Transcript Second Year Chemistry

Second Year Chemistry
• 1st semester: Organic
• 1st semester: Physical (2005-2006)
• December exams
• 2nd: Analytical & Environmental
• 2nd: Inorganic
• Summer exams
• Physical: 3 lecturers 8 topics
• Dónal Leech: four topics
• Thermodynamics
• Gases, Laws & Phases, Equilibrium
1
Introduction
Energetics and Equilibria
What makes reactions “go”!
This area of science is called THERMODYNAMICS
Thermodynamics is expressed in a mathematical language
BUT
Don’t, initially anyway, get bogged down in the detail of the equations: try to
picture the physical principle expressed in the equations
We will develop ideas leading to one important Law, and explore practical
applications along the way
The Second Law of Thermodynamics
 rG   RT ln K
0
 rG   r H  T r S
0
2
0
0
Lecture Resources
12 lectures leading to four exam questions
(section A, you must answer two from this section)
• Main Text:
“Elements of Physical Chemistry”
Atkins & de Paula, 4th Edition (Desk reserve)
http://www.oup.com/uk/booksites/content/0199271836/
OTHERS.
“Physical Chemistry” Atkins & de Paula, 7th Edition or any other
PChem textbook
These notes available on NUI Galway web pages at
http://www.nuigalway.ie/chem/degrees.htm
See also excellent lecture notes from James Keeler,
Cambridge, although topics are treated in a different running
order than here.
3
Thermodynamics: the 1st law
The internal energy of an isolated system is constant
Energy can neither be created nor destroyed only inter-converted
Energy: capacity to do work
Work: motion against an opposing force
System: part of the universe in
which we are interested
Surroundings: where we make our
observations (the universe)
Boundary: separates above two
4
System and Surroundings
Systems
• Open: energy and matter
exchanged
• Closed: energy exchanged
• Isolated: no exchange
• Diathermic wall: heat transfer
permitted
• Adiabatic wall: no heat transfer
5
Work and Heat
• Work (w): transfer of
energy that changes
motions of atoms in the
surroundings in a
uniform manner
• Heat (q): transfer of
energy that changes
motions of atoms in the
surroundings in a
chaotic manner
• Endothermic: absorbs heat
• Exothermic: releases heat
6
Work
• Mechanical work can generally be described by dw = -F.dz
•
•
•
•
Gravitational work (mg.dh)
Electrical work (.dq)
Extension work (f.dl)
Surface expansion work (.d)
As chemists we will concentrate on
EXPANSION WORK
(many chemical reactions produce gases)
Expansion against constant external
pressure
dw = -F.dz but pex = F/A
therefore
w = -pex.V
7
Expansion Work
Expansion against zero external pressure (free expansion)
w = -pex.V = 0 (external pressure = 0)
Reversible isothermal expansion
• In thermodynamics “reversible” means a process that can be
reversed by an infinitesimal change of a variable.
• A system does maximum expansion work when the external pressure is
equal to that of the system at every stage of the expansion
8
Isothermal reversible expansion
w   pex dV
Vf
Vf
Vi
Vi
w    pdV  nRT 
1
dV
V
w  nRT (ln V  constant ) V
Vf
i
w  nRT ( ln V f  constant) - (ln Vi  constant )
w  nRT (ln V f - ln Vi )
w  nRT ln
Vf
Vi
Exercise: Calculate the work done when 1.0 mol Ar(g) confined in a cylinder of volume
1.0 dm3 at 25°C expands isothermally and reversibly to 2.0 dm3.
9
10
1st Law of Thermodynamics
The internal energy of an isolated system is constant
Energy can neither be created nor destroyed only inter-converted
U = q+w
Exercise: A car battery is charged by supplying 250 kJ of energy to it as electrical work,
but in the process it loses 25kJ of energy as heat to the surroundings. What is the change
in internal energy of the battery?
INTERNAL ENERGY is a State Function
Use calorimetry. If we enclose our system in a
How do we
constant volume container (no expansion),
measure heat?
provided no other kind of work can be done,
then w = 0.
U = qV
11
Bomb calorimetry
• By measuring the change in Temperature of the
water surrounding the bomb, and knowing the
calorimeter heat capacity, C, we can determine
the heat, and hence U.
Heat Capacity
 dU 
CV  

