NAME _____________________________________ UNIT 5 (2): THERMODYNAMICS: HESS’S LAW &
ENTHALPIES OF FORMATION AND REACTION
I-IV) The basics to Thermochemistry and Trivedi assignment
In this section of notes, I am taking us back (temporarily) to enthalpy ….as it has a number of facets…
V) A LITTLE BONDING 101 on BOND STRENGTH and HESS’S LAW
A) Generally: Chemical stability (the resistance to reacting) increases as bond strength increases.
Checkout: Covalent Bonding: http://www.bozemanscience.com/ap-chem-019-covalent-bonding
B) When it comes to bond strength, there are two ideas to explore: bond energy and bond length.
1) Bond Energy: measure of bond strength in a chemical bond. It is the heat required to break
one mole of molecules into their individual atoms.
a) always a positive value as it applies to energy required to break a bond (endothermic)
b) as the number of bonds between two species increases …the bonds get shorter and
stronger.
c) We study the bond strength of ionic compounds and molecular compound differently.
We will explore these in our bonding unit, more fully … but for now…
2) When dealing with ionic compounds it is all about the Lattice Energy
a) Lattice energy is heat of formation (∆H° f) released when 1 mol of ions in the gas
phase bond to produce 1 mol of ionic solid (crystal)
ii) Factors affecting lattice energy (and thus the bond strength of ionic compounds) are:

the magnitude of the charges … As the magnitude of the charges
on the ion species increases …
meaning: The greater the charges, *the greater the increase in
lattice energy because more energy is released upon formation
and thus the stronger the ionic bond

the radius of the ion species … As the radius (as the size) of the
ion species increases, *the lattice energy decreases.
iii) lattice energy has implications regarding the solubility of ionic compounds
in water …Yes, Ksp rears its slightly soluble head again!!!
464
3) When dealing with molecular substances… (for which the upcoming Hess’s Law is most concerned)
There is a bond length … and an associated bond energy too – although we think of these
factors a bit differently from ionic compounds. Reflect back on Bozeman #19 and the
determination of the bond length as the “sweet spot” where the repulsive nuclear
forces and the attractive electron-nuclear forces balance out.
http://images.tutorvista.com/cms/images/81/potential-energy-curve-H2-molecule.png
a) For any given pair of atoms, the bond is stronger, the greater the overlap of atomic
orbitals. This translates into …a shorter bond tends to be a stronger bond …
b) There is a maximum level of overlap for bonding atoms. Once the maximum level of
overlap occurs,* the repulsive forces between the 2 positive nuclei come into play.
i) this brings us back to the diagram
re: interaction of atoms
c) This maximum or optimal overlap is called the bond length.
i) Bond Length: is the average distance between nuclei of two bonded atoms
in a molecule
d) For example: With dihydrogen we say that the H─H covalent bond has a
bond length of 0.074 nanometers & a bond energy of 4.52 eV or 7.24 x 10-19 Joules
465
4) Thus, with bond strength, there is a correlation between bond energy and bond length.
a) There are tables listing average bond strengths
https://wilenskychemistry.wikispaces.com/Thermodynamics+PowerPoint+3+(Hess's+Law+Part+2+breaking+and+making+bonds)
i) The table lists average values. Absolute values may vary from the above
values. For instance, The C – H absolute bond energy in methane may be
slightly different than the absolute bond energy of the C – H bond in
chloroform
vs.
methane
chloroform
Using electronegativity values, partial charge distribution, and / or ionic
character of the bonds suggest a reason why the absolute values of the C – H
bonds in the two molecules are slightly different.
466
b) There are tables which list bond lengths
i) There are a number of factors which influence/affect bond length
⍟ The smaller the atoms * the shorter the bond length
⍟ The greater the effective nuclear charges of the bonded species,
[the greater the attractive coulombic forces] …the shorter the bond
length
⍟ The larger the electronegativity difference [ionic character] of the
bonded species, …the shorter the bond length
⍟ The greater the bond multiplicity (single vs. double vs. triple)
….the shorter the bond length.
