Download chem16 part2 lect1 thermodynamics

1/6/2009
Chemistry 16
Thermodynamics
Thermodynamics
• Forget everything
except these:
Lecture by Jim G. Minglana
– Chemical and
physical processes
involve the transfer
of energy.
– Thermodynamics
governs the
direction of physical
and chemical
processes.
Note: This lecture is intended solely for use by
students of Chem 16 Section THY (T,Th 4-530), Room
1100
2nd Semester 2008-2009
– study of the changes in energy and transfers of
energy that accompany chemical and physical
processes.
• Spontaneous process
– the process that occurs at a given condition.
– compare this with “non-spontaneous process”
1.
2.
3.
4.
5.
The First Law of Thermodynamics
Enthalpy Changes
Calorimetry (to be taken in the lab)
Themochemical Equations
Standard States and Standard Enthalpy
Changes
6. Standard Molar Enthalpy of Formation, ∆Hf°
7. Hess's Law
8. Bond Energies
9. Changes in Internal Energy, ∆E
10. Relationship of ∆H and ∆E
Part 2: Spontaneity of Physical
and Chemical Changes
1.
2.
3.
4.
Entropy, S, and Entropy Change, ∆S
The Second Law of Thermodynamics
Free Energy Change, ∆G, and Spontaneity
The Temperature Dependence of
Spontaneity
– to understand how
energy is involved in
any process
– to be able to identify
the conditions for
such process to
proceed
spontaneously
Part 1: Heat Changes and
Thermochemistry
Some terms
• Thermodynamics
• Main goals:
Thermodynamics
The First Law of Thermodynamics:
- The combined amount of energy in the
universe is constant.
- also known as the Law of Conservation of
Energy:
Energy is neither created nor destroyed in chemical
reactions and physical changes.
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Exothermic Reactions or Processes
- release energy in the form of
heat
• System - anything under study
– In the chemistry lab, the system is the chemicals
inside the beaker.
• Surroundings - the environment around the
system.
– The surroundings are outside the beaker.
• The system plus the surroundings is called
the universe.
Endothermic Reactions or
Processes
C3H8(g) + 5O2(g) → CO2(g) + 4H2O(l) + 2.22×103 kJ
H2O(g) → H2O(l) + 44 kJ
Or
C3H8(g) + 5O2(g) → CO2(g) + 4H2O(l)
H2O(g) → H2O(l)
∆H = -2.22×103 kJ
∆H = -44 kJ
Thermochemical equations.
∆H = amount of heat involved (in kJ) in the process.
Note: The states of all compounds are indicated.
Energy diagrams
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) + 890 kJ
- absorb energy in the form of
heat
CO2(g) + 4H2O(l) + 2.22×103 kJ → C3H8(g) + 5O2(g)
H2O(l) + 44 kJ → H2O(g)
Or:
CO2(g) + 4H2O(l) → C3H8(g) + 5O2(g)
∆H = +2.22×103 kJ
H2O(l) → H2O(g)
∆H = +44 kJ
Reverse reactions: exothermic and endothermic
processes.
Signs for ∆H values also reverse.
Endothermic if ∆H > 0
Exothermic if ∆H < 0
Basic Ideas Important to
Thermodynamic Systems
•
Chemical systems tend toward a state of
minimum potential energy.
H2O flows downhill.
Objects fall when dropped.
•
Chemical systems tend toward a state of
maximum disorder.
A mirror shatters when dropped and does not reform.
It is easy to scramble an egg and difficult to unscramble
it.
Food dye when dropped into water disperses.
• Energy (PE) vs progress of reaction
• Energy difference between products and reactants
• Energy of activation
Enthalpy Change, ∆H or ∆Hrxn
- quantity of heat transferred into or out of a
system as it undergoes chemical or physical
change at constant T and P.
- applies to most chemical reactions
- determines whether a process is exothermic or
endothermic
- an extensive property
C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) ∆H = -3523 kJ
1 mol
8 mol
5 mol
6 mol
∆H = -3523 kJ for every mole of C5C12(l) that undergoes
combustion.
∆H = -3523 kJ for 6 mol of H2O(l) that is formed from the
reaction or
∆H = -587.2 kJ for every mole of H2O(l) formed in the
reaction.
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Exercise
• How much heat is liberated when 2.60 g
of sodium metal reacts with excess
water according to the following
thermochemical equation?
2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq)
∆H = -368 kJ/mol rxn
Enthalpy Change, ∆H or ∆Hrxn
• also referred to as the heat of reaction.