 dT V
U  CV T  qV 
12
Amount of energy required to raise the
temperature of a substance by 1°C
(extensive property)
For 1 mol of substance: molar heat capacity
(intensive property)
For 1g of substance: specific heat capacity
(intensive property)
Calorimeter calibration
Can calibrate the calorimeter, if its heat capacity is unknown, by
passing a known electrical current for a given time to give rise to a
measured temperature change.
q  IVt
Amperes.Volts.Sec = Coulombs.Volts = Joules
Exercise: In an experiment to measure the heat released by the combustion of
a fuel, the compound was burned in an oxygen atmosphere inside a calorimeter
and the temperature rose by 2.78°C. When a current of 1.12 A from an 11.5 V
source was passed through a heater in the same calorimeter for 162 s, the
temperature rose by 5.11°C. What is the heat released by the combustion
reaction?
13
Enthalpy
Most reactions we investigate occur under
conditions of constant PRESSURE (not Volume)
ENTHALPY: Heat of reaction at constant pressure!
H  U  pV
 H  U  pV
but w  - pV
H  qP
Heat capacity
Use a “coffee-cup”
calorimeter to measure it
 dH 
CP  

 dT  P
H  CP T  qP 
Exercise: When 50mL of 1M HCl is mixed with 50mL of 1M NaOH in a coffeecup calorimeter, the temperature increases from 21°C to 27.5°C. What is the
enthalpy change, if the density is 1g/mL and specific heat 4.18 J/g.K?
14
Perfect gas enthalpy
• Use intensive property of
molar enthalpy and
internal energy
• At 25°C, RT = 2.5 kJ/mol
H m  U m  pVm
H m  U m  RT
Thermicity-Revision
Endothermic reaction (q>0) results in an increase in enthalpy
(H>0)
Exothermic reaction (q<0) results in an increase in enthalpy
(H<0)
NB: Internal energy and Enthalpy are STATE FUNCTIONS
15
Temperature variation of enthalpy
Convenient empirical expression
to use for heat capacity is:
C p ,m
c
 a  bT  2
T
Exercise: What is the change in molar enthalpy of N2 when it is heated from
25°C to 100 °C, given that:
C p ,m
16
50000
 28.58  0.00377T 
T2
Relation between heat capacities
H m  U m  RT
H m  U m  RT
H m  U m  RT
H m U m

