Chapter 14
Heat, Work, Energy, Enthalpy
1 The Nature of Energy
2 Enthalpy
3 Thermodynamics of Ideal Gases
4 Calorimetry
5 Hess's Law
6 Standard Enthalpies of Formation
7 Present Sources of Energy (skip)
8 New Energy Sources (skip)
James Prescott Joule (1818-1889)
Highlights
– English physicist and !!brewer!!
– Studied the nature of heat, and discovered
its relationship to mechanical work
– Challenged the "caloric theory”l that heat
cannot be created nor destroyed.
– Led to the theory of conservation of energy,
which led to the development of the first
law of thermodynamics
– Worked with Lord Kelvin to develop the
absolute scale of temperature
– Pupil of John Dalton (Atomic Theory)
– Joule effect: found the relationship between
the flow of current through a resistance and
the heat dissipated, now called Joule's law
Moments in a Life (Death)
– Gravestone is inscribed with the number
“772.55” the amount of work, in ft lb, he
determined experimentally to be required to
raise the temperature of 1 lb of water by 1°
Fahrenheit.
“772.55”
Thermite Reaction
Reactants → Products
2 Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (s)
Energy
• Definition of Energy: Capacity to do work.
• The study of energy and its interconversions is
called Thermodynamics.
• 1St Law of Thermodynamics: Conservation of
energy.
– Energy can be converted from one form to another but
cannot be created of destroyed.
– The energy of the universe is constant.
– The energy of a closed system is constant.
– Heat and work are interconvertable
Forms of Energy
•
•
•
•
•
•
Chemical Energy
Potential energy (e.g., ΔEpotential= mgΔh)
Kinetic energy (e.g., ΔEKinetic= Δ (1/2mv2)
Electrical Energy
Nuclear Energy
• Energy transfer
– Through heat
– Through work
Energy
• Energy is a state property, which means it
depends on the initial and final state, not the
path between them.
ΔE = Efinal – Einitial
ΔE = Eproducts – Ereactants
What is a “State Property”?
Thermodynamics:
Systems and Surroundings
A thermodynamic system is the part of the universe that is under consideration. A
boundary separates the system from the rest of the universe, which is called the
surroundings. A system can be anything that you wish it to be, for example a
piston, and engine, a brick, a solution in a test tube, a cell, an electrical circuit, a
planet, etc.
Heat (q) and Work (w)
ΔE = q + w
•
•
•
q = Heat absorbed by the system
 If q > 0 , heat is absorbed
 If q < 0 , heat is given off
w = Work done on the system
Increase the energy of a system by heating it (q>0)
or by doing work on it (w>0).
Work
 w = Force x distance
= (Pressure x area) x distance
= pressure x ΔV = PΔV
Work is Pressure-Volume work (P-V)
Exothermic reactions release energy to
surroundings (by heating the surroundings)
CH4 (g) + 2O2 (g) → CO2 (g) +2 H2O (g) + heat
In an exothermic process, the energy stored in
chemical bonds/molecular interactions is converted
to thermal energy (random kinetic energy).
Endothermic reactions absorb energy
N2 (g) + O2 (g) + Energy (heat) → 2 NO (g)
An endothermic reaction consumes heat and stores
energy in the chemical bonds/molecular interactions
of the products.
Work
w = F x Δh
P=F/A
P x A x Δh
ΔV = A Δh
W=P ΔV
• Force exerted by heating = P1A
– Where P1 is the pressure inside the vessel
– Where A is the Area of the piston
Pext = P1 if balanced
w = F x d = F x Δh = P x A x Δh
ΔV = A Δh
ratio of Gas law constants
w = −Pext ΔV
– Units are atm·L or L atm
– Where 1 L atm = 101.3 Joules
8.3145 J
101.3J
K mol
=
=
0.08206L atm L atm
K mol
• Expansion (w<0)
– The system does work on the surroundings.
• Compression (w>0)
– The surroundings do work on the system.
Energy
• Energy is the capacity to do work.
• Energy is a state property, which means it depends on
the initial and final state, not the path between them.
ΔE = Efinal – Einitial
ΔErxn = Eproducts – Ereactants
Enthalpy
• Heat is a means by which energy is transferred.
• For processes carried out at constant pressure, the heat
absorbed equals a change in Enthalpy, ΔH.
qp= ΔH
Enthalpy Change
(p denotes constant pressure)
• Enthalpy is a state property, which means it depends on
its initial and final state, not the path between them.
