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Ch. 6
Thermochemistry
Unit 2 Exam Analysis
 20 points total
10 points Multiple Choice (MC)
10 points Free Response Questions (FRQ)
 -Analyze areas of strength and those in need of assistance.
 -Utilize Learning Objectives and Topic Sheets (tan colored)
 -Reflect on
*labs
*practice packets
Unit 2 Exam Analysis
 30 minutes individual review.
 10 minutes peer review….for half credit (based on reflection)!
 Learning is a PROCESS!
 For credit you must correct your answer and provide a brief
reflection of why your answer choice was incorrect and how
you corrected it. (3 to 4 sentences per question)
Question
Number
(MC or
FRQ)
Incorrect Corrected REFLECTION of LEARNING
Answer
Answer
Course Support
 AP Chemistry Prep-book suggestions (4)
 Suggested practice problems are a minimum.
 Form study groups!
 Support before and after school
 Readings done prior to class (class notes are a
refresher prior to application).
 YOU CAN DO THIS!!
Ch. 6 Thermochemistry Topics
Ch. 6 Thermochemistry
Internal energy, heat and work-keep track of signs and units…1 kJ is 1000 J

Exothermic and endothermic reactions. What is happening at molecular level and how
is this reflected with H and the signs?

Energy profiles

Hess’s Law can be used to calc heat of reaction

Heat of formation (Hf) can be used to calc heat of reaction (in prdts-reactants eqn)

Heat of formation-all elements in natural state are zero. Why?

Know what (Hf) values mean and what they are used for.

Bond energies can be used to calc heat of reaction

Know all heats of reaction symbols and what they mean

Hand warmer lab report. Know how to do the calcs both with a calorimeter constant
(qrxn = -(q soln + q cal) and where heat lost to calorimeter is 0…..(q soln = - q rxn) Watch signs
and units!. A change in temperature is not a Joule!

Heat capacity-describe it. What does it mean to have high or low heat capacity?

