Download Chapter 6. Therrnochemistry

Chapter 6. Therrnochemistry
The Nature & Types of Energy (Section 6.1)
Energy Changes in Chemical Reactions (Section 6.2)
Introduction to Thermodynamics (Section 6.3)
Enthalpy of Chemical Reactions (Section 6.4)
Calorimetry (Section 6.5)
Standard Enthalpy of Formation and Reaction (Section 6.6)
Heat-of Solution and Dilution (Section 6.7)
SUMMARY
Truly balanced reactions include the energy changes that accompany chemical reactions.
This chapter describes these energy changes, techniques used to measure them
experimentally and methods used to predict them quantitatively.
The Nature & Types of Energy (Section 6.1)
Energy. Energy is one of those terms that is difficult to understand because it is an abstract
concept that is used in many different ways. The standard scientific definition is a system’s
capacity to do work (move an object against an opposing force). There are three principal
types of energy, some with a number of sub-classes. The types of energy that are important
in our discussion of chemical reactions are
kinetic (dynamic) energy:
potential (static) energy:
radiant (electromagnetic) energy:
sub-classes:
energy an object possesses by virtue of its motion
energy an object possesses by virtue of its position
energy a wave possesses by virtue of its frequency
chemical energy (kinetic and potential energy E
associated with bond breakage & formation)
thermal energy (kinetic and potential energy E
associated with random particle motion)
heat (thermal energy transferred between objects
at different temperatures)
The law of conservation of energy states that energy is neither destroyed nor created; it is
transferred from one part of the universe to another or it is converted from one form to
another.
Energy Changes in Chemical Reactions (Section 6.2)
In order to study energy changes in chemical reactions we divide the universe into
the system (where the reaction occurs) and the surroundings. In an open system, mass and
energy in the system are exchanged with the surroundings. For example, a burning fire is an
open system. The system consists of the reactants (wood and oxygen) and products (CO2,
water vapor, ashes and so on). The surroundings are everything else. In a closed system,
108
Chemistry, Ch. 6: Thermochemistry 109
only energy from the system is exchanged with the surroundings. In an isolated system,
neither mass nor energy from the system is exchanged with the surroundings.
When a reaction in the system produces heat that is released to the surroundings,
that is when energy is a product, the reaction is called exothermic. When a reaction in the
system consumes heat from the surroundings (heat is a reactant) the reaction is called
endothermic.
Introduction to Thermodynamics (Section 6.3)
The theories that allow us to describe energy changes quantitatively are part of
thermodynamics, the study of the interconversion of different types of energy. The first law
of thermodynamics (conservation of E) limits the energy absorbed or released by a reaction
to two.#arts: work done on or by the system, w, and heat absorbed or released by the
system, q. By definition, work done on the system is positive (work done by the system is
negative) and heat absorbed by the system is positive (heat released from the system is
negative).. For the hypothetical reaction: A~nit~ -) Ann~, the conservation law is
AEsys --&Esurr
&Esys =q+w
where AEsys is the change in the internal energy of the system and AEsu,r is the change in the
energy of the surroundings. If we could relate q and w directly to the reactants and products
(Ainitiat and A,o~=), we’d have the quantitative description we’re looking for. The problem is
that q and w are path dependent properties, in other words their values change as the
conditions and procedures used to carry out the reaction change. We need pathindependent properties that only depend on Aioitia= and A, na~ and stay the same regardless of
how the reaction is carried out. It is a little more work, but we can relate q and w to state
functions, which is the technical term for path-independent properties.
Work done by gases is -PV (work = force x distance; pressure = force/area; volume
= area x distance), so we can define an energy state function especially for reactions in
which the pressure is constant. The state function is called the enthalpy, H. The formula for
the enthalpy is
H=E+PV
The following equations show that enthalpy changes (&H) equal the heat transferred from
the system by the reaction at constant pressure, q~.
