Chapter 6

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Chapter 6
Energy
Thermodynamics
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Energy is...

Conserved

Made of heat and work.
– Work is a force acting over a distance
– Heat is energy transferred between objects
because of temperature difference.

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A state function which means that the
result is independent of the path, or how
you get from point A to B.
The universe
Is divided into two halves: the system
and the surroundings.
 The system is the part you are
concerned with.
 The surroundings are the rest.

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Unfortunately, it is easier to measure
the effect on the surroundings than the
system directly
The universe
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
Exothermic reactions release energy to
the surroundings.

Endothermic reactions absorb energy
from the surroundings.
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Potential energy
CH 4 + 2O 2  CO 2 + 2H 2 O + Heat
CH 4 + 2O 2
Heat
CO 2 + 2 H 2 O
N 2 + O 2 + heat  2NO
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Potential energy
2NO
Heat
N2 + O2
Direction
Every energy measurement has three
parts.
1. A unit ( Joules or calories).
2. A number how many.
3. A sign to tell direction.
 negative – exothermic

–
positive- endothermic
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–
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System to the surrounding
Surroundings to the system
Surroundings
System
Energy
DE <0
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Surroundings
System
Energy
DE >0
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Same rules for heat and work
Heat given off is negative.
 Heat absorbed is positive.
 Work done by system on surroundings
is negative.
 Work done on system by surroundings
is positive.
 Thermodynamics- The study of energy
and the changes it undergoes.
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First Law of Thermodynamics
The energy of the universe is constant.
 Law of conservation of energy.
 q = heat
 w = work
 DE = q + w
 Take the systems point of view to
decide signs.
 Punch Line: DE = q (At constant P)
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What is work?
Work is a force acting over a distance.
 w= F x Dd
 P = F/ area
 d = V/area
 w= (P x area) x D (V/area)= PDV
 Work can be calculated by multiplying
pressure by the change in volume at
constant pressure.
 units of liter - atm L-atm
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Work needs a sign
If the volume of a gas increases, the
system has done work on the
surroundings.
 work is negative
 w = - PDV
 Expanding work is negative.
 Contracting, surroundings do work on
the system w is positive.
 1 L atm = 101.3 J
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Example #1

What amount of work is done when 15
L of gas is expanded to 25 L at 2.4 atm
pressure?
– volume increase – work is being done by the
system on the surroundings (-)
w = - PDV
w = -(2.4atm)(25L-15L)
w = -24L atm
-24 L atm x (101.3J/Latm) = -2431 J = 2.4kJ
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Example #2
If 2.36 J of heat are absorbed by the gas
above. what is the change in energy?
DE = q + w
DE = 2.36 J + -2431 J
DE = -2429 J
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Enthalpy (H)

H = E + PV (that’s the definition)
– at constant pressure.

DH = DE + PDV
– the heat at constant pressure qp can be
calculated from

DE = qp + w = qp – PDV (w=-PDV)

qp = DE + P DV = DH
– PUNCH LINE qp = DH
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Calorimetry
Measuring heat using a calorimeter.
 Two kinds – the first kind being a

– Constant pressure calorimeter (called a coffee
cup calorimeter)
heat capacity for a material, C is
calculated (heat required to change a
substances temperature)
 C= heat absorbed/ DT = DH/ DT
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Calorimetry
Specific heat capacity = C/mass
 Molar heat capacity = C/moles

heat = specific heat x mass x DT
 heat = molar heat x moles x DT
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Make the units work and you’ve done
the problem right.
Calorimetry
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
A coffee cup calorimeter measures DH.
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The specific heat of water is 1 cal/gºC
or 4.184 J/gC
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Heat of reaction= DH = SH x mass x DT
Examples

The specific heat of graphite is 0.71
J/gºC. Calculate the energy needed to
raise the temperature of 75 kg of
graphite from 294 K to 348 K.
DH = SH x mass x DT
 DH = 0.71J/gC (75000g)(54C)
 DH = 2876 kJ
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Examples

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A 46.2 g sample of copper is heated to 95.4ºC
and then placed in a calorimeter containing
75.0 g of water at 19.6ºC. The final
temperature of both the water and the copper
is 21.8ºC. What is the specific heat of copper?
DHsurr = -DHsys
75.0g (4.184J/g C)(2.2C) = -[46.2g(SH)(-73.6C)]
690J = 3400g C (SH)
.203 J/g C = SH
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Calorimetry
Constant volume calorimeter is called a
bomb calorimeter.
 Material is put in a container with pure
oxygen. Wires are used to start the
combustion. The container is put into a
container of water.
 The heat capacity of the calorimeter is
known and tested.
 Since DV = 0, PDV = 0, DE = q
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Bomb Calorimeter
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
thermometer

stirrer

full of water
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ignition wire
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Steel bomb
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sample
Bomb Calorimeter
A bomb calorimeter works in a similar
manner as the coffee cup calorimeter,
but there is one significant difference.
 In a coffee cup calorimeter, the reaction
takes place in the water. In a bomb
calorimeter, the reaction takes place in a
sealed metal can, which is then placed
in the water (contained in an insulated
container).
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Bomb Calorimeter
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Analysis of the heat flow is a bit more
complex than it was for the coffee cup
calorimeter, because the heat flow absorbed
by the metal parts of the calorimeter (the
“bomb” part) must be taken into account:

qrxn = - (qwater + qbomb)
– where qwater = 4.18 J/(g·°C) x mwater x ΔT
– The heat flow of the bomb is:
– qbomb = Ccal x ΔT
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One More Twist – Phase Changes
Molar Heat of Fusion – Energy required
to change 1 mol from a solid to a liquid
(visa versa)
6kJ/mol for water
Molar Heat of Vaporization – Energy
required to change 1 mol from a liquid to
a gas (vise versa)
40.7kJ/mol for water
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Phase Changes
If 20g of ice at 0C is placed in a calorimeter
containing 100ml of water at 60C, what is the
final temperature of the system?
qphasechange + qheating = -qwater
20g (1mol/18.02g) = 1.10mol
1.10mol(6000J/mol) + 20g(4.184J/gC)(Tf-0) = -100g(4.184J/gC)(Tf-60)
6600 + 83.68Tf = -418.4Tf + 25104
502.88 Tf = 18504
Tf = 36.8 C
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Problem Solving Strategies

The 1st step is to ascertain whether the
process is constant pressure (open to the
atmosphere) or constant volume.
– If it’s constant pressure, use ΔH = −ΔEsurr;
– for constant volume it’s ΔErxn = −ΔEsurr.
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In many problems
– ΔEsurr = (mass)(specific heat)ΔT.
– To use this equation, you must determine the part of the
overall system that is changing temperature.
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Problem Solving Strategies


For example, if we add 1.0 g of Mg to 100.0
mL of 1.0 M HCl, it is not the Mg that is
changing temperature, but rather the 100.0
mL of acidic solution in which the Mg is
reacting.
In this example, then,
– ΔEsurr = (100.0 g)(4.184 J/[g oC])ΔT,
– where we have used the usual assumptions stated above. We
would not use the mass of Mg (1.0 g) and the specific heat of Mg
(1.02 J/[g oC]) in the ΔEsurr equation because it’s not the Mg that
is changing temperature.
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Problem Solving Strategies
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Always ask yourself this question: “What is
actually changing temperature in this
process?”
Wherever the thermometer goes, that’s what
is changing temperature.
*Note: in some problems – the heat capacity
is given, so you’ll use ΔEsurr = CΔT, where C
is heat capacity)
Problem Solving Strategies
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
In the equations ΔH = −ΔEsurr or ΔErxn = −ΔEsurr,
the units on both sides are joules (J)!

Therefore, if the problem asks you to find an answer
in J or kJ per mole (or g), you need to find J first and
then divide by the appropriate moles (or g)

Likewise, if the problem gives you ΔH (or ΔErxn)
and asks you to find one of the terms in ΔEsurr, you
first need to make sure that the units on ΔH (or
ΔErxn) are J and not J per mole or J per gram
Hess’s Law

Enthalpy is a state function.
We can add equations to to come up
with the desired final product, and add
the DH
 Two rules

– If the reaction is reversed the sign of DH is
changed
– If the reaction is multiplied, so is DH
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H (kJ)
O2 NO2
-112 kJ
180 kJ
N2 2O2
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NO2
68 kJ
Standard Enthalpy
The enthalpy change for a reaction at
standard conditions (25ºC, 1 atm , 1 M
solutions)
 Symbol DHº
 When using Hess’s Law, work by
adding the equations up to make it look
like the answer.
 The other parts will cancel out.
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Example
Given
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C 2 H 2 (g) + O 2 (g)  2CO 2 (g) + H 2 O( l)
2
DHº= -1300. kJ
C(s) + O 2 (g)  CO 2 (g)
DHº= -394 kJ
1
H 2 (g) + O 2 (g)  H 2 O(l)
2
DHº= -286 kJ

calculate DHº for this reaction
2C(s) + H 2 (g)  C 2 H 2 (g)
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Example
Given
O 2 (g) + H 2 (g)  2OH(g) DHº= +77.9kJ
O 2 (g)  2O(g) DHº= +495 kJ
H 2 (g)  2H(g) DHº= +435.9kJ
Calculate DHº for this reaction
O(g) + H(g)  OH(g)
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Standard Enthalpies of Formation
Hess’s Law is much more useful if you
know lots of reactions.
 Made a table of standard heats of
formation. The amount of heat needed
to for 1 mole of a compound from its
elements in their standard states.
 Standard states are 1 atm, 1M and 25ºC
 For an element it is 0
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Standard Enthalpies of Formation
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
Need to be able to write the equations.

What is the equation for the formation
of NO2 ?

½N2 (g) + O2 (g)  NO2 (g)

Have to make one mole to meet the
definition.
Since we can manipulate the
equations

We can use heats of formation to figure
out the heat of reaction.

Lets do it with this equation:
ΔH = ∑ nΔHf (products) – ∑ nΔHf (reactants)
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Example

Use the table of standard enthalpies of formation at
25°C to calculate ΔH for the reaction
4NH3(g) + 5O2(g) → 6H2O(g) + 4NO(g)
ΔH = ∑ nΔHf (products) – ∑ nΔHf (reactants)
= [6 ΔHf (H2O) + 4 ΔHf (NO)] – [4 ΔHf (NH3) + 5 ΔHf (O2)]
= 6(–241.8) kJ mol–1 + 4(90.3) kJ mol–1 – 4(–46.1 kJ mol–1) – 5 × 0
= –1450.8 kJ mol–1 + 361.2 kJ mol–1 + 184.4 kJ mol–1
= –905.2 kJ mol–1
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