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Heat, Enthalpy and Ionic Bonding
Notes and images from instructor’s resource CD for Chemistry: The Central Science, 9 th Edition, Brown, LeMay, Bursten & Burdge, © 2003, Pearson Education, One Lake Street, Upper Saddle River, NJ 07485
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5.4 Enthalpies of Reaction
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For a reaction, Hrxn = Hproducts – Hreactants.
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The enthalpy change that accompanies a reaction is called the enthalpy of
reaction or heat of reaction (Hrxn).
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Consider the thermochemical equation for the production of water:
2H2(g) + O2(g)  2H2O(g) H = –483.6 kJ
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The equation tells us that 483.6 kJ of energy are released to the
surroundings when water is formed.
H noted at the end of the balanced equation depends on the
number of moles of reactants and products associated with the H
H  H final  H initial
value.
 H products  H reactants
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These equations are called thermochemical equations.
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Enthalpy diagrams are used to represent enthalpy changes associated with
a reaction.
In the enthalpy diagram for the combustion of H2(g), the reactants,
2H2(g) + O2(g), have a higher enthalpy than the products 2H2O(g); this
reaction is exothermic.
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Enthalpy is an extensive property.
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Therefore, the magnitude of enthalpy is directly proportional to the
amount of reactant consumed.
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Example: If one mol of CH4 is burned in oxygen to produce CO2
and water, 890 kJ of heat is released to the surroundings. If two mol
of CH4 is burned, then 1780 kJ of heat is released.
The sign of H depends on the direction of the reaction.
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The enthalpy change for a reaction is equal in magnitude but
opposite in sign to H for the reverse reaction.
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Example: CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = –890 kJ,
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But CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)
H = +890 kJ.
Enthalpy change depends on state.
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2H2O(g)  2H2O(l)
H = –88 kJ
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When we reverse a reaction, we change the
sign of DH:
CO2(g) + 2H2O(g)  CH4(g) + 2O2(g)
DH = +802 kJ
• Change in enthalpy depends on state:
H2O(g)  H2O(l)
DH = -88 kJ
Sample Exercise:
How much heat is released when 4.50g of methane gas is burned in a constant-pressure
system?
5.6
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Hess’s Law
Hess’s Law: If a reaction is carried out in a series of steps, H for the
reaction is the sum of H for each of the steps.
The total change in enthalpy is independent of the number of steps.
Total H is also independent of the nature of the path.
Example:
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CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = –802 kJ
2H2O(g)  2H2O(l)
H = –88 kJ
_______________________________________________
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = –890 kJ
Therefore, for the reaction CH4(g) + 2O2(g)  CO2(g) + 2H2O(l), H= –890 kJ.
Note that H is sensitive to the states of the reactants and products.
Hess’s law allows us to calculate enthalpy data for reactions which are
difficult to carry out directly: C(s) + O2(g) produces a mixture of CO(g) and
CO2(g).
• Hess’s law: if a reaction is carried out in a number of steps,
H for the overall reaction is the sum of H for each
individual step.
• For example:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -802 kJ
2H2O(g)  2H2O(l)
H = -88 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ
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Sample Problem: The enthalpy of combustion of C to CO2 is -393.5KJ/mol C, and the enthalpy
of combustion of CO to CO2 is -283.0KJ/mol CO:
1) C(s) + O2(g)  CO2(g)
H = -393.5KJ
2) CO(g) + 1/2O2(g)  CO2(g)
H = -283.0KJ
Using this data, calculate the enthalpy of combustion of C to CO:
3) C(s) + ½ O2(g)  CO2(g)
5.7
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Enthalpies of Formation
Hess’s law states that if a reaction is carried out in a number of steps, H for
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Consider the formation of CO2(g) and 2H2O(l) from CH4(g) and 2O2(g).
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If the reaction proceeds in one step:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l), then H1 = –890 kJ.
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However, if the reaction proceeds through a CO intermediate:
CH4(g) + 2O2(g) CO(g) + 2H2O(l) + ½O2(g)
H2 = –607 kJ
CO(g) + 2H2O(l) + ½O2(g)  CO2(g) + 2H2O(l)
H3 = –283 kJ,
Then H for the overall reaction is:
H2 + H3 = –607 kJ – 283 kJ = –890 kJ = H1
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If a compound is formed from its constituent elements, then the enthalpy
change for the reaction is called the enthalpy of formation, Hf.