Do not forget PV=nRT

Stoichiometry and thermochem. Know what a Hrxn has to do with reactants/prdts and
be able to manipulate simple stoich with heat.
The Nature of Energy
Key Terms
 Energy – the capacity to do work or produce
heat
 Law of conservation of energy – energy can be
converted from one form to another but can
be neither created nor destroyed. (The total
energy content of the universe is constant.)
 Potential energy – energy due to position or
composition
 Lifting a pencil in the air
 Propane burned (attractive and repulsive forces between charged
particles)
 Kinetic energy – energy due to motion
 KE = ½ mv2
 Energy conversion
 Heat – involves transfer of energy between two objects due to a
temperature difference (different from Temperature)
 Work – force acting over a distance
Force x distance
 State function (property): a property that is independent of
pathway.
It does not matter how you get there, the difference in the
value is the same. Home to school. It does not matter if you
walk, bike, or drive the actual distance between the two
locations is the same.
Energy is a state function – work and heat are NOT.
Thus, the amount of work and heat ARE dependent on the
pathway
The universe is composed of the system and
the surroundings.
 System
 Part of the universe we are focusing on.
 Surroundings
 Everything else in the universe, besides the system.
 Exothermic
 Energy(as heat) flows out of the system (feels warmer).
 “heat” in product-side of chemical equation
 Endothermic
 Energy (as heat) flows into the system (feels colder).
 “heat” in the reactant-side of chemical reaction
Thermodynamics
 Thermodynamics- the study of energy and its interconversions
 First Law of Thermodynamics
 The energy of the universe is constant.
 Internal energy – sum of the kinetic and potential energy of all
the “particles” in the system.
 Change in internal energy can be by flow of work, heat, or
both.
ΔE = q + w
 Thermodynamic quantities always have a magnitude (number)
and a sign.
The SI unit for Energy is the Joule (J)
 The sign is from the system’s perspective.
 Energy flows into a system (endothermic) + q system is increasing in heat/energy.
 Energy flows out of a system (exothermic) –q system is decreasing in heat/energy
 1000 J = 1 kJ
Units of Work
 When the system expands, it is doing positive work on the
surroundings, therefore it is doing negative work on the system.
 When the system contracts, the surroundings have done work
on the system, therefore there is positive work done on the
system.
 Thus, from the point of view of the system,
w = - PΔV
1 L atm = 101.3 J
 For an ideal gas, work can occur only when its
volume changes. Thus, if a gas is heated at
constant volume, the pressure increases but no
work occurs.
Work Example
 Calculate the work associated with the contraction of a gas from 75 L to
30. L (work is done “on the system”) at a constant external pressure of 6.0
atm in:
*Hint: system compression is positive work, and system expansion is
negative work.
L atm
J
Work Practice
 Consider a mixture of air and gasoline vapor in a cylinder with a piston. The
original volume is 40. cm3. If the combustion of this mixture releases 950. J of
energy, to what volume will the gases expand against a constant pressure
of 650. torr if all the energy of combustion is converted into work to push
back the piston?
* Hint: watch units!
First Law of Thermodynamics Example
 Calculate the change in energy of the system if 38.9 J of work is done BY
the system with an associated heat loss of 16.2 J.
1st Law Practice
 Calculate ΔE for each of the following.
 q = -47 kJ, w = +88 kJ
 q = +82 kJ, w = -47 kJ
 Write it out!!
In each of the situations above describe if heat is lost or gained by
the system. Also, describe if work is being done by or on the system.
 ΔE = “-” means the system loses energy.
 ΔE = “+” means the system gains energy.
Combining Heat and Work
Example
 A piston is compressed from a volume of 8.3 L to 2.8 L against a constant
pressure of 1.9 atm. In the process, there is a heat gain by the system of 350
J. Calculate the change in energy of the system.
Heat and Work Practice
 A piston expands against 1.00 atm of pressure,
from 11.2 L to 29.1 L. This is done without any
transfer of heat.
 Calculate the change in energy of the system.
 Calculate the change in energy for the above change if, in
addition, the system absorbs 1,037 J of heat from the surroundings.
 A: -1.81 kJ ; -773J
Enthalpy
 A property of a system equal to E + PV, where E is the internal
energy of the system, P is the pressure of the system, and V is
the volume of the system. At constant pressure the change in
enthalpy equals the energy flow as heat.
H = E + PV
or
ΔH = ΔE + PΔV
 Enthalpy is a state function. A change in
enthalpy does not depend on the pathway
between two states.
 ΔH = qp
 Change in enthalpy = Heat of reaction (used interchangeably)
POGIL – Heats of Formation
 Group 1:
 Group 2:
 Group 3:
 Group 4:
 Group 5:
Building meaning/understanding TOGETHER!
You will need the following
- textbook
-calculator
-team work mindset!
Enthalpy Continued…
 The change in enthalpy of a system has no easily interpreted
meaning except at constant pressure, where ΔH = q (heat).
 For a chemical reaction, the enthalpy change is given by the
equation:
ΔH = Hproducts – Hreactants
(Think Δ ; Final – Initial)
 At constant pressure,
 exothermic means ΔH is negative (heat given off by system);
 endothermic means ΔH is positive (heat absorbed by system)
Enthalpy Example
When solid sodium hydroxide pellets are added to water, the
following reaction occurs:
NaOH(s)
NaOH(aq)
For this reason at constant pressure, ΔH = -43kJ/mol. Answer
the following questions regarding the addition of 14 g of
NaOH to water:
 Does the beaker get warmer or colder?
 Is the reaction exo- or endothermic?
 What is the enthalpy change for the dissolution?
Enthalpy Practice
The overall reaction in a commercial heat pack can be
represented as
4Fe(s) + 3O2(g)
2Fe2O3(s)
ΔH= -1652kJ
 How much heat is released when 4.00 mol of iron (Fe) is reacted
with excess O2?
 How much heat is released when 1.00 mol of Fe2O3 is
produced?
 How much heat is released when 1.00 g iron is reacted with
excess O2?
 How much heat is released when 10.0 g Fe and 2.00 g O2 are
reacted?
Calorimetry
 The science of measuring heat.
 Based on observing (experimental) the temperature change when
a body absorbs or discharges energy as heat (q).
 At constant pressure, q= ΔH
 At constant volume, q = ΔE (work is zero)
 Heat is either gained or lost. The amount of heat exchange in a
reaction is dependent upon:
1.
The net temperature change during the reaction.
2.
The amount of substance. The more you have, the more heat can be
exchanged (direct relationship).
3.
The heat capacity (C) of a substance.
Calorimetry
 Some substances can absorb more heat than others for a given
temperature change.
 Three ways of expressing heat capacity:
1. Heat capacity = J/°C
2. Specific heat capacity = heat capacity per gram of substance
(J/°C g) or (J/K g)
3. Molar heat capacity = heat capacity per mole of substance
(J/°C mol) or (J/K mol)
Conceptual Understanding
See Table 6.1 on page 254 in textbook.
 Based on the values for specific heat capacity, which conducts heat
better, water or aluminum? Why is this important in cooking?
 Share out responses.
Calorimetry under constant
pressure…
 If two reactants at the same temperature are mixed and the resulting
solution gets warmer, this means the reaction taking place is
exothermic. An endothermic reaction cools the solution.
 Energy released/absorbed by the reaction
 = energy absorbed/released by the solution
 = specific heat capacity x mass of solution x change in temperature
= s x m x ΔT
 Heat of reaction is an extensive property.
 What does this mean?
 Temperature is an intensive property.
 What does this mean?
Calorimetry
 Mentally keep track of the direction of the energy flow and
assign the correct sign at the end of the problem.
 Exothermic
 Released
 Evolved
 Given off
 Temperature increase
 Endothermic
 Absorbed
 Gained
 Temperature decrease
 Enthalpies often expressed in terms of moles of reacting
substances. kJ/mol or J/mol (determine moles of sample and
divide by energy)
Calorimetry Example
 If 596 J of heat are added to 29.6 g of water at 22.9°C in
a coffee cup calorimeter, what will be the final
temperature of the water?
 A: 27.7°C
Calorimetry Practice
 The specific heat of aluminum is 0.89 J/g°C. How much
energy is required to raise the temperature of a 15.0 g
aluminum can 18°C ?
 A:240J
Constant Volume Calorimetry
 “Bomb calorimeter”
 ΔV = 0 thus, ΔE = qv (zero work)
 Energy released (gained) by the reaction
 = temperature increase(decrease) x energy required to change temperature by 1
degree C
 =ΔT x heat capacity of bomb calorimeter
 Enthalpies often expressed in terms of moles of reacting substances. kJ/mol or
J/mol (determine moles in sample and divide by energy)
Calorimetry Example
 The heat of combustion of glucose, C6H12O6, is 2800 kJ/mol. A sample of
glucose weighing 5.00 g was burned with excess oxygen in a bomb
calorimeter. The temperature of the bomb rose 2.4°C . What is the heat
capacity of the calorimeter?
 A 4.40 g sample of propane (C3H8) was then burned with excess oxygen in
the same bomb calorimeter. The temperature of the bomb increased
6.85°C. Calculate ΔEcombustion of propane.
 A: 32.4 kJ/°C ; -2200 kJ/mol
Calorimetry Practice
The heat capacity of a bomb calorimeter was determined by burning
6.79 g of methane (heat of combustion = -802 kJ/mol) in the bomb. The
temperature changed by 10.8°C.
 What is the heat capacity of the bomb?
 A 12.6 g sample of acetylene, C2H6, produced a temperature
increase of 16.9°C in the same calorimeter. What is the heat of
combustion of acetylene (kJ/mol)?
Hand-warmer Lab
 Green Lab Book – Investigation 12 page 97
 Central Challenge
 Simulation/Questions
 With lab group
 Materials
 Procedure
 Works best to use 75-100mL (NOT 45.0mL) of water to 5.0g of solid
 Skip Calorimeter Constant (#1 in Data Collection) 7%
Pre-Lab Assignment
 Purpose
 Procedure
 Data Tables
Lab Partners