Checkout: Bond Length and Bond Energy: http://www.bozemanscience.com/ap-chem-052-chemical-potential-energy
C) Recall: Enthalpy is a state function… it is what it is … and does not depend upon how it became
what it is … or how it evolved.
1) Thus, ∆H of a chemical process depends only on the amount of matter that undergoes
the chemical changes and on the nature of the initial state of the reactants and the final
state of the products.
2) So, the number of steps from reactants to products does not matter …. The sum of the
enthalpy changes of those individual steps must be equal to the enthalpy changes associated
with the direct synthesis (or a one-step process.) …
467
The visual learner in you may appreciate an Enthalpy Diagram. For instance, consider
the combustion of methane in oxygen as a one step process or as a 2-step process, with
the formation of CO as an intermediary product:
CH(g) + 2 O2(g) → CO2(g) + H2O(𝓁)
Combustion
producing
liquid water: 1
step process
Combustion
producing
liquid water: 2
step process
http://www.mikeblaber.org/oldwine/chm1045/notes/Energy/HessLaw/Energy04.htm
The net reaction is the same … as long as we can write a series of equations that add up
to the equation we need, and as long as we know a value for ∆ H for all intermediate
reactions, we can calculate the overall ∆H.
a) In a direct synthesis (a formation), the ∆H = ∆H° f with ∆H° f being the heat of
formation … the energy exchange for the production of 1 mol of product from
its constituent elements. We will see this heat of formation throughout this
packet.
D) Hess’s law: (really a very sweet concept) When a reaction occurs in a series of steps, ∆H (the change
in enthalpy, a.k.a., the heat of reaction) for the overall reaction should equal the sum of the
enthalpy changes for the individual steps.
That is: When a reaction is the sum of two or more other reactions, the ΔH for the
overall process is the sum of the enthalpy changes for the constituent reactants
1) Hess’s Law is a neat means of calculating enthalpy changes which may be fleeting or difficult
to measure.
a) per your text (p187) “It is impossible to measure directly the enthalpy for the
combustion of carbon to form carbon monoxide. Combustion of 1 mol of carbon with
0.5 mol of O2 produces both CO and CO2, leaving some carbon unreacted. … But,
solid carbon and carbon monoxide can both be completely burned in O2 to produce
CO2. We can therefore us the enthalpy changes of these reactions to calculate the
heat of combustion of carbon”
468
b) We are going to look at 3 ways of employing Hess’s Law… We will find the:
i) overall ΔHrxn when given synthesis reactions and heats of reaction
ii) overall ΔHrxn when given a complete reaction equation and heats of
formation. We will use: ΔH°rxn = Σ n ΔH°f (products) – Σ m ΔH°f (reactants)
iii) overall ΔHrxn when given a complete reaction equation and bond
energies. We will use:
H = (bond energies of broken bonds) - (bond energies of formed bonds)
2) Here’s How I see It….A Summery for the 3 Applications of Hess’s Law
HESS'S LAW
Hess's Law: given reactions
and Heats of Reaction
This is all about rearranging synthesis
equations and adding the changes in enthalpy
Strategy 1: Arrange the given equations to
get the reactants and the products of the
master equation. (Reverse the signs of ∆H
where necessary.)
Hess's Law given standard molar enthalpies
This approach uses the sum of the heats of
formation of the products minus the sum heats of
formation of the reactants:
ΔH°rxn = Σ n ΔH°f (products) – Σ m ΔH°f (reactants)
Strategy 2: Add up the molar enthalpies of
formation (Hf°) of the products (each multiplied
by the stoichiometric value for "n"
Get the correct # of moles of the substances
on each side per the master reaction.
Make sure other substances in the equations
Subtract from that sum, the sum of the molar
enthalpies of formation of the reactants,
multiplied by their value of "m"
cancel when the equations are added.
