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Hess’s Law and Standard
Enthalpies of Formation
Thermochemistry
Pages 242 - 252
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Hess’s Law
 Enthalpy
changes are state functions.
does not matter if ΔH for a reaction is calculated
in one step or a series of steps = Hess’s Law
 It
using values of ΔH of known reactions, we can
use Hess’s law to solve for enthalpies of reactions
whose values we do not know.
 By
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Hess’s Laws Problem-Solving

Manipulate equations so that they add up to the desired
equation.

There are 3 ways we can manipulate equations:
 1. We can reverse the entire equation (switch
reactants and products).
 2. We can multiply the entire equation by a factor
(such as 3, 2, 1/2 )
 3. We can do both #1 and #2.

**When you manipulate an equation you must
manipulate the ΔH value in the EXACT same
way!!**
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 Work
backward from the required reaction, using
the reactants and products to decide how to
manipulate the other given reactions at your
disposal.
 Reverse
any reactions as needed to give the
required reactants and products
 Multiply
reactions to give the correct numbers of
reactants and products
 (Trial
and error- allow the final reaction to guide
you)
 **Start
by finding a substance that only appears
once in the reactants.
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Hess’s Law Example
Calculate ΔH for
N2 (g) + O2 (g)
2NO(g)
Given:
a.
N2(g) + 2O2 (g)
2NO2(g)
ΔH = 66.4 kJ/mol
b.
2NO(g) + O2(g)
2NO2(g)
ΔH = -114.1kJ/mol
A: ΔH = 180.5 kJ/mol
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Hess’s Law Practice Together
Given
a.
2H2 (g) + C(s)
CH4(g)
ΔH = -74.81 kJ/mol
b.
2H2(g) + O2(g)
2H2O (l)
ΔH = -571.66 kJ/mol
c.
C(s) + O2(g)
CO2(g)
ΔH = -393.52 kJ/mol
Calculate ΔH for
CH4(g) + 2O2(g)
A: ΔH = -890.37 kJ/mol
CO2(g) + 2H2O(l)
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
Hess’s Law Practice Problem
Given the following thermochemical data, calculate the ΔH° for:
Ca(s) + 2H2O(l)
Ca(OH)2(s) + H2(g)
a.
H2(g) + ½ O2(g)
H2O(l)
ΔH° = -285kJ
b.
CaO(s) + H2O(l)
Ca(OH)2(s)
ΔH° = -64 kJ
c.
Ca(s) + ½ O2(g)
CaO(s)
ΔH° = -635 kJ
A: -414 kJ
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Hess’s Law Practice Problem

Calculate ΔH for
4NH3 (g) + 3O2(g)
2N2(g) + 6H2O(l)
Given the following reactions and ΔH values,
a. 2N2O(g)
b. 2NH2(g) + 3N2O(g)
A: ΔH= -1532 kJ
O2(g) + 2N2(g)
4N2(g) + 3H2O(l)
ΔH= -164 kJ
ΔH= -1012 kJ
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Hess’s Law Practice Problem
For the reaction:
H2O (l)
H2O(g)
ΔH = +44kJ
How much heat is evolved when 9.0 grams of water
vapor is condensed to liquid water?
A: 22kJ evolved or -22kJ
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Standard Enthalpy of Formation
ΔH°f
 The
change in enthalpy that accompanies the
formation of one mole of a compound from its
elements with all substances in their standard states.
 ΔH°f
 The
degree symbol (°) on a thermodynamic function
indicates the process has been carried out under
standard conditions.
 Standard state is not the same as standard
temperature and pressure (STP)
+ ΔH°f Key Ideas
1.
2.
3.
ΔH°f is always given per mole of compound formed.
ΔH°f involves formation of a compound from its elements with
the substances in their standard states.
1.
For an element:
1.
It is the form which the element exists in at 25°C and 1
atm.
2.
For a compound:
1.
For a gas it is a pressure of exactly 1 atm (IUPAC – 1bar )
2.
For a substance in solution, it is a concentration of exactly
1M.
3.
For a pure solid or liquid, is the pure solid or liquid.
ΔH°f for an element in its standard state, such as Ba(s) or
N2(g), equals 0.
+ The enthalpy change for a given reaction can be calculated by
subtracting the enthalpies of formation of the reactants from the
enthalpies of formation of the products.

When a reaction is reversed, the magnitude of ΔH remains the
same, but its sign changes.

When the balanced equation for a reaction is multiplied by an
integer, the value of ΔH for that reaction must be multiplied by
the same integer.

Elements in their standard states are not included in the ΔHreaction
calculations. That is, ΔH°f for an element in its standard state is
zero.
ΔH°reaction = Σnp ΔH°f (products) - Σnr ΔH°f (reactants)
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Working with Standard Enthalpy of
Formation…
 Consulting
your textbook Appendix (4) and
knowledge on standard states, list the
standard enthalpy of formation for each of the
following substances.
 Al2O3(s)
 Ti(s)
 P4(g)
 SO42-(aq)
 F2 (g)
+ Using the standard heats of formation,
calculate ΔH for the following reactions.
HCl (g)
H+(aq) + Cl-(aq)
2NO2(g)
N2O4 (g)
C2H2(g) + H2(g)
C2H4(g)
2NaOH(s) + CO2(g)
Na2CO3(s) + H2O(g)
+ Standard Enthalpy of Formation
Practice
 The
heat released when HNO3 reacts with NaOH is
56 kJ/mole of water produced. How much energy is
released when 400.0mL of 0.200M HNO3 is mixed
with 500.0mL of 0.150 M NaOH?
 The
enthalpy of neutralization for the reaction of a
strong acid with a strong base is -56kJ/mol of water
produced. How much energy will be released
when 200.0mL of 0.400M HCl is mixed with
150.0mL of 0.500M NaOH? How does this compare
with the answer in the first part? Why?
+ Bond Energies – Calculating Heat of
Reaction
 All
substances must be in the gaseous form.
 Breaking
 Forming
 ΔH
bonds requires energy (Endothermic)
bonds releases energy (Exothermic)
= ΣEbonds broken - ΣEbonds formed
+ Bond Energies and Heat of
Reaction Example
Calculate the value of ΔH for the reaction below
using the average bond energies.
H2(g) + F2(g)
2HF(g)
Bond
Average Bond Energy (kJ/mol)
H-H
432
H-F
565
F-F
160
+ Bond Energies and Heat of
Reaction Practice
Calculate the value of ΔH for the reaction below
using the average bond energies.
H2(g) + ½ O2(g)
H2O(g)
Bond
Average Bond Energy (kJ/mol)
H-H
432
O=O
496
H-O
463
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Individual Practice
 57, 58, 59, 60, 61, 63, 65, 67, 71
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Reminders
 Hand-warmer
Lab Report Due ___________!
 Practice
and Review Ch. 6 Concepts and
Problems
 Begin
reading Ch. 7