R
T
T
C p ,m  CV ,m  R
17
Thermochemistry
Chemists report data for a set of standard conditions:
The standard state of a substance (°) is the pure substance at exactly 1 bar
It is conventional (though not obligatory) to report data for a T of 298.15K
Standard enthalpies of phase transition
Energy that must be supplied (or is evolved) as heat, at constant
pressure, per mole of molecules that undergo the phase transition
under standard conditions (pure phases), denoted H°
Note: the enthalpy change of a
reverse transition is the negative of
the enthalpy change of the forward
transition
18
H°
Substance
Freezing point, Tf/K fusHo/(kJ mol ) Boiling point, Tb/K vapHo/(kJ mol )
Ammonia, NH3
195.3
5.65
239.7
23.4
83.8
1.2
87.3
6.5
Benzene, C6H6
278.7
9.87
353.3
30.8
Ethanol, C2H5OH
158.7
4.60
351.5
43.5
3.5
0.02
Argon, Ar
Helium, He
Mercury, Hg
1
1
4.22
0.08
234.3
2.292
629.7
59.30
90.7
0.94
111.7
8.2
175.5
3.16
337.2
35.3
Propanone, CH3COCH3 177.8
5.72
329.4
29.1
Water, H2O
6.01
373.2
40.7
Methane, CH4
Methanol, CH3OH
273.15
* For values at 298.15 K, use the information in the Data section.
19
Sublimation
l
Direct conversion of a solid to a vapour
The enthalpy change of an overall
process is the sum of the enthalpy
changes for the steps into which it may
be divided
20
Enthalpies of ionisation (kJ/mol)
1
2
13
14
15
16
17
18
H
He
1312
2370
5250
Li
Be
B
C
N
O
F
Ne
519
900
799
1 090
1400
1310
1680
2080
7300
1760
2 420
2 350
2860
3390
3370
3950
14 800
3 660
P
S
Cl
Ar
1060
1000
1260
1520
25 000
Na
Mg
Al
Si
494
738
577
4560
1451
1 820
7740
2 740
786
11 600
ionH°(T)= Ionisation energy(0) + (5/2)RT (see Atkins & de Paula, Table 3.2)
21
Problems
 Ethanol is brought to the boil at 1 atm. When the electric
current of 0.682 A from a 12.0 V supply is passed for 500 s
through a heating coil immersed in the boiling liquid, it is
found that the temperature remains constant but 4.33 g of
ethanol is vapourised. What is the enthalpy of vapourisation
of ethanol at its boiling point at 1 atm?
 Calculate the standard enthalpy of sublimation of ice at 0°C
given that fusH° is 6.01 kJ/mol and vapH° is 45.07 kJ/mol,
both at 0°C.
 subH° for Mg at 25°C is 148 kJ/mol. How much energy as
heat must be supplied to 1.00 g of solid magnesium metal to
produce a gas composed of Mg2+ ions and electrons?
22
Bond enthalpies (kJ/mol)
H
H
436
C
412
C
N
O
F
Cl
Br
I
S
P
Si
348 (1)
612 (2)
838 (3)
518 (a)
N
O
388
463
305 (1)
163 (1)
613 (2)
409 (2)
890 (3)
945 (3)
360 (1)
157
743 (2)
146 (1)
497 (2)
F
565
484
270
185
155
Cl
431
338
200
203
254
Br
366
276
219
193
I
299
238
210
178
S
338
259
250
212
P
322
Si
318
496
242
264
200
374
466
Values are for single bonds except where otherwise stated (in parentheses).
(a) Denotes aromatic.
23
151
226
Problem

24
Estimate the
standard reaction
enthalpy for the
formation of liquid
methanol from its
elements as 25°C
Enthalpies of combustion
Enthalpies (heats) of combustion: complete reaction of compounds with oxygen.
Measure using a bomb calorimeter. H  U  PV
 nRT 
 RT 
V  
  ng 

 P 
 P 
 H  U  ng RT
Most chemical reactions used for the production of heat are combustion reactions.
The energy released when 1g of material is combusted is its Fuel Value. Since
all heats of combustion are exothermic, fuel values are reported as positive.





Most of the energy our body needs comes from fats and carbohydrates.
Carbohydrates are broken down in the intestines to glucose. Glucose is
transported in the blood to cells where it is oxidized to produce CO2, H2O and
energy:
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l) cH°=-2816 kJ
The breakdown of fats also produces CO2 and H2O
Any excess energy in the body is stored as fats
25
Heats of formation
If one mole of the compound
is formed under standard
conditions from its elements
in their reference state then
the resulting enthalpy
change is said to be the
standard molar enthalpy
(Heat) of formation, fH°
where the subscript indicates
this.
The reference state is the
most stable form under the
prevailing conditions.
26
Element
Reference state
Arsenic
Grey arsenic
Bromine
Liquid
Carbon
Graphite
Hydrogen
Gas
Iodine
Solid
Mercury
Liquid
Nitrogen
Gas
Oxygen
Gas
Phosphorus
White phosphorus
Sulfur
Rhombic sulfur
Tin
White tin, -tin
Hess’s Law
To evaluate unknown heats of
reaction
The standard enthalpy of a
reaction is the sum of the
standard enthalpies for the
reactions into which the overall
reaction may be divided
rxnHo = nfHom(products) - nfHom(reactants)
27
Variation of rH° with T
rH°(T2) = rH°(T1) + rCp°(T2-T1)
Kirchoff’s Law
rCp° = nCp,m°(products) - nCp,m°(reactants)
If heat capacity is temperature
dependent, we need to integrate over
the temperature range
T2
 r H (T2 )   r H (T1 )    rC p dT
o
o
T1
28
o
Thermodynamics: the 2nd law
Deals with the direction of
spontaneous change
(no work required to bring it about)
Kelvin Statement
No process is possible in which the sole
result is the absorption of heat from a
reservoir and its complete conversion
into work
Impossible!
29
Entropy
The apparent driving force for spontaneous change is the dispersal
of energy
A thermodynamic state function,
Entropy, S, is a measure of the
dispersal of energy (molecular
disorder) of a system
2nd Law: The Entropy of
an isolated system
increases in the course of
spontaneous change
Stot>0
30
Thermodynamic definition of S