Hess’s Law given a reaction and bond strengths
Only an approximation for ΔH: The difference between the energy absorbed to
BREAK the bonds of reactants minus the energy released to MAKE the bonds
of the products, is really:
Strategy 3: Account for the moles of each reactant and multiply the moles of
bonds by bond energies. Do the same for the products. Subtract the two sums
using: H = (bond energies of broken bonds) - (bond energies of formed bonds)
Checkout: Hess’s Law: http://www.bozemanscience.com/ap-chem-053-net-energy-change
469
Before moving on:
What effect does reversing a reaction have on the value of ∆H?
* You keep the numeric value the same, but reverse its sign.
What effect does multiplying the coefficients of the reaction equation
by “2” have on the value of ∆H?
* It doubles the value of the heat of reaction.
3) Strategy 1: Hess’s Law: given reaction equations and given heat of reaction (ΔH) values
e.g. Calculate the heat of reaction (change in enthalpy, ∆H) of the synthesis of methane
from solid carbon and hydrogen gas.
C(s) + 2 H2(g)  CH4(g)
Given:
E1
C(s) + O2(g)  CO2(g)
ΔHᵒ1 = -393.5 kJ/mol rxn
E2
H2(g) + ½ O2(g)  H2O(g)
ΔHᵒ2 = -285.8 kJ/mol rxn
E3
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
ΔHᵒ3 = -890.3 kJ/mol rxn
****************************************************************
We see methane is a reactant in E3 ... so reverse it to make it a product (to match the above master rxn)
CO2 + 2 H2O  CH4 + 2 O2
ΔHᵒ3 = +890.3 kJ/mol rxn
Next, double E2 ... this will give you 2 mol of hydrogen gas ...just as in the above master rxn
2 H2 + O2  2 H2O
ΔHᵒ2 = -571.6 kJ/mol rxn
Keep E1 intact, for it give you C as a reactant... NOW you have:
E1
C + O2  CO2
ΔHᵒ1 = -393.5 kJ/mol
E2
2 H2 + O2  2 H2O
ΔHᵒ2 = -571.6 kJ/mol
E3
CO2 + 2 H2O  CH4 + 2 O2
ΔHᵒ3 = +890.3 kJ/mol
Cross out common species on the reactant and product side & sum the enthalpies…
E1
C + O2  CO2
ΔHᵒ1 = -393.5 kJ/mol
E2
2 H2 + O2  2 H2O
ΔHᵒ2 = -571.6 kJ/mol
E3
CO2 + 2 H2O  CH4 + 2 O2
ΔHᵒ3 = +890.3 kJ/mol
C + 2 H2  CH4
ΔH = -74.8 kJ
470
TRY THIS:
Calculate the change in enthalpy for the formation of CS2(s) from Cs and S(s)
C + 2 S  CS2
ans: ∆Hf = 116.8 kJ
Notice you’re given an overall “target” reaction
Notice you’re given a series of other equations
Notice you’re given heats of reaction …hint: use strategy 1 …rearrange and add enthalpies
C(s) + O2(g)  CO2(g)
S(s) + O2(g)  SO2(g)
CS2(s) + 3 O2(g)  CO2(g)+ 2SO2(g)
E1
E2
E3
ΔHᵒ1 = -393.5 kJ/mol
ΔHᵒ2 = -296.8 kJ/mol
ΔHᵒ3 = -1103.9 kJ/mol
*remove …. Re-write the correct reaction equations and each one’s change in enthalpy …before checking ….
* E1
C + O2  CO2
ΔHᵒ1 = -393.5 kJ/mol rxn
* E2
2S + 2O2  2 SO2
ΔHᵒ2 = -593.6 kJ/mol rxn
* E3
CO2 + 2SO2  CS2 + 3 O2
ΔHᵒ3 = +1103.9 kJ/mol rxn
TRY THIS:
Calculate ∆H for the synthesis reaction: C(s) + ½ O2(g) → CO (g) ans: ∆Hf = -110.5 kJ
Notice you’re given an overall “target” reaction
Notice you’re given a series of other equations
Notice you’re given heats of reaction …hint: use strategy 1 …rearrange and add enthalpies
E1
E2
C(s) + O2(g)  CO2(g)
CO(g) + ½ O2(g)  CO2(g)
ΔHᵒ1 = -393.5 kJ/mol
ΔHᵒ2 = -283.0 kJ/mol
*remove …. Re-write the correct reaction equations and each one’s change in enthalpy …before checking ….