• is a state function: path-independent
∆Hrxn = Σ Hfinal
- Σ Hinitial
∆Hrxn = Σ Hproducts - Σ Hreactants
Thus, if Σ Hfinal < Σ Hinitial , then ∆Hrxn < 0 (exo.)
if Σ Hfinal > Σ Hinitial , then ∆Hrxn > 0 (endo.)
∆Hf°or the Standard Molar Enthalpy of
Formation
• enthalpy for the reaction in which one mole of a
substance is formed from its constituent elements in
their standard states.
• ∆Hf°for MgCl2 is:
Mg(s) + Cl2(g) → MgCl2(s) + 641.8 kJ
or
Mg(s) + Cl2(g) → MgCl2(s) ∆Hf°= -641.8 kJ/mol
Exercise
•
Which will provide a greater amount of heat, the
combustion of 1.00 g of propane, C3H8(g) or 1.00
g of pentane, C5H12(l)?
• Which of the two hydrocarbons above will
produce more CO2(g) with each kJ of heat
evolved?
• How are the problems above related to
(a) the use of hydrocarbons as energy source?
(b) environmental concerns vs energy
consumption?
Determination of ∆H of Reactions
Using Calorimetry (in the lab)
• A coffee-cup
calorimeter is used to
measure the amount of
heat produced (or
absorbed) in a reaction
at constant P
• Thermochemical standard state conditions
T = 298.15 K
P = 1.0000 atm.
• Thermochemical standard states of matter
• For pure substances in their liquid or solid phase the
standard state is the pure liquid or solid.
• For gases the standard state is the gas at 1.00 atm of
pressure.
• For gaseous mixtures the partial pressure must be 1.00 atm.
• For aqueous solutions the standard state is 1.00 M
concentration.
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∆Hf°Values
• have been determined for many substances
• are tabulated in Table 15-1 and Appendix K in
the text.
• Standard molar enthalpies of elements in their
most stable forms at 298.15 K and 1.000 atm
are zero.
Set up the equation for the formation of 1 mole of
glucose, C6H12O6(s) from its elements in the standard
states. What is ∆Hf°of glucose?
Recall: ∆H is a state function
∆Hrxn = Σ Hfinal
- Σ Hinitial
∆Hrxn = Σ Hproducts - Σ Hreactants
In the same way,
∆H°rxn = Σ H°f, products - Σ H°f, reactants
Exercise
• Calculate the enthalpy change for the
reaction in which 15.0 g of aluminum
reacts with oxygen to form Al2O3 at
25oC and one atmosphere.
(Ans. -466 kJ)
Exercises
The ∆Hf°phosphoric acid is -1281 kJ/mol. Write the
thermochemical equation for the reaction for
which ∆H
∆ orxn = -1281 kJ.
P in standard state is P4(s)
Phosphoric acid in standard state is H3PO4(s)
• Which is more exothermic, the combustion
of one mole gaseous benzene, C6H6, or the
combustion of one mole of liquid benzene?
Why? (No calculations are necessary.)
Exercises
• Calculate the enthalpy change for the
reaction of one mole of H2(g) with one mole
of F2(g) to form two moles of HF(g) at 25oC
and one atmosphere. (Ans. -542 kJ)
• Is the conversion of graphite to diamond
exothermic or endothermic?
Hess’s Law
• Hess’s Law of Heat Summation states that the
enthalpy change for a reaction is the same
whether it occurs by one step or by any
(hypothetical) series of steps.
– Hess’s Law is true because ∆H is a state function.
• If we know the following ∆H°’s
[1] 4FeO(s) + O2(g) → 2Fe2O3(s)
∆H°= -560 kJ
[2] 2Fe(s) + O2(g) → 2FeO(s)
∆H°= -544 kJ
[3] 4Fe(s) + 3O2(g) → 2Fe2O3(s) ∆H°= -1648 kJ
Then we can calculate the ∆Ho for reaction [1] by
properly adding (or subtracting) the ∆Ho’s for
reactions [2] and [3].