AH : H~.., - HinitiaI : (Efina, + PVfinal) - (Ei.,ti~’ + PVinitiaI )
Examples
6.1 -6.3
: (Efinal - EinitiaI) + P (Vfinal - VinitiaI )
= AE + PAV = AE - w = qp
Exercises
6.1 -6.5
Enthalpy of Chemical Reactions (Section 6.4)
Most reactions are run at constant pressure, so the enthalpy ~s equal to the heat transferred
by the reaction. When the enthalpy is included in a chemical reaction it is called a
thermochemical reaction. For exothermic reactions, (heat is released by the system) the
enthalpy change is negative:
CH4(g)+ 2 O2(g)--,’. CO2(g)+ 2 H20(g)
AH =-802.4 kJ/mol
~ ~u ~neml~[ry, ~n. o, iRermocnemls[ry
For endothermic reactions, (heat is absorbed by the system) the enthalpy change is
positive:
CaCO~(s) --+ CaO(s) + CO2(g)
z~H = 177.8 k J/
When a reaction is reversed, the sign of ,&H is reversed
H20(s) --> H20(£)
H20(~’) --~ H20(s)
&H = 6.01 kJ/mol
AH = -6.01 kJ/mel
The enthalpy is an extensive property, its value depends on the amount of material in the
system. Therefore, scaling the reaction (multiplying by a constant) also scales
H20(s) --> H20(~)
2 H20(s) -~ 2 H2Oi#)
AH = 6.01 kJ/mol
AH = 12.0 kJ/reel
The physical states of all reactants and products are crucial to enthalpy values
CH4(g)+2 O2(g)--> CO2(g)+2 H20(g) AH =-802.4 k J/reel
CH4(g)+ 2 O2(g) -~ CO2(g)+ 2 H20(#) AH =-890.4 kJ/mol
Since 80.0 g, of CH4 equals nearly 5 moles, if all the gas was burned under conditions
Example
that produce gaseous products, approximately 4 MJ (5 mol x 0.8 M J/reel) of energy 6.4
would be released.
80.0gx lm°lCH4 x 802.4kJ released = 4.00 MJ released
16.032 g CH4
moICH4
Exercises
6-6 - 6-7
Calorimetry (Section 6.5)
Heat Capacity and Specific Heat. A calorimeter is a device for measuring the heat
changes associated with chemical reactions. All calorimetry experiments involve carrying
out a reaction in a vessel immersed in a medium such as water and measuring the
temperature change (At) of the medium. The heat capacity, C, of a substance is the amount
of heat, q, required to raise the temperature of a given mass of substance by 1 °C, The heat
capacity of 100 g H20 is 418.4 J/°C. This means that 418.4 J of heat will raise the
temperature of 100 g water 1 degree Celsius. The heat capacity is a proportionality
constant that relates the amount of heat absorbed or released by a material to its change in
temperature:
The specific heat, s, is a related term which is the amount of heat required to raise the
temperature of one gram of a substance by 1 °C. In terms of the specific heat, the heat
absorbed or released is
q = msAt.
Chemistry, Ch. 6: Thermochemistry 111
Constant-Volume Calorimetry. Bomb calorimeters, as shown in Figure 6.8 in the
textbook, are used to measure the heat evolved in combustion reactions. A high-pressure
steel vessel, called a bomb, is loaded with a small amount of a combustible substance and
02 at 30 atm of pressure. The loaded bomb is immersed in a known amount of water. The
heat evolved during combustion is absorbed by the water and the calorimeter:
q~, : -(qcal + qwater )
The change in the temperature of the wate~ and calorimeter are related to the heat evolved
by the reaction, q, and to the enthalpy, AH.
AH = q~xn = -(CcalAt + mswaterAt)
Reactions in a bomb calorimeter occur under constant volume rather than constant pressure
conditiops, and so the heat released does not equal AH exactly. For most reactions, the
difference is small and can be neglected. For instance, for the combustion of 1 mole of
pentane, the difference is only 7 kJ out of 3500 kJ.
Constant-Pressure Calorimeter. The heat evolved in non-combustion reactions is
measured in a constant-pressure calorimeter. Coffee cup calorimeters, shown in Figure 6.9
in the textbook, are very inexpensive examples of these devices. Water-soluble reactants
are added to water in the calorimeter. The reaction occurs in the solution, absorbing or
liberating heat that changes the temperature of the solution (mostly water), and the
calorimeter:
qrxn - -(q~.l + qwater )
Tl~ese calorimeters are constructed of low heat capacity materials, such as Styrofoam, so
the heat absorbed by the calorimeter can be neglected:
qrxn =-qwater ’
Examples
6.5 - 6.7
Reactions in a coffee cup calorimeter occur under constant pressure conditions,
Exercises
and so the heat released is equal to AH.