Concentrates on the change in entropy:
S = qrev/T
Can use this equation to quantify entropy changes.
We will see later (3rd & 4th year) a statistical description
of entropy
S = k lnW (Boltzmann formula)
31
Heat Engines
All heat engines have a hot region “source” and a cold region
“sink”: some energy must be discarded into the cold sink as
heat and not used to do work
qh qc
Stot  

Th Tc
Tc
qc 
qh
Th
qc
work qh  qc
Tc


 1
 1
heat
qh
qh
Th
32
Expansion entropy


Intuitively can guess that entropy increases with gas
expansion.
Thermodynamic definition allows us to quantify this
increase
Recall that: w = -nRT ln (Vf/Vi)
BUT qrev = -w (U = 0 for isothermal processes)
S = nR ln (Vf/Vi)
Note: independent of T
Also: Because S is a state function, get the same value for an
irreversible expansion
33
Heating Entropy
dq
dS 
for infinitesi mal change in T
T
q
dq
C
or for infinitesi mal change in T C 
T
dT
dqrev  CdT
S  
Tf
Ti
CdT
dS 
T
T f dT
Tf
CdT
 C
 C ln
Ti T
T
Ti
for constant heat capacity
34
Entropy of phase transition

Entropy of fusion
 fusS 

Tf
Entropy of vapourisation
 vap S 
35
 fus H (T f )
 vap H (Tb )
Tb
Trouton’s rule
The entropy of vapourisation is
approximately the same (85
J/K.mol) for all non-polar liquids
Phase transitions
To evaluate entropies of transition at T other
than the transition temperature
Entropy of vapourisation of water at 25°C?
 Sum of S for heating from 25°C to 100°C,
S for vapourisation at 100°C, and S for
cooling vapour from 100°C to 25°C. Try it!

(+118 J/K.mol).
36
Entropy changes in the
surroundings
S
tot =
qsur,rev
S sur 
T
qsur
S sur 
T
q
S sur  
T
Ssys + Ssur
Stot = Ssys – q/T
Example: Water freezing to ice.
Entropy change of system is -22 J/K.mol,
and heat evolved is -6.01 kJ/mol.
Entropy change in surroundings must be
positive for this process to occur
spontaneously.
Check this for different temperatures.
Note that Stot = 0 at equilibrium
37
Spontaneity of water freezing
Stot = Ssys - qsys/T

Stot = -22 JK-1mol-1 – (6,010Jmol-1/278K)
At 5°C:


At -5°C:


38
Stot = -22 JK-1mol-1 – (6,010Jmol-1/268K)
= +0.43 JK-1mol-1
Stot = -22 JK-1mol-1 – (6,010Jmol-1/273K)
At 0°C:
= 0.01 JK-1mol-1
To find transition temperature, set Stot = 0 and solve for T.
273.18 K (slight error because of rounding of entropy and
heats).