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[1] 4FeO(s) + O2(g) → 2Fe2O3(s)
∆H°= -560 kJ
[2] 2Fe(s) + O2(g) → 2FeO(s)
∆H°= -544 kJ
[3] 4Fe(s) + 3O2(g) → 2Fe2O3(s) ∆H°= -1648 kJ
∆H°
2 {2FeO(s) → 2Fe(s) + O2(g) } +544 kJ × 2
4Fe(s) + 3O2(g) → 2Fe2O3(s)
-1648 kJ
4FeO(s) + O2(g) → 2Fe2O3(s)
-560 kJ
0
0
Exercise
• Given the following equations and ∆H° values
[1] 2N2(g) + O2(g) → 2N2O(g)
∆H°= +164.1 kJ
[2] N2(g) + O2(g) → 2NO(g)
∆H°= +180.5 kJ
[3] N2(g) + 2O2(g) → 2NO2(g)
∆H°= +66.4kJ
calculate ∆Ho for the reaction below. (Ans. +155.5 kJ)
N2O(g) + NO2(g) → 3NO(g)
0
∆H rxn = ∑ n ∆H f products− ∑ n ∆H f reactants
n
n
Exercise
n = stoichiometric coefficients
• Given the following information, calculate ∆Hf°for
H2S(g).
2H2S(g) + 3O2(g) → 2SO2(g) + 2N2O(l) ∆H°298 = -1124
kJ
Ans. -20
20..6 kJ
Exercise
• A 8.10 g sample of magnesium is
burned in excess nitrogen at constant
atmospheric pressure to form Mg3N2.
The reaction mixture is brought back to
25°C. In this provess 76.83 kJ of heat is
given off. What is the standard molar
enthalpy of formation of Mg3N2?
Exercise
From the following enthalpies of reaction,
CaCO3(s) → CaO(s) + CO2(g)
∆H
∆ = -202.4 kJ
CaO(s) + H2O(l) → Ca(OH)2(s)
∆H
∆ = -65.3 kJ
Ca(OH)2(s) → Ca2+ (aq) + 2OH–(aq)
∆H
∆ = -16.2 kJ
calculate ∆H for
Ca2+ (aq) + 2OH–(aq) + CO2(g) → CaCO3(s) + H2O(l)
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Bond Energy
Exercise
Aluminum reacts vigorously with many
oxidizing agents. For example,
4Al(s) + O2(g) → 2Al2O3(s)
∆H
∆ = -3352 kJ
4Al(s) + 3MnO2(s) → 3Mn(s) + 2Al2O3(s)
∆H
∆ = -1792 kJ
Use this information to determine the enthalpy
of formation of MnO2(s).
• the amount of energy needed to break a
bond in the gas phase into separate atoms
also in the gas phase.
• applicable to covalent bonds only.
• also known as bond dissociation energy.
A–B(g) + bond energy → A(g) + B(g)
H–Cl(g) + 432 kJ
→ H(g) + Cl(g)
or H–Cl(g) → H(g) + Cl(g) ∆H = +432 kJ
NOTE:
Breaking a bond always requires an absorption of energy.
Forming a bond always involves a release of energy.
Some Average Bond Energies
• in kJ per mole of the bond
• taken from several experimental values (see page 564
Tables 15-2 and 15-3)
• give us an idea of relative bond strengths
Single Bonds
C–H
N–H
O–H
H–H
F–F
413
391
463
436
155
C–C
N–N
O–O
C–N
C–O
346
163
143
305
358
Double
Bonds
C=C 602
N=N 418
O=O 498
C=N 615
C=O 732*
Triple
Bonds
C≡C 835
N≡N 945
C≡N 887
C≡O 1072
Bond Energies, BE
• In gas phase reactions ∆H°values may
be related to bond energies of all
species in the reaction.
∆H°298 = Σ BEreactants
+ Σ BEproducts
Another way is
∆H°298 = Σ BEbonds broken + Σ BEbonds
formed
Always positive
Always negative
*Except in CO2, where it is 799
Exercise
Exercise
• Estimate the value of ∆H for the reaction
below.
C3H8(g)
+ Cl2(g) → C3H7Cl(g)
+ HCl(g)
• Use the bond energies listed in the
Table to estimate the heat of reaction at
25°C for the combustion of methane,
CH4.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
H
H
H
H C
C
C
H
H
H
H
+ Cl
Cl
H
H
H
H C
C
C
H
H
H
∆H = ?
Ans. -686 kJ
Cl
+
H Cl
• This time, solve for ∆H using ∆Hf°
values. Is there any discrepancy?
Where could the difference come from?
Ans. -116 kJ/mol
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Different Enthalpy Values (kJ/mol)
Exercises
• Calculate the bond energy for hydrogen
fluoride, HF(g).
HF(g) → H(g)
+ F(g)
Ans. +568 kJ/mol
NH3(g) → N(g)
+ 3H(g)
∆Hatomization
+108.7
Na(g) → Na+(g) + e–
(g)
∆Hionization = IE =
ionization energy
+495.4
Cl2(g) → 2Cl(g)
BE = bond
dissociation
energy
EA = electron
affinity
+241.8
Cl(g) + e–(g) → Cl–(g)
• Do the same for ammonia, NH3(g).