6-8 - 6-13
AH = qrxn = -mSwaterz~t
Standard Enthalpy of Formation and Reaction (Section 6.6)
Recapping the information presented so far, we can calculate AH for any reaction for
which areactants and Hproducts are known and measure AH for most reactions. The problem is
that there is no way to measure Hreactants and aproducts. Absolute energy values, such as
Hreactants and Hproducts, can’t be measured because there, is no way to define absolute zero
energy. Relative energy values can be measured because the reference energy is arbitrary.
The reference point for enthalpy values is the standard of enthalpy of formation of the
elements which is defined as zero at 1 atm (g) or 1 M and 25 °C (in their standard state).
The standard enthalpy of atlotropes and single atoms of diatomic elements have non-zero
values. For example,
112 Chemistry, Ch. 6: Thermochemistry
AH~(O2 ) = 0 kJ / mol
°
AHf(O3)=142
kJ/ mol
AH~ (Cgraphite ) = 0
kJ / mol
°
AHf(Cdiamond)=l.9
kJ/mol
AH~ (H2 ) = 0 kJ / mol
AH~(H)=218.2 kJ/mol
Direct Calculation ofz~H,~,. Now we can calculate AH for any reaction using A/-!) of the
reactants and products. The enthaipy change for any reaction is the difference of the SUM
of the standard enthalpies of formation of the products multiplied by their stoichiometric
coefficients and the SUM of the standard enthaipies of formation of the reactants multiplied
by their stoichiometric coefficients. For e_xample, the enthaipy of the decomposition of
CaCO3, CaCO3(s) --> CaO(s) + CO2(g), is
AH~xn = ~produotsnproduclsAH~(products)-’~reactantsnreactantsAH~(reactants)
= AHf(CaO) + AHf (CO2)- AHf(CaCO3)
= (1 mol)( 635.6 kJ / mol) + (1 mol)(-393.5 kJ / mol) - (1 mol)(-1206.9 kJ / tool) = 177.8 kJ / tool
Indirect Calculation ofz~H~,. Some reactions don’t proceed as written. For example, the
reaction may proceed too slowly or generate side-products. In other cases, the enthalpy of
formation of one or more reactant or products may not be available. The alternative comes
from Hess’s law of heat summation. When a reaction is the sum of several reaction steps,
then the enthalpy change for the reaction is equal to the sum of the enthalpy changes of the
reaction steps. For example, it’s not convenient to measure the enthalpy of the conversion
of graphite to diamond. This reaction takes thousands of years at high pressure in the
earth. However, it is easy to measure the enthalpies of combustion of graphite and
diamond:
Cgraphite -}- 02 ~ CO2
AHoor~b :-393.51 kJ/mol
Cdiamond + 02 --~ 002
AHoomb :-395.40 kJ/mol
Notice that when we reverse the diamond thermochemical reaction, the sum of the Examples
combustion reactions is the graphite to diamond conversion:
6.8-6.9
Cgraphite ÷ 02 --)- CO2
AHcomb =-393.51 kJ/mol
002 --). Cdiamond + 02
AHcomb = 395.40 kJ/mol
AH~,n = 1.89 kJ / mol
Cgraphite ~)" Cdiarnond
Exercises
6-14-6-17
Heat of Solution and Dilution (Section 6.7)
Heat of Solution. Dissolving ionic solutes in a solvent often produces temperature changes
as the extent of solute hydration changes. We can think of solute dissolution as requiring 2
steps: solute disassociation to gaseous ions followed by ion hydration to form the solution.
This is not what actually happens, but these are steps for which AH values are available.
The energy required to separate 1 mole of an ionic compound to gaseous ions (step 1) is
called the lattice energy, Etattice. The enthalpy change of the hydration of 1 mole of gaseous
Chemistry, Ch. 6: Thermochemistry 113
ions (step 2) is the enthalpy of hydration, AHhydr. The enthalpy of solution is the sum of
these two terms. For the dissolution of LiCI:
LiCl(s)+energy-->Li+(g)+CI (g) Elattice =853 kJ/mol
Li+(g) + Cl-(g) --> Li+(aq) + CI (aq) AH,ydr = -890.1 kJ / mol
LiCl(s) --> Li+(aq) + CI (aq)
AHso~n -- Elattice + AH~dr = -37.1 kJ/mol
Heat of Dilution. When Elat~ice, which is ~lways positive, is smaller than AHhydr, the
solvation reaction is exothermic. If more water is added to such a solution, more
heat will be released by the system. (This is why you never add water to acid, but
always add acid to water with vigorous stirring. If the water boils it is much less
dangerous than if concentrated acid boils!) Using the same reasoning, if the
solvation reaction is endothermic, dilution will cause more heat to be absorbed
from the.surroundings.