Standard state (standard conditions) refer to the substance at:
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1 atm and 25oC (298 K).
Standard enthalpy, Ho, is the enthalpy measured when everything is in its
standard state.
Standard enthalpy of formation of a compound, Hof, is the enthalpy change
for the formation of 1 mol of compound with all substances in their standard
states.
If there is more than one state for a substance under standard conditions, the
more stable one is used. Example: When dealing with carbon we use graphite
because graphite is more stable than diamond or C60.
The standard enthalpy of formation of the most stable form of an element is
zero.
Using Enthalpies of Formation to Calculate Enthalpies of Reaction
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Use Hess’s law!
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Example: Calculate H for
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
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We start with the reactants, decompose them into elements, then rearrange the
elements to form products. The overall enthalpy change is the sum of the
enthalpy changes for each step.
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Decomposing into elements (note O2 is already elemental, so we concern
ourselves with C3H8(g):
C3H8(g)  3C(s) + 4H2(g)
H1 = –Hof[C3H8(g)]
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Next we form CO2 and H2O from their elements:
3C(s) + 3O2(g)  3CO2(g)
H2 = 3Hof[CO2(g)]
4H2(g) + 2O2(g)  4H2O(l)
H3 = 4Hof[H2O(l)]
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We look up the values and add:
o
H rxn = –1(–103.85 kJ) + 3(–393.5 kJ) + 4(–285.8 kJ) = –2220 kJ
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In general:
Horxn =nHof (products) – mHof (reactants)
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Where n and m are the stoichiometric coefficients.
H rxn   nH f products    mH f reactants 
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Sample Problem:
Calculate the standard enthalpy of formation of CaCO3(s) from the following
reaction: CaCO3(s)  CaO(s) + CO2(g)
8.1
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Chemical Bonds, Lewis Symbols, and the Octet Rule
The properties of many materials can be understood in terms of their
microscopic properties.
Microscopic properties of molecules include:
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The connectivity between atoms and
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The 3D shape of the molecule.
When atoms or ions are strongly attracted to one another, we say that there is
a chemical bond between them.
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In chemical bonds, electrons are shared or transferred between atoms.
Types of chemical bonds include:
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Ionic bonds (electrostatic forces that hold ions together, e.g., NaCl);
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Covalent bonds (result from sharing electrons between atoms, e.g., Cl2);
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Metallic bonds (refers to metal nuclei floating in a sea of electrons, e.g.,
Na
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Lewis Symbols
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The electrons involved in bonding are called valence electrons.
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Valence electrons are found in the incomplete, outermost shell of an
atom.
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As a pictorial understanding of where the electrons are in an atom, we
represent the electrons as dots around the symbol for the element.
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The number of valence electrons available for bonding are indicated by
unpaired dots.
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These symbols are called Lewis symbols or Lewis electron-dot symbols.
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We generally place the electrons on four sides of a square around the
element’s symbol.
The Octet Rule
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Atoms tend to gain, lose or share electrons until they are surrounded by eight
valence electrons; this is known as the octet rule.
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An octet consists of full s and p subshells.
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We know that s2p6 is a noble gas configuration.
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We assume that an atom is stable when surrounded by eight electrons
(four electron pairs).
The Octet Rule
• All noble gases except He has an s2p6 configuration.
• Octet rule: atoms tend to gain, lose, or share electrons until they are
surrounded by 8 valence electrons (4 electron pairs).
• Caution: there are many exceptions to the octet rule.
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Sample Problem:
Write the electron dot symbols for the following:
K, Ba, Si, Br, Xe Ge, B
8.2 Ionic Bonding
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Consider the reaction between sodium and chlorine:
Na(s) + ½Cl2(g)  NaCl(s);
Hof = –410.9 kJ/mol
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The reaction is violently exothermic.
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We infer that the NaCl is more stable than its constituent elements.
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Sodium has lost an electron to become Na+ and chlorine has gained
the electron to become Cl–.
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Note Na+ has an Ne electron configuration and Cl– has an Ar
configuration.
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That is, both Na+ and Cl– have an octet of electrons.
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NaCl forms a very regular structure in which each Na+ ion is surrounded by
six Cl– ions.