= -0.38 JK-1mol-1
Problem



39
Typical person heats the surroundings
at a rate of 100W (=J/s). Estimate
entropy change in one day at 20°C.
qsur = 86,400 s × 100 J/s
Ssur = qsur/T = (86,400 × 100 J)/293 K
= 2.95 × 104 J/K
3rd Law



Entropy of sulfur phase transition
is 1.09 J/K.mol.
Consider plot at left. Subtract
entropy for phase transition (to
give plot at right)
T=0 intercept is the same.
Entropies of all perfectly
crystalline substances are
the same at T=0.
40
Absolute and standard molar
entropies (S and S0m)
Absolute entropies can be
determined by integration of
areas under heat capacity/T as
a function of T, and including
entropies of phase transitions.
Standard molar entropies are
the molar entropies of
substances at 1bar pressure
(and usually 298 K)
41
Standard molar entropies
Substance
ө
-1
S m/JK mol
Gases
-1
Substance
ө
Solids
Ammonia, NH3
192.5
Calcium oxide, CaO
39.8
Carbon dioxide, CO2
213.7
Calcium carbonate, CaCO3
92.9
Helium, He
126.2
Copper, Cu
33.2
Hydrogen, H2
130.7
Diamond, C
2.4
Neon, Ne
146.3
Graphite, C
5.7
Nitrogen, N2
191.6
Lead, Pb
64.8
Oxygen, O2
205.1
Magnesium carbonate, MgCO3
65.7
Water vapour, H2O
188.8
Magnesium oxide, MgO
26.9
Sodium chloride, NaCl
72.1
Liquids
Benzene, C6H6
173.3
Ethanol, CH3CH2OH
160.7
Water, H2O
42
69.9
Sucrose, C12H22O11
Tin,
360.2
Sn (white)
51.6
Sn (grey)
44.1
See the Data section for more values.
-1
S m/JK mol
-1
Standard reaction entropies

Difference in molar entropy between
products and reactants in their standard
states is called the standard reaction
entropy and can be expressed (like
enthalpy) as:
rxnSo = nSom(products) - nSom(reactants)

43
Note: absolute entropies, S, and standard molar
entropies, S0m, are discussed in section 4.7 of the
textbook
Spontaneity of reactions
Consider the reaction:
2H2(g) + O2(g)  2H2O (l)
rS0 = 2(70 J/K.mol) –[2(131 J/K.mol) + (205 J/K.mol)]
= -327 J/K.mol
But this reaction is spontaneous (explosive even!)
When considering the implications of entropy, we must
always consider the total change of the system and its
surroundings
rH0 = -572 kJ/mol. Therefore rSsur = +1920 J/K.mol
rStot is positive +1590 J/K.mol (spontaneous reaction!).
44
Gibbs Energy





Introduced by J.W. Gibbs to combine the
calculations of 2 entropies, into one.
Because Stot = S – H/T (constant T and P)
Introduce G = H – TS (Gibbs “free” energy)
Then G = H – TS (constant T)
So that G = – TStot (constant T and P)
G = H – TS
In a spontaneous change at constant temperature
and pressure, the Gibbs energy decreases
45
Maximum non-expansion work



Can derive (see box 4.5 in textbook) that G = w’max
Example: formation of water: enthalpy -286kJ, free
energy -237kJ
Example: suppose a small bird has a mass of 30 g.
What is the minimum mass of glucose that it must
consume to fly to a branch 10 m above the ground?
(G for oxidation of glucose to carbon dioxide and water is -2828 kJ at 25°C)
Exercise: A human brain operates at about 25 W (J/s). What mass of glucose
must be consumed to sustain that power for 1 hour?
46
Problems solved
w’ = (30 × 10-3 kg) × (9.81 m s-2) × 10 m
Note 1J = 1kg m2 s-2
n = 2.943 J/(2828 × 103 J/mol)
m = nM = (1.04 × 10-6 mol) × 180 g/mol
= 1.9 × 10-4 g
Answer 2: 5.7g
47
Phase Equilibria
Phase transitions
Changes in phase without a change in chemical composition
Gibbs Energy is at the centre of the discussion of transitions
Molar Gibbs energy
Gm = G/n
Depends on the phase of the substance
A substance has a spontaneous tendency to change into a phase
with the lowest molar Gibbs energy
48
Variation of G with pressure