Example
6.10
Exercises
6-18-6-19
Chemistry, Ch, 6: Thermochemistry 115
GLOSSARY LIST
kinetic energy
potential energy
chemical energy
thermal energy
heat
radiant energy
energy conservation law
work
heat
state function
enthalpy
endothermic
exothermic
thermochemical equation
thermodynamics
system
surroundings
open system
closed system
Isolated system
enthalpy of reaction
standard state
heat capacity
specific heat
calorimeter
standard enthalpy
of formation
Hess’ law
lattice energy
heat of hydration
heat of solution
enthalpy of solution
heat of dilution
EQUATIONS
Algebraic Equation
AE-q+w
w -- -PAV
H=E+PV
AH - AE + PAV
AE = AH - RTAn
C=ms
q = CAt
q = msAt
AH~o = ~nAH~(products)- ~mAH~(reactants)
English Translation
The energy of a system consists of heat
and work
118 Chemistry, Ch. 6: Thermochemistry
WORKED EXAMPLES
EXAMPLE 6.1 Work Done on the System
Calculate the work done on the system when 6.0 L of a gas is compressed to 1.0 L by a
constant external pressure of 2.0 atm,
¯ Solution
The work done is:
W = -P AV = -P(V2 - Vl)
= -2.0 atm (I.0 L - 6.0 L) = + 10 L.atm
The answer can be converted to joules.
w = 1.01 × 103 J
¯ Comment
Obtaining a positive value for work means that work is done on the system by the
surroundings in a compression. A positive work value means the system gains energy.
EXAMPLE 6.2 Gas Expansion Work
A gas, initially at a pressure of 10.0 arm and having a volume of 5.0 L, is al{owed to expand
at constant temperature against a constant external pressure of 4,0 arm until the new
volume is 12.5 L. Calculate the work done by the gas on the surroundings.
o Solution
In this problem the system does work on the surroundings as it expands, and by convention
(Table 6.5 in the text) the sign is negative:
w = -Pz~V = -P(V2 - V1)
P is the pressure opposing the expansion, ~V is the change in volume of the system,
w = -4,0 atm (12.5 L - 5.0 L) = -30 L.atm
This quantity can be expressed in units of joules.
w = -30 L,atm x
101.3 J
= -3.0 x 103 J
t L. atm
Chemistry, Ch. 6: Thermochemistry 119
EXAMPLE 6.3 The First Law of Thermodynamics
A gas is allowed to expand at constant temperature from a volume of 10.0 L to 20.0 L
against an external pressure of 1.0 atm. If the gas also absorbs 250 J of heat from the
surroundings, what are the values of q, w, and AE?
Solution
The work done by the system is
w = -P AV = -P(V2 - V1)
=-1.0 atm (20.0 L- 10.0 L) =- 10 L.atm
w=-10L.atmx 101.3J =_1.0x103j
1 L ¯ atm
The amdunt of heat absorbed was 250 J, and so q -- 250 J.
Substituting into the first law of thermodynamics gives the energy change, AE.
AE=q+w
= 250 J - 1000 J
AE = -750 J
o Comment
tn this example, the system did more work than the energy absorbed as heat, therefore the
internal energy E decreased.
EXAMPLE 6.4 Thermochemical Equations
The thermochemical equation for the combustion of propane is:
C3H8(g) + 5 O2(g) -~ 3 CO2(g) + 4 H20 AH~n = -2220 kJ/mol
a.
b.
How many kJ of heat are released when 0.50 mole of propane reacts?