ΔH = Hfinal – Hinitial
ΔHrxn = Hproducts – Hreactants
Enthalpy
• ΔE=q+w = ΔH + w (at constant P)
• If a chemical reaction occurs in solution with
no PV work (no change in V), then w=0 and
ΔE = ΔH.
• ΔH for a reaction may be either positive
(absorbs heat from surroundings) or negative
(disperses heat to the surroundings).
• For a gas phase reaction, where ΔV is
allowed, then ΔE ≠ ΔH.
Phase Changes
Phase changes (are not chemical reactions) but involve enthalpy
changes
Melt ice: H2O(s) → H2O(l)
ΔHfusion= + 6 kJ at 0C
ΔHfusion is positive means endothermic (add heat)
if reversed
H2O(l) → H2O(s)
ΔHfreeze= – 6 kJ
GAS
ΔHfreeze= – ΔHfusion
ΔHvaporization = – ΔHcondensation
ΔHsublimation = – ΔHdeposition
SOLID
Melting/
Fusion
Freezing
LIQUID
Thermodynamics of Ideal Gases
(volume changes)
(KE)avg is the average translational energy of 1 mole of a
perfect gas.
(KE)ave = 3 RT
2
The energy (“heat”) required to change the translational
energy of 1 mole of an ideal gas by ΔT is
3
Energy ("heat") required = RT
2
Heat Capacity (C)
• How much heat is required to raise the temp of one mole
of something by 1 degree? C is a proportionality constant
linking T and q.
CΔT=q
If no work is allowed (ΔV = 0)
CvΔT=q
If work is allowed (ΔP=0)
CpΔT=q
Cp>Cv
Heating an Ideal Gas
Molar Heat Capacity (C) of a substance is defined as the heat (q)
required to raise the temperature of 1 mole of a substance by 1K.
• Molar Heat Capacity: Cv = 3/2 R (constant volume)
(for a perfect monatomic gas)
• Molar Heat Capacity: Cp = 5/2 R (constant pressure)
(for a perfect monatomic gas)
• polyatomic
molecules have larger
Cv‘s & Cp’s due to
vibrational and
rotational degrees of
freedom.
Thermodynamic Properties of an Ideal Gas
ΔE = q + w = ΔH + w (const P)
ΔH = ΔE + (PΔV) = qp
ΔH = ΔE + PΔV = ΔE + nRΔT = qp
Example
Consider 2.00 mol of a monoatomic ideal gas that
is take from
state A (PA = 2.00 atm, VA = 10.0 L) to
state B (PB = 1.00 atm, VB = 30.0 L)
Calculate q, w, ΔE, and ΔH
Know: PV=nRT, and Cp, Cv, for monatomic gas.
Conversion: L-atm to Joules = 1 L atm = 101.3 Joules
State A
V=10 L
P=2 atm
State B
V=30 L
P=1 atm
Two pathways from A to B
I)
Step 1. A to C (constant P)
Step 2. C to B (constant V)
II)
Step 3. A to D (constant P)
Step 4. D to B (constant V)
State A
State B
Vi = 10L
Vf = 30L
Pi = 2 atm
Pf = 1atm
Calculate q, w, ΔE, ΔH
q
q1, q4: Steps 1 and 4 are
constant pressure, use Cp
q2, q3: Steps 2 and 3 are
constant volume, use Cv
P(ΔV)
q p = nCp ΔT where ΔT =
nR
 5  P(ΔV) 
q p = n  R 

 2  nR 
(ΔP)V
q v = nC v ΔT where ΔT =
nR
 3  (ΔP)V 
q v = n  R 

2
nR



State A
State B
Vi = 10L
Vf = 30L
Pi = 2 atm
Pf = 1atm
Calculate q, w, ΔE, ΔH
w
q2, q3: Step 1 and 4 are
constant volume, w2=w3=0
w = - PV
w1= -2(20) = -40 l-atm
w2 = 0 (ΔV = 0)
w3 = 0 (ΔV = 0)
w4= -1(20) = -20 l-atm
State A
State B
Vi = 10L
Vf = 30L
Pi = 2 atm
Pf = 1atm
Calculate q, w, ΔE, ΔH
ΔE
Use
ΔE1 = q1+w1
ΔE2 = q2+w2
ΔE3 = q3+w3
ΔE4 = q4+w4
ΔEAB
Use
ΔEAB = ΔE1 + ΔE2
or
ΔEAB = ΔE3 + ΔE4
Heat Capacity
• Heat Capacity, Cp, is the amount of heat required to
raise the temperature of a substance by one degree at
q
heat absorbed
constant pressure.