We can derive (see
derivation 5.1 in textbook)
that Gm = Vmp
Therefore Gm>0 when
p>0
Can usually ignore
pressure dependence of G
for condensed states
Can derive that, for a gas:
Gm = RT ln(pf/pi)
49
Proof-go back to fundamental
definitions
G = H – TS; H = U + pV; dU = dq + dw
For an infinitesimal change in G:
G + dG = H + dH – (T + dT)(S + dS)
= H + dH – TS – TdS – SdT – dTdS
dG = dH – TdS – SdT
Also can write: dH = dU + pdV + Vdp
dU = TdS – pdV (dS = dqrev/T and dw = -pdV)
dG = Vdp – SdT
Master Equations
50
Variation of G with temperature
Gm = -SmT
Can help us to understand why transitions occur
The transition temperature is the temperature when the
molar Gibbs energy of the two phases are equal.
The two phases are in EQUILIBIRIUM at this temperature
51
Phase diagrams


52
Map showing conditions
of T and p at which
various phases are
thermodynamically
stable
At any point on the
phase boundaries, the
phases are in dynamic
equilibrium
Location of phase boundaries

Clapeyron equation (see derivation 5.4)
 trsH
p 
T
T trsV

Clausius-Clapeyron equation
(derivation 5.5)
 ln p  
 vap H
RT
2
T
 vap H  1 1 
    constant
ln p2  ln p1 
R  T2 T1 
53
Constant is
vapS/R
Derivations
dGm = Vmdp – SmdT
dGm(1) = dGm(2)
Vm(1)dp – Sm(1)dT = Vm(2)dp – Sm(2)dT
{Vm(2) – Vm(1)}dp = {Sm(2) – Sm(1)}dT
trsV dp = trsS dT
T trsV dp = trsH dT
dp/dT = trsH/(T trsV)
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Derivations: liquid-vapour transitions
dp/dT = vapH/(T vapV)
≈ vapH/{T Vm(g)} = vapH/{T (RT/p)}
(dp/p)/dT = vapH/(RT2)
(d lnp)/dT = vapH/(RT2)
d ln p 

ln p2
 vap H
RT
d ln p  
T2
2
dT
 vap H
2
dT
RT
 vap H
p2  vap H T2 1
ln

dT  
2

p1
R T1 T
R
ln p1
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T1
1 1
   + constant
 T2 T1 
Using the equation
The vapour pressure of mercury is 160 mPa at 20°C.
What is its vapour pressure at 50°C given that its
enthalpy of vapourisation is 59.3 kJ/mol?
 The vapour pressure of pyridine is 50.0 kPa at 365.7 K
and the normal boiling point is 388.4 K. What is the
enthalpy of vapourisation of pyridine?
 Estimate the normal and standard boiling point of
benzene given that its vapour pressure is 20.0kPa at
35°C and 50.0kPa at 58.8°C.
Remember:
 BP: temperature at which the vapour pressure of the
liquid is equal to the prevailing atmospheric pressure.
 At 1atm pressure: Normal Boiling Point (100°C for water)
 At 1bar pressure: Standard Boiling Point (99.6°C for
water; 1bar=0.987atm, 1atm = 1.01325bar)

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Summary
l
Thermodynamics tells which way a process will go
• Internal energy of an isolated system is constant (work and
heat). We looked at expansion work (reversible and
irreversible).
• Thermochemistry usually deals with heat at constant
pressure, which is the enthalpy.
• Spontaneous processes are accompanied by an increase in
the entropy (disorder?) of the universe
• Gibbs free energy decreases in a spontaneous process
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