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Similarly each Cl– ion is surrounded by six Na+ ions.
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There is a regular arrangement of Na+ and Cl– in three dimensions.
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Note that the ions are packed as closely as possible.
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Note that it is not easy to find a molecular formula to describe the ionic
lattice.
Energetics of Ionic Bond Formation
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The heat of formation of NaCl(s) is exothermic:
Na(s) + ½Cl2(g)  NaCl(s)
Hof = –410.9 kJ/mol
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Separation of the NaCl into sodium and chloride ions is endothermic:
NaCl(s)  Na+(g) + Cl–(g) H = +788 kJ/mol
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The energy required to separate one mole of a solid ionic compound into
gaseous ions is called the lattice energy, Hlattice.
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Lattice energy depends on the charge on the ions and the size of the ions.
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The stability of the ionic compound comes from the attraction between
ions of unlike charge.
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The specific relationship is given by Coulomb’s equation:
F = k(Q1Q2)/d2
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Where Q1 and Q2 are the charges on the particles, d is the distance
between their centers, and k is a constant.
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As Q1 and Q2 increase, E increases, and as d increases, E
decreases.
Calculation of Lattice Energies: The Born-Haber Cycle
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The Born-Haber cycle is a thermodynamic cycle that analyzes lattice energy
precisely.
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Consider a Born-Haber cycle for the formation of NaCl(s) from Na(s) and
Cl2(g).
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The direct route is:
Na(s) + ½Cl2(g)  NaCl(s)
Hof = –410.9 kJ
Alternatively, we can form:
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Sodium gas (108 kJ; endothermic), then
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Chlorine atoms (122 kJ; endothermic), then
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Sodium ions (ionization energy for Na, 496 kJ; endothermic), then
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Chloride ions (electron affinity for Cl, –349 kJ; exothermic), then
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Form the ionic lattice (exothermic).
The sum of the above enthalpies is –410.9 kJ.
The lattice energy can be solved as follows:
Na(s)  Na(g)
½ Cl2(g)  Cl(g)
Hs = +108kJ/mol x 1 mol = 108 kJ
Hd = 244 kJ/mol x ½ mol = 122 kJ
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Na(g)  Na+(g) + eCl(g) + e-  Cl-(g)
Na+(g) + Cl-(g)  NaCl(s)
Na(s) + ½Cl2(g)  NaCl(s)
HIE = 496 kJ/mol x 1 mol = 496 kJ
HEA = -349 kJ/mol x 1 mol = -349 kJ
Hlattice = x
Hf = –410.9 kJ
-410.9 kJ = (108 + 122 + 496 + (-349) + x) kJ; x = 788 kJ = Hlattice
Notice that the sum of the equations equals the overall equation for the
formation of NaCl in a single step process.
Electron Configuration of Ions of the Representative Elements
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These are derived from the electron configuration of elements with the
required number of electrons added or removed from the most accessible
orbital.
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Electron configuration of ions can predict stable ion formation:
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Na: [Ne]3s1
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Na+: [Ne]
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Cl: [Ne]3s23p5
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Cl–: [Ne]3s23p6 = [Ar]
• Electron configurations can predict stable ion formation:
• Mg: [Ne]3s2
• Mg+: [Ne]3s1
not stable
2+
• Mg : [Ne]
stable
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• Cl: [Ne]3s 3p
• Cl-: [Ne]3s23p6 = [Ar]
stable
Sample Problems:
Write the stable electron configuration for :
a. K b. Ca
c. Al d. P e. S
f. I
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Transition-Metal Ions
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Lattice energies compensate for the loss of up to three electrons.
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We often encounter cations with charges of 1+, 2+ or 3+ in ionic compounds.
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However, transition metals can’t attain a noble gas conformation (>3 electrons
beyond a noble gas core).
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Transition metals tend to lose the valence shell electrons first and then as
many d electrons as are required to reach the desired charge on the ion.
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Thus electrons are removed from 4s before the 3d, etc.
Polyatomic Ions
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Polyatomic ions are formed when there is an overall charge on a compound
containing covalent bonds.
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Examples: SO42–, NO3–
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In polyatomic ions, two or more atoms are bound together by predominantly
covalent bonds.
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The stable grouping carries a charge.