E1
C(s) + O2(g)  CO2(g)
ΔHᵒ1 = -393.5 kJ/mol
E2
CO2(g)  ½ O2(g) + CO(g)
ΔHᵒ2 = +283.0 kJ/mol
C(s) + ½ O2(g) → CO (g)
Note that only ½ mol of O2(g) is cancelled
471
TRY THIS:
Calculate the change in enthalpy for the reaction: C(s) + H2O(g) → CO(g) + H2(g)
This one has a small twist to the process … can you figure out that twist???
ans: ∆Hrxn = +131.3 kJ
Notice you’re given an overall “target” reaction
Notice you’re given a series of other equations
Notice you’re given heats of reaction …hint: use strategy 1 …rearrange and add enthalpies
E1
E2
E3
C(s) + O2(g)  CO2(g)
2 CO(g) + O2(g)  2 CO2(g)
2 H2(g) + O2(g)  2H2O(g)
ΔHᵒ1 = -393.5 kJ/mol
ΔHᵒ2 = -566.0 kJ/mol
ΔHᵒ3 = -483.6 kJ/mol
*remove …. Re-write the correct reaction equations and each one’s change in enthalpy …before checking ….
C(s) + O2(g)  CO2(g)
ΔHᵒ1 = -393.5 kJ/mol
CO2(g) 
CO(s) + ½ O2(g)
ΔHᵒ2 = +283.0 kJ/mol
H2O(g)  H2(g) + ½ O2(g)
ΔHᵒ3 = +241.8 kJ/mol
C(s) + H2O(g) → CO(g) + H2(g)
Twist to process: The 2 boldfaced reactions were reversed and multiplied by -1/2
TRY THIS! Consider the following heats of combustion:
Chang AP Achiever p 92
CH3OH(ℓ) + 3/2 O2(g) → CO2(g) + 2H2O(ℓ)
∆H° = -730 kJ/mol
O2(g) → CO2(g)
∆H° = -390 kJ/mol
½ O2(g) → H2O(ℓ)
∆H° = -290 kJ/mol
C(graphite) +
H2(g) +
Determine the enthalpy of formation of methanol CH3OH(ℓ)
1)
2)
3)
4)
-240 kJ/mol
+50 kJ/mol
-1400 kJ/mol
-680 kJ/mol
CO2(g) + 2 H2O(ℓ) → CH3OH(ℓ) + 3/2 O2(g)
∆H = +730 kJ/mol
C(graphite) + O2(g) → CO2(g)
∆H = -390 kJ/mol
2H2(g) +
O2(g) → 2 H2O(ℓ)
C(graphite) + 2 H2(g) + ½ O2(g) → CH3OH(ℓ)
∆H° = -580 kJ/mol
Hint 1:* You have reactions
and heats of reaction. Have
you considered manipulating
the equations so as to create
the balanced reaction equation
to form methanol, (make
methanol the product)
Hint 2 * As you manipulated
the reaction equations, did you
change the signs of delta H
appropriately?
Hint 3: * Add up the heat of
reaction values ….
ans:* choice 1 -240 kJ/mol
472
And now for something completely different …
E)
1) Hess’s Law given Standard Enthalpies of Formation
The magnitude of any enthalpy change depends on the temperature, pressure and
state (gas, liquid or solid crystalline form) of the reactants and products.

To compare enthalpies of different reactions, we must define a set of conditions,
called a standard state, at which most enthalpies are tabulated. The standard
state of a substance is its pure form at atmospheric pressure (1 atm) and the
temperature of interest, which we usually choose to be 298 K (25°C)

Recall: The standard enthalpy change of a reaction is defined as the enthalpy
change when all reactants and products are in their standard states. We denote a
standard enthalpy change as ∆H°, where the superscript, °, (pronounced: naught)
indicates the standard state.

