Ch. 6
Thermochemistry
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!!
2014 AP Chem Exam
Exam 2 Analysis
• Averages
• Grab a different colored pen/pencil.
• Grab out a sheet of paper for notes.
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 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
Ch. 6 Section 1 Practice
Problems
• Beginning on page 266
• # 9, 19, 21, 22, 23, 25, and 27
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)
Enthalpy Continued…
• The change in enthalpy of a system has no easily interpreted
meaning except at constant pressure, where ΔH = 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 237 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)?
Independent Practice Continued…
• Page 267 #35, 37, 51, 52, 53, 55, 56
Hand-warmer Lab
• Green Lab Book – Investigation 12 page 97
• Central Challenge
• Simulation/Questions
• With lab group
• Materials
• Procedure
• Works best to use 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:
• **Assigned Solids to Investigate**
• Will share class data.
Post – Lab Assessment
• Safety!
• Complete Data Collection and Computation Questions
• Argumentation and Documentation
• Complete numbers 1, 2, 3, 4b, and 5 (omit 4a)