Heat Capacity  Cp 

T
T
where Cp denotes Heat Capacity at constant pressure
T  Tf -Ti
 Cp is always a positive number
 q and ΔT can be both negative or positive
 If q is negative, then heat is evolved or given off
and the temperature decreases
 If q is positive, then heat is absorbed and the
temperature increases
Heat Capacity (continued)
• Cp units are energy per temp. change
• Cp units are Joule/oK or JK-1
• Molar heat capacity
• Units are J/K/mole or JK-1mol-1
Specific Heat Capacity
• The word Specific before the name of a physical
quantity very often means divided by the mass
• Specific Heat Capacity is the
amount of heat required to raise the
temperature of one gram of
material by one degree Kelvin (at
constant pressure)
Standard State Enthalpies
• Designated using a superscript ° (pronounced naught)
• Is written as ΔH°
• OFB text appendix D lists standard enthalpies of
formation of one mole of a variety of chemical species
at 1 atm and 25°C.
• Standard State conditions
– All concentrations are 1M or 1 mol/L
– All partial pressures are 1 atm
Standard Enthalpies of Formation
ΔH°rx is the sum of products minus
the sum of the reactants
For a general reaction
aA+bB→cC+dD
H  cH (C)  dH (D)  aH (A)  bH (B)
o
r
o
f
o
f
o
f
o
f
Enthalpy of Reaction: ΔHrxn
CO (g) + ½ O2 (g) → CO2 (g)
o
o
o
H reaction
 H formation
(products) H formation
(reactants)
H
o
formation
(CO 2 )  394kJ /m ol
o
H formation
(CO )  111kJ /m ol
o
H formation
(O 2 )  0kJ /m ol
0
o
H reaction  394 111  283kJ /m ol
2
CO (g) + ½ O2 (g) → CO2 (g)
ΔH°rxn= – 283kJ/mol
An exothermic reaction
If stoichiometric coefficients are doubled (x2), then double ΔH°
2CO (g) + 1 O2 (g) → 2 CO2 (g)
ΔH°rxn= (2)(-283kJ/mol)
If the reaction direction is reversed
CO2 (g) → CO (g) + ½ O2 (g)
Change the sign of ΔH°rxn
How much heat heat at constant pressure is
needed to decompose 9.74g of HBr (g) (mwt
=80.9 g/mol) into its elements?
H2 + Br2 → 2HBr (g)
ΔH°rxn = -72.8 kJmol-1
Strategy:
•
Decomposition is actually the reverse reaction
•
ΔH to decompose 2 moles of HBr is +72.8kJ/mol
Hess’s Law
• Sometimes its difficult or even impossible to
conduct some experiments in the lab.
• Because Enthalpy (H) is a State Property, it
doesn’t matter what path is taken as long as
the initial reactants lead to the final product.
• Hess’s Law is about adding or subtracting
equations and enthalpies.
• Hess’s Law: If two or more chemical
equations are added to give a new equation,
then adding the enthalpies of the reactions
that they represent gives the enthalpy of the
new reaction.
Principle of Hess's law
N2 (g) + O2 (g) → 2 NO
ΔH2 = +180 kJ
2 NO (g) + O2 (g) → 2 NO2
ΔH3 = –112 kJ
N2 (g) + 2 O2 (g) → 2 NO2
ΔH1 = +68kJ
Suppose hydrazine and oxygen react to give
dinitrogen pentaoxide and water vapor:
2 N2H4 (l) + 7 O2 (g) → 2 N2O5 (s) + 4 H2O (g)
Calculate the ΔH of this reaction, given that the
reaction:
N2 (g) + 5/2 O2 (g) → N2O5 (s)
has a ΔH of -43 kJ.
2 N2H4 (l) + 7 O2 (g) → 2 N2O5 (s) + 4 H2O (g)
ΔHr° = 2 x ΔHf°N2O5 (s) + 4 x ΔHf°H2O (g)
– 2 x ΔHf° N2H4 (l) – 7 x ΔHf°O2
(g)
What values to use?
ΔHf°N2O5 (s) given ΔH° = ─ 43 kJmol-1
ΔHf°H2O (g) look up in App. 4
ΔHf° N2H4 (l) look up in App. 4
ΔHf°O2 (g) look up in App. 4
=