The standard enthalpy of formation of a compound, (∆H°f) is the change in
enthalpy for the reaction * that forms only 1 mole of the compound from its
elements with all substances in their standard states.

By definition the standard enthalpy of formation of the most stable form of
* any element is zero!!!
i) due to the fact that there is no formation reaction needed when the element is
already in its standard state. e.g.) the ∆H°f for graphite, H2(g), O2(g) and all
other elements = 0 kJ/mol

When an element exists in more than one form under standard conditions
(allotropy), the more stable form is used. Thus, O2(g) is assigned a ∆H°f = 0,
whereas O3 and O are not. We can see this again and again in the literature, as
graphite is considered to be the more thermodynamically stable isotope over
diamond (for the forms of C). H2 is used in lieu of H, I2(s) is more stable than I2(g),
or Cl2(g) is more thermodynamically stable than Cl2(𝓁).

Generally speaking, the form of the element per its phase on the periodic table is
the more thermodynamically stable form/phase of the element … due in large part
to the 1 atm criterion.

unit = kJ/mol

Essentially: ΔH°rxn = Σ nΔH°f (products) - Σ mΔH°f (reactants) …seen with entropy
where ∑ is “sum of” and n and m are the relevant stoichiometric values
473
c) So, let’s kick it back for a minute and look at how the definition affects various
issues. In thermodynamics,
All formations are syntheses, but not all syntheses are formations
C(graphite) + ½ O2(g) + 2 H2(g) → CH3OH(g)
represents a formation reaction …
(All the criteria are met …. the reactants are all
elements and in the most thermodynamically stable
form/phase & only 1 mol of product is made.)
SO3(g) + H2O(𝓁) → H2SO4(𝓁)
does NOT represent a formation reaction
(the reactants are not elements)
3/2 O2(g) → O3(g)
does NOT represent a formation reaction
(Ozone is not the most stable form of the element
oxygen, thus in this case ∆H°f ≠ 0 kJ/mol) The
REVERSE rxn would represent a formation reaction.
Al(s) + 3/2 I2(s)  AlI3(s)
represents a formation reaction…the
reactants are in their most stable form/phase and
only 1 mol of product is produced.
TRY THIS! For which of these reactions at 25°C does the enthalpy change represent a standard enthalpy
of formation? For each that does NOT, what changes are needed to make it an equation whose ∆H is an
enthalpy of formation? (Brown and LeMay 191)
1) 2 Na(s) + ½ O2(g) → Na2O(s)
*This does represent a heat of formation… This corresponds to 2
important factors: only 1 mole of product it produced from its
elements and those elements are in their standard (most stable)
states.
2) 2 K(𝓁) + Cl2(g) → 2 KCl(s)
*This does NOT represent a heat of formation. It is accurate in
that a compound is produced from its elements. However it fails
to meet the definition, because 2 moles (not 1) are produced and
the potassium is not in its standard state (solid). Thus the reaction
would need to be re-written as: K(s) + ½ Cl2(g) → KCl(s)
3) C6H12O6(s) → 6 C(diamond) + 6 H2(g) + 3 O2(g) * This does NOT correspond to a heat of formation. It
is not a synthesis …it is a decomposition. C is not
most stable as diamond at room temp. and 1 atm