Ans. +391 kJ/mol
∆H°f
Na(s) + ½Cl2(g) →
KCl(s)
Na(s) → Na(g)
-411.0
-348.5
Na+(g) + Cl–(g) → KCl(s) LE = Lattice
Energy
?
• What is the LE of NaCl(s)?
Other Enthalpy Values
Na(s) → Na(l)
∆Hmelting =
-∆
∆Hfusion
Na(l) → Na(g)
∆Hvaporization
H+(aq) + OH–(aq) → H2O(l)
∆Hneutralization
(of a strong acid
and a strong base)
K+(g) → K+(aq)
∆Hhydration
C6H6(l) + CCl4 (l) →
C6H6-CCl4 solution
∆Hmixing
The Born-Haber Cycle
• allows us to calculate the LE of ionic
compounds
• LE indicates the strength of ionic bonds
(you will
determine this
experimentally
)
And many more!
Internal Energy, E
Changes in Internal Energy, ∆E
• Recall the 1st Law of TD:
• is all of the energy contained within a
substance.
– kinetic, potential, gravitational,
electromagnetic, etc.
– a state function (independent of path)
∆E = Efinal
- Einitial
∆E is negative when energy is released by a
system.
∆E is positive when energy is absorbed by a
system
the change in internal energy, ∆E, is
determined by the heat flow, q, and the work,
w.
∆E = q + w , where
q = heat flow
w = work done
• Thus, the following increase the internal
energy of a system
• heat absorbed by the system
• work done on the system
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Changes in Internal Energy,
∆E
∆E = Efinal - Einitial
∆E = q + w
q>0
heat is absorbed by the system
q<0
heat is released by the system
w>0
the surroundings do work on the system
w<0
the system does work against the
surroundings
Pressure-Volume Work
• the only kind of work in most chemical
and physical changes
Pressure is the force F per unit area A:
P = F / A or P = F / d2
Volume is distance cubed:
V = d3
P∆
∆V is a work term: the same units are used for
energy and work.
P∆
∆V = ( F /
When
Then
d2
)
(d3 )
=
F×d
This is work!
Examples
1) ∆ng =
0
V2 = V1
P∆
∆V = 0
w=0
CO(g) + H2O(g) → H2(g) + CO2(g)
2) ∆ng >
0
V2 > V1
P∆
∆V > 0
w<0
Zn(s) + H+(aq) → Zn2+(aq) + H2(g)
3) ∆ng <
0
V2 < V1
P∆
∆V < 0
w>0
N2(g) + 3H2(g) → 2NH3(g)
Problem 1
• If 1.200 kJ of heat is added to a system
in energy state E1, and the system does
0.800 kJ of work on the surroundings,
what is the energy change for the
system?
• Ans. +0.400 kJ
• What is the energy change for the
surrounding?
• What is the energy of the system in the
new state E2?
Work and P∆
∆V
• Work - force acting through a specified
distance.
w = F × d = -P∆
∆V = -( ∆ng )RT
where ∆ng = Σngaseous products - Σngaseous
reactants
n = moles
Thus, w = -( ∆ng )RT
at constant T and P
• Consequently, there are three possibilities for
volume changes:
Exercises
• Consider the following gas phase reactions
at constant pressure.Is work done on the
system by the surroundings?
• At 200oC:
2NO(g) + O2(g) → 2NO2(g)
• At 1000oC:
PCl5(g) → PCl3(g) + Cl2(g)
NOTE: All reactions are under constant T and P.
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∆H and q
Relationship of ∆H and ∆E
∆H = ∆E + P∆
∆V
• The total amount of heat energy that a
system can provide to its surroundings at
constant temperature and pressure is given
by
∆H = change in enthalpy of system
∆E = change in internal energy of system
P∆
∆V = work done by system
∆ H = qp
∆H is the change in heat content at constant
pressure.
• qp = heat content at constant pressure
∆H = ∆E + P∆
∆V and ∆E = q + w
Since w = P∆
∆V, then ∆E = q + w = q - P∆
∆V
And so, ∆H = q - P∆
∆V + P∆
∆V
∆H
at constant P
∆ =q
or
∆ H = qp
Special case: solids and
liquids
• volume change during reaction is very
small or equal to zero.
∆V ≈ 0 and P∆
∆V ≈ 0
• Since ∆H = ∆E + P∆
∆V,
then ∆H ≈ ∆E
• That is why only changes in the moles
of gaseous products and reactants are
considered.