Group 1:
Group 2:
Group 3:
Group 4:
Group 5:
Group 6:
Building meaning/understanding TOGETHER!
 **Assigned Solids to Investigate**
 Will share class data.
Post – Lab Assessment
 Safety!
 Complete Data Collection and Computation Questions
(omit 1a-d)
 Argumentation and Documentation
 Complete numbers 1, 2, 3, 4b, and 5 (omit 4a)
Hess’s Law and Standard
Enthalpies of Formation
Thermochemistry
Hess’s Law
 Enthalpy changes are state functions.
 It does not matter if ΔH for a reaction is calculated in
one step or a series of steps = Hess’s Law
 By 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.
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!!**
 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.
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
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)
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)
b.
CaO(s) + H2O(l)
Ca(OH)2(s) ΔH° = -64 kJ
c.
Ca(s) + ½ O2(g)
CaO(s)
A: -414 kJ
ΔH° = -285kJ
ΔH° = -635 kJ
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)
O2(g) + 2N2(g)
b. 2NH3(g) + 3N2O(g)
A: ΔH= -1532 kJ
ΔH= -164 kJ
4N2(g) + 3H2O(l) ΔH= -1012 kJ
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
Standard Enthalpy of Formation
ΔH°f - POGIL (learning experience)
 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.
ΔH°f is always given per mole of compound formed.
2.
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.
3.
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)
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 bonds requires energy (Endothermic)
 Forming bonds releases energy (Exothermic)
 ΔH = 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
Reminders
Hand-warmer Lab Report Due _________!!
Practice and Review Ch. 6 Concepts and
Problems
Begin reading Ch. 7