pressure. To be made correct the reaction should be
re-written as: 6 C(graphite) + 6 H2(g) + 3 O2(g) → C6H12O6(s)
474
E 2) Strategy 2: Using Hess’s Law Given ONLY Standard Enthalpies of Formation
Calculate the heat of reaction (∆Hrxn) for the complete combustion of propane in oxygen gas,
given by the balanced chemical reaction: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4H2O(𝓁)
Standard Enthalpies of Formation
Substance Hf ° kJ/mol
C3H8(g)
-103.85
CO2(g)
-393.5
-285.8
H2O(𝓁)
Why has O2(g) been omitted from the table of Hf°? * It is an element, and as such its heat of
formation = 0 kJ/mol
Use: ΔH°rxn = Σ nΔH°f (products) - Σ mΔH°f (reactants)
Account for Moles: *1 mol of reactant C3H8(g) = -103.85 kJ
4 mol of product H2O(𝓁) = -1143.2 kJ
3 mol of product CO2(g) = -1180.5 kJ
ΔH°rxn = *(-1180.5 kJ/mol) + (-1143.2 kJ/mol) - (-103.85 kJ/mol)
* -2323.7
- (-103.85) = -2,219.9 kJ/mol
TRY THIS! Calculate the standard enthalpy change for the reaction:
CH2N2(s) + O2(g) → CO(g) + H2O(𝓁) + N2(g)
Notice you’re given an overall “target” reaction
Notice you’re given heats of formation… Use strategy 2
Substance
CH2N2 (s)
CO(g)
H2O(𝓁)
∆Hf ° kJ/mol
+62.4
-110.5
-285.8
* Account for moles: 1 mol reactant CH2N2(s) = +62.4 kJ
1 mol product CO(g)
= -110.5 kJ
1 mol product H2O(𝓁) = - 285.8 kJ
* ΔH°rxn = Σ nΔH°f (products) - Σ mΔH°f (reactants)
(-110.5) + (-285.8) - (62.4)
-396.3 - 62.4 = 458.7 kJ
ans: -458.7 kJ
475
TRY THIS! Pentaborane-9, B5H9 is a colorless, highly reactive liquid that explodes when exposed
to oxygen gas. Using the provided heats of formation, calculate the energy exchange
for this combustion. The equation for the combustion is:
(Chang AP Achiever p. 89)
2 B5H9(𝓁) + 12 O2(g) → 5 B2O3(s) + 9 H2O(𝓁)
Notice you’re given an overall “target” reaction
Notice you’re given heats of formation… Use strategy 2
Substance
B5H9(𝓁)
B2O3(s)
H2O(𝓁)
∆Hf ° kJ/mol
73.2
-1263.6
-285.8
*Account for Moles: 2 mol reactant B5H9(𝓁) = 146.4 kJ
5 mol product B2O3(s) = -6318 kJ
9 mol product H2O(𝓁) = -2572.2 kJ
* ΔH°rxn = Σ n ΔH°f (products) - Σ mΔH°f (reactants)
* ΔH°rxn =
-8890.2 - 146.4 = -9036.6 kJ
ans: -9036.6 kJ are released when 2 moles of pentaborane-9 are combusted.
TRY THIS! Calculate the energy exchange when 25.00 grams of Pentaborane-9, are combusted in oxygen
gas. Using the provided heats of formation, calculate the energy exchange for this
combustion. The equation for the combustion is:
Substance ∆Hf ° kJ/mol
2 B5H9(𝓁) + 12 O2(g) → 5 B2O3(s) + 9 H2O(𝓁)
B5H9(𝓁) 73.2
B2O3(s)
-1263.6
Notice you’re given an overall “target” reaction
-285.8
H2O(𝓁)
Notice you’re given heats of formation… Use strategy 2
BUT then use stoich to apply the moles of compound actually combusted.
* Account for moles: 2 mol B5H9(𝓁) = 146.4 kJ
5 mol B2O3(s) = -6318 kJ
9 mol H2O(𝓁) = -2570.5 kJ
* ΔH°rxn = Σ n ΔH°f (products) - Σ mΔH°f (reactants)
* ΔH°rxn = -8888.5
- 146.4 = -9034.9 kJ which is for 2 moles of pentaborane-9
* Thus:
kJ = 25.00 grams | 1 mol | -9034.9 kJ | = -1789.1 kJ
63.126 g
2 mol
ans: -1789.1 kJ
476
TRY THIS! Nitroglycerin is a powerful explosive that forms four different gases when detonated:
Calculate the enthalpy change that occurs when 10.00 grams of nitroglycerin are detonated.