How much heat is released when 88.2 g of propane reacts?
o Solution for a
Let q = the heat absorbed or released by the reaction. The heat released by an exothermic
reaction is an extensive property. This means that q depends on the amount
of propane consumed. The equation indicates that 2220 kJ of heat is released per mole of
propane burned.
q = 0.50 mol C3H8 x
q =-1.1 x 103 kJ
-2220 kJ
1 mol C3H8
120 Chemistry, Ch. 6: Thermochemistry
o Solution for b
Since we know the he~t.of reaction per mole, we convert the number of grams of C3H8 to
moles.
q = 88.2 g C3H8 ×
1 mol C3H8 -2220 kJ
x
44.1 g C3H8 1 mol C3H8
q = -4440 kJ
EXAMPLE 6.5 Specific Heat
a.- What is the heat capacity of a block of lead if the temperature of a 425-g block increases
2.31 °C when it absorbs 492 J of heat?
b. What is the specific heat of lead?
¯ Solution
a. The heat capacity of the block of lead is the heat absorbed divided by the temperature
rise:
C - q - 492J =213J/°C
At (2,31 °C)
b. In terms of the specific heat (s), the amount of heat absorbed when an object of mass m
is heated from T~ to Tr is:
q = ms At
Substituting into the equation the given quantities for q, m, and/~t gives:
492 J = (425 g) s (2.31 °C)
Rearranging to solve for the specific heat s:
s-
492 J
= 0.501 J/g.°C
(425 g)(2.31 °C)
¯ Comment
Since C = ms we could have solved part b from s = C/m using our answer from part a.
Chemistry, Ch. 6: Thermochemistry 121
EXAMPLE 6.6 Determining The Calorimeter Constant
The combustion of benzoi~ acid is often used as a standard source of heat for calibrating
combustion bomb calorimeters. The heat of combustion of benzoic acid has been accurately
determined to be 26.42 kJ/g. When 0.8000 g of benzoic acid was burned in a calorimeter
containing 950 g of water, a temperature rise of 4.08 °C was observed. What is the heat
capacity of the bomb calorimeter (the calorimeter constant)?
¯ Solution
The combustion of 0.8000 g of benzoic acid produces a known amount of heat.
q~o = 0.8000 g ×
26.41 kJ =-21.14 kJ =-2.114 × 104j
lg
And since At and the amount of water are known, Cc,~, the heat capacity of the calorimeter.
can be calculated. All the heat from the combustion reaction is absorbed by the bomb
calorimeter and water.
q~n = -(q~,l + qwater)
qrxn = -(Cca~ At + ms At)
The heat absorbed by the water was qwater = ms At, where
m = 950 g water
s = 4.184 J/g-°C
At = 4.08 °C
qwater = 950 g (4.184 J/g’°C)(4.08 °C) = 1.620 x 104 J
The bomb calorimeter must have absorbed the difference between the 21.14 × 103 J
released, and the 16.20 x 103 J absorbed by the water.
qr~o = -(qo~l +
qc,I = -qr~o - q~,te~ = --(--2.114 X 104 J) - 1.620 x 104 J
qoal = 4.94 x 103 J
The heat capacity of the calorimeter is
Cc~- qbomb _ 4.94x10~ J
At 4.08 °C
Cc~ = 1210 J/°C
122 Chemistry, Ch. 6: Thermochemistry
EXAMPLE 6.7 Determining the Heat of Combustion
The thermochemical equation for the combustion of pentane (C5H12) is:
C5H12(~) + 8 O2(g) --> 5 CO2(g) +
6 H20(~) AH" : -3509 k J/mot
From the following information calculate the heat of combustion, AH°, and compare your
result to the value given above. 0.5521 g of C5H12 was burned in the presence of excess 02
in a bomb calorimeter. The heat capacity of the calorimeter was 1.800 x 103 J/°C, the
temperature of the calorimeter and 1.000 x 103 g of water rose from 21.22 °C to 25.70 °C.
¯ Solution
The heat evolved by the combustion reaction is absorbed by the water and the calorimeter
assembly.
qrxn = -[qoal +
where
qrxn = -(Cca=At + msAt)
At = 25.70 °C - 21.22 °C = 4.48 °C
Substituting:
1"800x103
qca~ =--x 4.48 J
1°C
qwater : 1.000 x 103 g x
°C = 8.06 x 103 J
4.184 J
x
4.48 oC = 1.87 x 104 J
qrxn = -[8,060 J + 18,700 J]
qr~n = --26,800 J = - 26,8 kJ (three significant figures)
AH° refers to the reaction of 1 mole of pentane. The 26,800 J was evolved by 0.5521 g
C5H12. The molar mass of pentane is 72.15 g.
