °C?

Solution:

The ionic compound lithium fluoride is made from Li⁺ and F^– ions, so the reaction is

LiF Li⁺ + F^–

The moles of lithium fluoride that dissolved are

1.893 g LiF

1 mol 25.94 g

=

0.07298 mol LiF

The heat evolved is determined from

q = (mass solution)(specific heat water)(T)

q = (251.9 g)(4.184 J/g•°C)(23.69 °C – 19.59 °C)

q = (251.9)(4.184)(4.10)

q = 4321 J (Note: only three digits are significant because of the value of T)

The heat of solution will be positive because the temperature of the solution decreased.

So

H

=

+q mol

=

–(4321 J) 0.07298 mol

=

+59,211 J/mol

=

+5.92 x 104 J/mol

If you prefer, you can convert J to kJ

+5.92 x 104 J/mol

1 kJ 1000 J

=

+59.2 kJ/mol

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The relationship between lattice energy (U), heat of hydration (H_hydration),and >H_soln is

H_soln = – U + H_hydration

Values for the heat of hydration can be determined from Table 13.1. For a salt, the heat of hydration is the sum of the heat of hydration of each ion.

>> Example 3

Based on Table 13.1, what is the heat of hydration of NaI and CaBr₂?

Solution:

The ions that make NaI are Na⁺ and I^–. The heat of hydration for Na⁺ is –410 kJ/mol and that for I^– is –247 kJ/mol. Thus for NaI, the heat of hydration is

H_hydration = –410 kJ/mol – 247 kJ/mol = –657 kJ/mol

The ions that make up CaBr₂ are Ca²⁺ and two Br^–. The heat of hydration for Ca²⁺ is –1591 kJ/mol and that for Br^–; is –284 kJ/mol.

H_hydration = –1591 kJ/mol + 2(–284 kJ/mol)= –2159 kJ/mol

Equation 13.2 can be used to determine whichever quantity is missing. Usually it is used to determine lattice energy.

>> Example 4

What is the lattice energy of KI (H_soln =+21.5 kJ/mol) and MgCl₂ (H_soln = –150 kJ/mol)? Use Table 13.1 to determine heat of hydration.

Solution:

The heat of hydration for KI is

–336 kJ/mol (K⁺) + –247 kJ/mol (I^–) = –583kJ/mol.

The heat of hydration for MgCl₂ is

–1903 kJ/mol (Mg²⁺) + 2(–313 kJ/mol) (Cl^–) = –2529kJ/mol

Using Equation 13.2,

H_soln = –U + H_hydration

For KI

H_soln = –U + H_hydration

+21.5 kJ/mol = –U + –583 kJ/mol

604.5 kJ/mol = –U

604 kJ/mol = U = lattice energy

Since this was addition and subtraction, the fewest decimal places were used to determine significant figures. For MgCl₂

H_soln = –U + H_hydration

–150 kJ/mol = –U + –2529 kJ/mol

2379 kJ/mol = –U

–2379 kJ/mol = U

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