2 C3H5N3O9(𝓁)  3 N2(g) + 6 CO2(g) +
1
2
O2(g) + 5 H2O(g)
*Account for moles: 2 mol reactant C3H5N3O9(𝓁) = -728 kJ
6 mol product CO2(g)
= -2361 kJ
5 mol product H2O(g)
= -1209 kJ
Substance ∆Hf ° kJ/mol
C3H5N3O9
-364
CO2(g)
-393.5
H2O(g)
-241.8
ans: -62.6 kJ
*Equation: ΔH°rxn = Σ n ΔH°f (products) - Σ mΔH°f (reactants)
* ΔH°rxn =
(- 3570) -
(-728) = -2842 for 2 mols of nitroglycerin..
* kJ = 10.0g | 1mol nitro| -2842 kJ |
227 grams 2 mol nitro
TRY THIS! A Twist: Consider the following reaction of the combustion of propane and the heat of
formation information:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4H2O(𝓁)
∆Hrxn =-2,220 kJ/mol
What is the heat of formation of the unknown hydrocarbon?
*Account for moles: 1 mol reactant C3H8 = ???
3 mol product CO2 = -393.5
4 mol product H2O = -1142
Substance ∆Hf ° kJ/mol
C3H8(g)
???
CO2(g)
-393.5
-285.5
H2O(𝓁)
* Equation: ΔH°rxn = Σ n ΔH°f (products) - Σ mΔH°f (reactants)
* -2,220 kJ = (3)(-393.5 kJ/mol ) + (4)(-285.5 kJ/mol) - x
*-2,220 kJ = (-1180.5 kJ) + (-1142 kJ) - x
*-2,220 kJ = -2322.5 kJ - x
ans: -102 kJ
477
TRY THIS! Consider the following reaction representing the combustion of sulfur dioxide gas:
(Chang AP Achiever p. 94)
a) Calculate ∆Hrxn for this process given the following information:
2 SO2(g) + O2(g) → 2 SO3(g)
*ΔH°rxn = Σ n ΔH°f (products) - Σ mΔH°f (reactants)
* 2 (-395.2) - 2(-296.1) + 0
*-790.4 - (-592.2) = -198.2 kJ
b) Is this an exothermic or an endothermic reaction?
Species ∆Hf °(kJ/mol)
SO2(g)
-296.1
O2(g)
0
SO3(g)
-395.2
Hint 1: * Think Hess’s
Law and Strategy 2
ans: -198.2kJ
ans: *exothermic, due to
a – (negative) ∆H value
c) Calculate the amount of heat liberated in the above process when 2.075 grams of SO3(g)
are produced.
* kJ = 2.075 grams | 1 mol
| -198.2 | = -2.568 kJ
80.063g
2 mol
ans: *-2.568 kJ
d) What volume in liters, of SO3(g) would be produced at 20 °C and 1 atm of pressure in the above
combustion reaction involving the transfer of 50.00 kJ of energy?
*PV = nRT
(1)(V) =n(0.08206)(293K)
*to convert for “n” …. mol = 50.00 kJ | 2 mol |
198.2 kJ
*n = 0.5045 mol of SO3 gas.
*V = (0.5045 mol)(0.08206 L atm mol-1 K-1)(293 K)
*= 12.1 Liters
Hint 1:* Hmmm You’re asked
about the volume of a gas with a
temperature, a pressure ….
sounds like an ideal gas law …
solving for V …. BUT! you need an
“n” or moles … .
Hint 2:* You are given an energy
transfer …and you know 2 mol
produce 198.2 kJ … You could use
this info saying : If 2 moles
produce 198.2 kJ …then how
many moles are produced with
only 50.00 KJ?
ans: 12. 1 Liters
478
F) Strategy 3: Using Hess’s Law given bond energies to approximate ΔH
If we know which bonds are broken and which bonds are made during a chemical reaction, we can
estimate the enthalpy change of the reaction (Hrxn) even if we don't know the enthalpies of
formation (Hf°) for the reactants and products:
H = (bond energies of broken bonds) - (bond energies of formed bonds)
This approach can be very helpful, especially with newly synthesized compounds, for which a complete
data file of chemical and physical characteristics has yet to be produced.