AH° = q
26.8 kJ
0.5521 g
72.15 g C5H12
1 mol
AH° = -3500 kJ/mol = -3.50 x 103 kJ/mol
¯ Comment
To three significant figures the result compares well with the value given in the
thermochemical equation. Small differences will arise because the heat evolved at constant
volume is not quite the same as the heat evolved at constant pressure (which is equal to
AE). Only the heat evolved at constant pressure exactly equals AH, the enthalpy change.
Chemistry, Ch. 5: Thermochemistry 123
EXAMPLE 6.8 Using Enthalpies of Formation
Using AH~ values in Appendix 3 of the textbook calculate the standard enthalpy change for
the incomplete combustion of ethane (C2H6).
5
C2H6(g) + ~O2(g)-~ 2 CO(g) + 3 H20(~)
¯ Solution
The enthalpy change for this chemical reaction in terms of enthalpies of formation is:
- AH~xn = ~’nAH~ (products) - ~ m AH~ (reactants)
AH~n = [2 x AH~ (CO)+ 3 x AH~ (H20)]- [1 x AH~ (C2H6)- ~- x AH~ (O2)]
To avoid cumbersome notation, the physical states of the reactants and products were
omitted from this equation. Be careful to obtain from Appendix 3 the appropriate value
ofAHf. Note that the coefficients from the chemical equation are equal to the number of
moles of each substance and that al~l the terms involving AH~ of the reactants are preceded
by a negative sign. Substituting the values from Appendix 3:
AH~o = (2 mol)(-110.5 kJ/mol) + (3 mol)(-285.8 kJ/mol)] - [(1 mol)(-84.68 kJ/mol)
+ (5/2 mol)(0 kJ/mol)]
= (-221.0 k J) + (-857.4 k J) - (-84.68 k J) = -221.0 kJ - 857.4 kJ + 84.68 kJ
AH~n = -993,7 kJ
EXAMPLE 6.9 Hess’s Law
The standard enthalpy change for the combustion of 1 mole of ethanol is:
C2HsOH(~) + 3 O2(g) --~ 2 CO2(g) + 3 H20([) AH~xn = -1367 kJ/mol
What is AH~n for the following reaction in which H20 is formed as a gas, rather than as a
liquid?
C2H5OH(~) + 3 O2(g) :~ 2 CO2(g) + 3 H20(g)
Given the heat of vaporization of water
H20(~) -+ H20(g) AHvap = 44 kJ/mol
124 Chemistry, Ch. 6: Thermochemistry
¯ Solution
We can imagine a two-step path for this reaction. In the first step, ethanol undergoes
combustion to form liquid H20, followed by the second step in which H20(I) is vaporized.
The sum of the two steps gives the desired overall reaction, and according to Hess’s law the
sum of the two AH’s gives the overall L~H.
C2HsOH(I) + 3 O2(g) -~ 2 CO2(g) + 3 H20(I)
3 H2O(I) -+ 3 H20(g)
C2HsOH(I) + 3 O2(g) --~ 2 C~)2(g) + 3 H20(g)
z~H~ = -1367 k J/tool
AH = 132 kJ/mol
AH=o = -1235 kJ/mol
EXAMPLE 6.10 Heat of Hydration
The heat of solution of NaBr is -1.0 k J/tool. The lattice energy of NaBr is 735 kJ/mol,
Determine the heat of hydration of NaBr and write the equation for the hydration reaction.
Solution
Recall that the heat of solution is the sum of the lattice energy (E~.~) and the heat of
hydration
~Hsoln --" Elattice + AHhydr.
Rearrranging for the heat of hydration:
= -1.0 k J/tool - 735 k J/tool
= -736 kJ/mol
~,Hhydr refers
to the heat liberated or absorbed when the gas phase ions are dissolved in
water: The equation is:
H20
Na*(g) + Br-(g) ~ Na+(aq) + Br-(aq)
¯ Comment
For NaBr, I::latt~ce and AH.y~ have essentially the same values, but are opposite in sign,
Therefore, when NaBr dissolves in water, the energy E~,~e required to break up the crystal
lattice is "paid back" by the hydration of the Na÷ and Br- ions. Also, NaBr does not dissolve
in nonpolar solvents. Molecules of nonpolar solvents can only interact weakly with ions in
the crystal lattice, and so "solvation" is not sufficient to compensate for the lattice energy.