Transcript Ch6.Thermochem - Mr. Fischer.com

Thermochemistry
Chapter 6
1
Energy and change
The study of the energy involved in a change is
THERMODYNAMICS
 In thermodynamics, the universe is divided into

System (what you are studying)
and
Surroundings (the rest of the universe!)
2
Systems

A system may be OPEN

both matter and energy can be
exchanged with the
surroundings
3
Systems

A system may be CLOSED

only energy can be exchanged
with the surroundings
4
Systems

A system may be ISOLATED

neither matter nor energy can
be exchanged with the
surroundings
5
Systems may be
open, closed, or isolated
6
Internal energy
The internal energy E of a system is all the energy
(both kinetic & potential) contained in the system
 Chemists are especially interested in

thermal energy (energy of random molecular motion)
 chemical energy (energy stored in chemical bonds and
intermolecular forces)

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First Law of Thermodynamics



The energy a system contains is its internal energy, E
Energy can move in or out of a system as heat (q) and/or
work (w).
Energy is conserved, so all energy lost or gained by a
system must be accounted for as heat and/or work:
E  q  w
8
E is a state function
A state function (aka function of state) is a
property whose value depends only on the state of
the system, not how it achieved that state
 The state of the system is specified by the pressure,
temperature, and composition of the system.
 P, V, & T are state functions. E is a state function.
 q and w are NOT state functions.

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

The elevation gain
is the same by
either path:
a state function
How ∆E is
distributed
between q and w
depends on which
path you take:
q and w are not
state functions
10
Path 1: short circuit the battery with a wrench.
Lots of heat, maybe even sparks and a fire, but no work!
fully charged
battery
fully discharged
battery
∆E
the same
by
either
path
Path 2: connect the battery to a motor.
Some heat, but also some work.
11
Heat

Heat is energy transferred
because of a temperature
difference
A system does not contain heat;
it contains ENERGY
 Heat is just a form by which energy is transferred
 The other form by which energy is transferred is work

Whether q or w, energy IN is positive, energy OUT is negative!
12
Heat

The amount of energy transferred as
heat, q, is related to
James Prescott Joule
the amount of matter gaining/losing energy (m)
 the specific heat of matter gaining/losing energy (c)
 the amount of the temperature change (∆T)

q  mcT
13
Heat capacity and specific heat

Specific heat capacity = c
Intrinsic capacity to gain/lose energy as heat
 unit = J/g K or J/g °C
 Molar specific heat capacity = J/mol K or J/mol °C


Heat capacity of a system = C
Quantity of energy to change temperature by 1 °C
 unit = J/K or J/°C

14
Examples
How much heat energy (in kJ) is required to raise
the temperature of 237 g ice water from 4.0 °C to
37.0 °C? cwater = 4.18 J/g °C
 How much heat energy (in kJ) is required to raise
the temperature of 2.50 kg Hg from –20.0 °C to –
6.0 °C? cHg = 28.0 J/mol °C

16
Determination of specific heat




Heat sample of metal to
temperature of boiling
water
Measure temperature of
measured amount of
water in insulated beaker
Put hot metal in cold water and measure final temperature
All energy is assumed to stay in the insulated container

qmetal + qwater = 0 or qmetal = – qwater
17
Examples
When 1.00 kg Pb (specific heat = 0.13 J/g °C) at
100.0 °C is added to some water at 28.5 °C, the
final temperature is 35.2 °C. What is the mass of
the water?
 100.0 g Cu (specific heat 0.385 J/g °C) at 100.0 °C
is added to 50.0 g water at 26.5 °C. What is the
final temperature of the mixture?

18
Work
Work is done when a force acts for a distance:
w=Fxd
 Energy is the capacity to do work


kinetic energy is the energy of motion


the random motion of molecules (thermal energy) is kinetic
potential energy is stored energy, that can do work
when it is released

the energy in chemical bonds is potential energy
19
Pressure-volume work
Imagine a gas trapped in
a cylinder with a
moveable lid (a piston)
 If the piston is not
moving, how does the
gas pressure inside
compare to the external
pressure?
 If the piston is not
moving, Pgas = Pexternal

Pexternal


Pgas
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Pressure-volume work

What if Pgas increases, so
Pgas > Pexternal?
Pexternal


Pgas
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Pressure-volume work
The piston rises and the
gas expands, until once
again Pgas = Pexternal
 The gas has moved the
piston some distance
against the opposing Pext
 The gas has done work!