1) Use the following table of average bond energies and the reaction between 1 mol of
chlorine and 1 mol methane, approximate the change in enthalpy for the reaction.
http://www.mikeblaber.org/oldwine/chm1045/notes/Bonding/Strength/Bond09.htm
Average Bond Energies
Bond
(kJ/mol)
C-H
413
C-C
348
C ≡C
839
C=O
799
C-Cl
328
H–H
432
H – Cl
431
O–H
467
O=O
495
Cl – Cl
242
Bonds broken: *1 mol of Cl-Cl bonds, 1 mol of C-H bonds
Bonds formed: *1 mol of H-Cl bonds, 1 mol of C-Cl bonds
Notice that I am taking a shortcut… I accounted for only 1 mol of C
– H bonds as broken. You can change this all up …And if you found
it easier, you could account for 4 mol of C – H bonds broken … but
be sure to account for 3 mol of those C – H) as being formed on the
product side. You’ll get the same approximation of change in
enthalpy.
Equation: H = (bond energies of broken bonds) - (bond energies of formed bonds)
H = [(Cl-Cl) + (C-H)] - [(H-Cl)+(C-Cl)]
= *[242 kJ + 413 kJ] - [431 kJ + 328 kJ]
= -104 kJ
Thus, the reaction is exothermic (because the bonds in the products
are stronger than the bonds in the reactants)
479
2) Approximate the ΔHrxn for the complete combustion of ethane.
Average Bond Energies
Bond
(kJ/mol)
C-H
413
C-C
348
C ≡C
839
C=O
799
C-Cl
328
H–H
432
H – Cl
431
O–H
467
O=O
495
Cl – Cl
242
Bonds broken: *1 mol C-C bonds
6 mol of C-H bonds
7/2 or 3.5 mol O=O bonds
Bonds made:
*4 mol C=O bonds
6 mol of O-H bonds
H = (bond energies of broken bonds) - (bond energies of formed bonds)
= *[348 + (6 * 413) + (3.5 *495)] - [(4*799) + (6*467)]
*4559 (rounded)
-
5998
=-1439 kJ ans.
3) Hydrogenation of double and triple bonds is an important industrial process. Approximate
in kilojoules the standard enthalpy change ΔH for the hydrogenation of ethyne (acetylene)
to ethane:
H−C≡C−H + 2 H2 → C2H6
or rather….
H−C≡C−H +
2 H-H →
H H
| |
H–C–C–H
| |
H H
Bonds Broken: *2 mol H-H bonds
1 mol C≡C bonds
Bonds Made:
*1 mol C – C bonds
4 mol H – H bonds
quation: H = (bond energies of broken bonds) - (bond energies of formed bonds)
= *[(2 * 432) + 839] - [348 + (4 * 413)]
1703
- 2000
-297 kJ ans
480
TRY THIS:
1) Which of the following is a statement of Hess’s Law?
1) If a reaction is carried out in a series of steps, the ΔH for the reaction will equal the sum
of the enthalpy changes for the individual steps.
2) If a reaction is carried out in a series of steps, the ΔH for the reaction will equal the product of the
enthalpy changes for the individual steps
3) The ΔH for a process in the forward direction is equal in magnitude and opposite in sign
to the ΔH for the process in the reverse direction.
4) The ΔH for a process in the forward direction is equal to the ΔH for the process in the reverse
direction
5) The ΔH of a reaction depends on the physical states of the reactants and the products.
2) Which of the following statements is true?
1)
2)
3)
4)
5)
Enthalpy is an intensive property
The enthalpy change for a reaction is independent of the state of the reactants and products
Enthalpy is a state function
H is the value of q measured under conditions of constant volume
The enthalpy change of a reaction is the reciprocal of the ΔH of the reverse reaction.
3) Which of the following is a statement of the first law of thermodynamics?
1)
2)
3)
4)
5)
EK = ½ mv2
A negative ΔH corresponds to an exothermic process
ΔE = Efinal – Einitial
Energy lost by the system must be gained by the surroundings.
Water has one of the higher specific heat values at 4.184 J/g•K
Ans: 1) 1 2) 3 3) 4
481