Relationship of ∆H and ∆E
∆H
amount of heat absorbed or released when a
reaction occurs at constant pressure
∆E
amount of heat absorbed or released when a
reaction occurs at constant volume.
How much do the ∆H and ∆E for a reaction
differ?
The difference depends on the amount of work
performed by the system or the surroundings.
Exercise
• For the combustion of n-pentane, n-C5H12,
the ∆H°= -3523 kJ/mol. Combustion of 1
mol of n-pentane at constant pressure
releases 3523 kJ of heat. What are the
values of the work term and ∆E for this
reaction?
Ans.-3516 kJ
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Free Energy Change, ∆G,
Dictates Spontaneity
Spontaneity of Physical and Chemical
Changes
• Spontaneous changes happen without any
continuing outside influences.
– the maximum useful energy obtainable in the form
of work from a process at constant temperature
and pressure.
– a state function
– A spontaneous change has a natural direction.
• Examples:
– the rusting of iron occurs spontaneously.
• Have you ever seen rust turn into iron metal without man
made interference?
– The melting of ice at room temperature occurs
spontaneously.
• Will water spontaneously freeze at room temperature?
∆G = ∆H - T∆
∆S
at constant T and P
• a reliable indicator of spontaneity of a
physical process or chemical reaction
• does not tell us how quickly the process
occurs
– Chemical kinetics is concerned with rates of
reactions.
• Sign conventions for ∆G:
• ∆G > 0
• ∆G = 0
• ∆G < 0
∆G = Gibbs Free Energy
∆G = ∆H - T∆
∆S
at constant T and P
∆S is the change in entropy or disorder of a
system
Entropy, S
• is a measure of the disorder or
randomness of a system.
• a state function
• have been measured and tabulated in
Appendix K as So298.
• When ∆S > 0 disorder increases
∆S < 0 disorder decreases
reaction is nonspontaneous
system is at equilibrium
reaction is spontaneous
Exercises
• Arrange the following in increasing
order of entropy: a solid, a liquid, a
vapor (of the same substance).
• Enumerate processes or reactions that
will lead to greater entropy.
The 2nd Law of Thermodynamics
• In spontaneous changes the universe tends
towards a state of greater disorder
∆Suniverse > 0 for a spontaneous process to
occur.
∆Suniverse = ∆Ssystem + ∆Ssurroundings
• Entropy changes for reactions can be
determined similarly to ∆H for reactions.
∆S°298 = Σ nS°products - Σ nS°reactants
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The 3rd Law of
Thermodynamics
Spontaneous processes have two
requirements:
•
– The free energy change ∆G of the system must
be negative.
– The entropy of universe must increase.
• The entropy of a pure, perfect, crystalline
solid at 0 K is zero.
• This law permits us to measure the absolute
values of the entropy for substances.
– To get the actual value of S, cool a substance to 0
K, or as close as possible, then measure the
entropy increase as the substance heats from 0 to
higher temperatures.
– Appendix K has values of S not ∆S.
– S values are usually given in J rather than in kJ.
•
Fundamentally, the system must be capable of doing
useful work on surroundings for a spontaneous
process to occur.
∆G = ∆H - T∆
∆S
• An exothermic reaction does not ensure spontaneity.
freezing of water is exothermic but spontaneous only below
0oC.
• An increase in disorder (entropy) of the system also
does not insure spontaneity.
• It is a proper combination of exothermicity and
disorder that determines spontaneity.
Problem
For 2NO2(g) → N2O4(g)
• Calculate the entropy change for the
reaction at 25oC.
• Is the reaction spontaneous in the
direction it is written?
• At what temperature is equilibrium
between NO2(g) and N2O4(g) attained?
2NO2(g) → N2O4(g)
←
∆G is a state function. Therefore,
∆G°298 = Σn∆
∆G°products - Σn∆
∆G°reactants
• Use this equation to solve for ∆G in the
previous problem.
Temperature Dependence of Spontaneity
∆H
∆S
∆G = ∆H -T∆
∆S
1) ∆H < 0 ∆S > 0 ∆G < 0 at any value of T
(+)
Spontaneous at all temperatures.
(–)
2) ∆H < 0 ∆S < 0 ∆G < 0 at lowT
(–)
(–)
Spontaneous at low temperatures.
3) ∆H > 0 ∆S > 0 ∆G < 0 at highT
(+)
(+)
Spontaneous at high temperatures.
Problem
• Use thermodynamic data to estimate the
normal boiling point of water.
• What is the percent error in this problem?
4) ∆H > 0 ∆S < 0 ∆G > 0 at lowT
(+)
(–)
Nonspontaneous at ALL temperatures.
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