Pexternal


Pgas
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Pressure-volume work

How much work did the
gas do?
Pexternal

F
P
or F  PA
A
w  Fd  PAd
d  h
∆h

Pgas
w  PAd  PAh  PV
23
Pressure-volume work

How much work did the
gas do?
Pexternal

w  PV

When the gas did work
by expanding, it lost
energy
∆h

Pgas
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Pressure-volume work

Now what if we increase
the external pressure, so
Pexternal > Pgas ?
Pexternal


Pgas
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Pressure-volume work
The gas is compressed
until once again Pgas =
Pexternal
 The piston has been
moved some distance
against the opposing Pgas
 This time work was done
on the gas
 How much work?

Pexternal

∆h

Pgas
w  PV
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Pressure-volume (P∆V) work

When the gas expands, it expends energy to do
work = Pext∆V
∆V is positive
 energy leaves the system (the gas)


When the gas is compressed, it gains the energy
used to compress it = Pext∆V
∆V is negative
 energy enters the system (the gas)

27
Sign conventions for work


When energy goes into the system as work, w is positive
When energy leaves the system as work, w is negative
w = –P∆V
when a gas expands ∆V is + and w is – (energy leaves the system)
when a gas is compressed ∆V is – and w is + (energy enters the system)
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Units

w = –P∆V
work is in joules (J)
 P is in atm and V is in L, so P∆V is in atm L


The relationship between J and atm L is
1 atm L = 101.325 J
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Example

What is the work done on a gas (in J) when the gas
is compressed from an initial volume of 35.0 L to a
final volume of 23.5 L under a constant pressure of
0.987 atm?
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Calorimetry
The process for measuring the amount of heat
energy exchanged by system and surroundings is
CALORIMETRY
 Assume total energy is constant: heat lost by
system is gained by surroundings
qsystem + qsurroundings = 0
qsystem = – qsurroundings

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Heat of reaction, qrxn




If the reaction releases
energy, qrxn is negative
(the reaction loses energy)
The calorimeter gains that
energy and gets hotter
The reaction is
EXOTHERMIC
qrxn = – qcalorimeter




If the reaction absorbs
energy, qrxn is positive (the
reaction gains energy)
The calorimeter loses that
energy and gets colder
The reaction is
ENDOTHERMIC
qrxn = – qcalorimeter
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Bomb calorimeter
Combustion reaction occurs
in sample compartment
 Reaction releases energy to
water in calorimeter
 qrxn = – qcalorimeter = – C∆T


C = heat capacity of calorimeter
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Examples
The combustion of 1.013 g vanillin (C8H8O3) in a
bomb calorimeter with heat capacity = 4.90 kJ/°C
causes the temperature to rise from 24.89 °C to
30.09 °C. What is the heat of combustion of
vanillin, in kJ/mol?
 Combustion of 1.176 g benzoic acid (HC7H5O2,
heat of combustion –26.42 kJ/g) causes the
temperature in a bomb calorimeter to increase by
4.96 °C. What is C for that calorimeter?

34
Coffee cup calorimeter
Reactants (usually in aqueous
solution) mix and react in the
insulated cup
 Reaction releases energy to (or
absorbs energy from) liquid in
which it is dissolved
 qrxn = – qcal = – (mc∆T)cal


Calorimeter = liquid in the cup
(cup assembly is usually ignored)
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Example

100.0 mL 1.00 M AgNO3 (aq) and 100.0 mL 1.00
M NaCl (aq), both initially at 22.4 °C, are mixed in
a coffee cup calorimeter. The temperature rises to
30.2 °C. What is the heat of reaction, in kJ per mol
AgCl, for the reaction:
Ag1+ (aq) + Cl1– (aq)  AgCl (s)
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Example

100.0 mL 1.020 M HCl and 50.0 mL 1.988 M
NaOH, both initially at 24.52 °C, are mixed in a
coffee cup calorimeter. If qneutralization = –56 kJ/mol
H2O, what will be the final temperature in the
mixture?

Hint: this is a limiting reactant problem
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Energy and Enthalpy

∆E = q + w
w = –P∆V
 At constant volume, ∆V = 0 so w = 0
and ∆E = qV


We define H = E + PV
H is called enthalpy
 At constant pressure, ∆H = ∆E + P∆V
 But ∆E = qP – P∆V, so ∆H = qP – P∆V + P∆V
 ∆H = qP

38
∆E and ∆H
At constant volume, measured heat = ∆E
 At constant pressure, measured heat = ∆H

Constant pressure is the more common lab condition
 ∆H is the heat of reaction for a chemical change carried
out at constant pressure

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How different are ∆E and ∆H?


C (s) + 1/2 O2 (g)  CO (g)
∆H = – 110.5 kJ
∆H = ∆E + ∆(PV) or ∆E = ∆H – ∆(PV)
 ∆(PV) = ∆(nRT) = ∆ngasRT (solids are not significant)
 ∆ngas = 1/2 mol
 At
298 K,
atmL
1kJ
PV  12 mol  0.08206 molK
298K  101.325J

1atmL
1000J
PV 1.2kJ
∆E
is – 111.7 kJ, only 1% difference from ∆H
40
Enthalpy
Like E, absolute values of H cannot be measured,
but changes in H can be measured: ∆H = qP
 ∆H is part of the chemical reaction stoichiometry

∆H is extensive (depends on amount of material)
 ∆H is directional (sign reverses if direction of reaction
reverses)


∆H is a state function

Value of ∆H same whether reaction occurs in a single
step or a series of steps, if final result is the same
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Reactions add together
Look at these reactions:
C (s) + O2 (g)  CO (g) + 1/2 O2 (g)
CO (g) + 1/2 O2 (g)  CO2 (g)
 The reactions add to give
C (s) + O2 (g)  CO2 (g)
 The CO and 1/2 O2 terms on both sides cancel

42
Heats of reaction add, too

H values add just like reactions:
∆H (kJ/mol)

C (s) + O2 (g)  CO (g) + 1/2 O2 (g)
–111 kJ/mol

CO (g) + 1/2 O2 (g)  CO2 (g)
–283 kJ/mol

C (s) + O2 (g)  CO2 (g)
–394 kJ/mol
43
Hess’ Law
Germain Hess discovered that heats of reaction add
together in 1840
 The additivity of ∆H values is called Hess’ Law
 Hess’ Law establishes enthalpy as a state function

State functions depend only on the state of the system
 It doesn’t matter how the system achieved that state

44
Hess’ Law
Hess’ Law allows us to calculate ∆H for a reaction
from measured ∆H values for other reactions
 The rules of the game

Multiply reaction by a factor to get desired coefficients
 Multiply ∆H by the same factor
 Reverse a reaction to get a substance on the desired side
 If you reverse the reaction, reverse the sign of ∆H

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Example

Given these heats of reaction,




H2 (g) + 1/2 O2 (g)  H2O (l)
2 C2H6 (g) + 7 O2 (g)  4 CO2 (g) + 6 H2O (l)
2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2O (l)
∆H = –286 kJ
∆H = –3123 kJ
∆H = –2602 kJ
Find ∆H for this reaction:
C2H2 (g) + 2 H2 (g) ––> C2H6 (g)
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Example
First look at the target reaction:
C2H2 (g) + 2 H2 (g)  C2H6 (g)
 You want 1 C2H2 on the left. We need to multiply
the reaction with C2H2 by 1/2:

1/2 [2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2O (l)]
 C2H2 (g) + 5/2 O2 (g)  2 CO2 (g) + H2O (l)
Multiply ∆H by 1/2: ∆H = 1/2[–2602 kJ] = –1301 kJ


47
Example
Another look at the target reaction:
C2H2 (g) + 2 H2 (g)  C2H6 (g)
 You need 2 H2 on the left. We need to multiply the
reaction with H2 by 2:

2 [H2 (g) + 1/2 O2 (g)  H2O (l)]
 2 H2 (g) + O2 (g)  2 H2O (l)
Multiply ∆H by 2: ∆H = 2[–286 kJ] = –572 kJ


48
Example
The target reaction:
C2H2 (g) + 2 H2 (g)  C2H6 (g)
 Finally, you want 1 C2H6 on the right. Multiply
the reaction with C2H6 by 1/2, and reverse it:

1/2 [2 C2H6 (g) + 7 O2 (g)  4 CO2 (g) + 6 H2O (l)]
 reverse C2H6 (g) + 7/2 O2 (g)  2 CO2 (g) + 3 H2O (l)
 2 CO2 (g) + 3 H2O (l)  C2H6 (g) + 7/2 O2 (g)
Multiply ∆H by 1/2, and reverse its sign:
∆H = 1/2[+3123 kJ] = +1561 kJ


49
Example

Now you have





C2H2 (g) + 5/2 O2 (g)  2 CO2 (g) + H2O (l)
∆H = –1301 kJ
2 H2 (g) + O2 (g)  2 H2O (l)
∆H = –572 kJ
2 CO2 (g) + 3 H2O (l)  C2H6 (g) + 7/2 O2 (g) ∆H = +1561 kJ
Now add the reactions and cancel items identical on both sides
The total is
C2H2 (g) + 2 H2 (g)  C2H6 (g)
312 kJ
∆H = –
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Formation reactions

To organize collections of ∆H values for reactions,
chemists defined a formation reaction


A formation reaction is the equation to form one mole
of compound from its elements in their most stable state
The formation reaction for H2O is
H2 (g) + 1/2 O2 (g)  H2O (l)
Notice that the reactants are elements in their most
common, stable form
 Notice that only one mole of product forms

51
Standard enthalpy of formation

The enthalpy of a formation reaction is called the standard
heat of formation (symbol ∆Hf°).



The little f means a formation reaction
The ° means standard conditions, 1 atm and 1 M solutions
∆Hf° values are collected in reference books (see
Appendix II in the back of your text)



∆Hf° for an element in its most stable state is zero
The state (liquid vs. gas or solid) is important
Equations are not given (you are supposed to know how the
write the appropriate formation reaction)
52
Heats of formation combine
You can combine formation equations and their
∆Hf° values just like any other reactions
 Same rules of the game:

If you reverse the reaction, reverse the sign of ∆Hf°
 If you multiply the reaction by some factor, multiply
∆Hf° by the same factor

53
But, of course, there’s a SHORTCUT

∆H° = ∑n∆Hf° products – ∑n∆Hf° reactants


n is the coefficient for the substance, in moles
C2H2 (g) + 2 H2 (g)  C2H6 (g)


∆Hf° values: C2H2 = +227 kJ/mol, H2 = 0 kJ/mol (element),
C2H6 = –85 kJ/mol
∆H° = [1(∆Hf° C2H6)] – [1(∆Hf° C2H2) + 2(∆Hf° H2)]

∆H° =
[1 mol (–85 kJ/mol)] – [1 mol (+227 kJ/mol) + 2 mol (0 kJ/mol)]

∆H° = –312 kJ
54
A shortcut for the shortcut

An easy way to set up the shortcut is
Write the target equation
 Beneath each substance, write its ∆Hf° (watch signs!)
 Beneath ∆Hf°, write the coefficient for that substance
 Multiply each ∆Hf° by its coefficient
 Add the answers for the products, add the answers for
the reactants
 ∆H° = products – reactants

55
The short shortcut

C2H2 (g) + 2 H2 (g)  C2H6 (g)
+227 kJ
0 kJ
–85 kJ
x1
x2
x1
+227 kJ
0 kJ
–85 kJ
+227 kJ
 –85 kJ
∆H° = (–85 kJ) – (+227 kJ) = –312 kJ
 The same result we got earlier by adding reactions

56
Reactions in aqueous solution

Additivity also applies to reactions in solution
∆Hf° values for ions are given in the Appendix
 The baseline for aqueous ions is H1+ (aq) = 0 kJ/mol
instead of the element form
 All rules apply as before

57