1 second year chemistry 1 st semester: organic 1 st semester: physical (2008-2009) december exams 2...
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Second Year Chemistry• 1st semester: Organic• 1st semester: Physical (2008-
2009)• December exams
• 2nd: Analytical & Environmental• 2nd: Inorganic
• Summer exams• Physical: 4 lecturers 8 topics• Dónal Leech: two topics
• Thermodynamics• Gases, Laws
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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 Thermodynamics000
0 ln
STHG
KRTG
rrr
r
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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
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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
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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
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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)
w = -F.z but pex = F/A
therefore w = -pex.V
Expansion against constant external pressure
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Expansion Work
• 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
Expansion against zero external pressure (free expansion)
w = -pex.V = 0 (external pressure = 0)Reversible isothermal expansion
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Isothermal reversible expansion
Come to the lecture to see what is on this slide!
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1st Law of ThermodynamicsThe internal energy of an isolated system is
constant
Energy can neither be created nor destroyed only inter-converted U =
q+wExercise: 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?
How do we measure
heat?
Use calorimetry. If we enclose our system in a constant volume container
(no expansion), provided no other kind of work can be done, then w = 0.
U = qV
INTERNAL ENERGY is a State Function
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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 CapacityAmount 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)
VV
VV
qTCU
T
UC
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Calorimeter calibrationCan 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.
IVtq 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?
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EnthalpyMost reactions we investigate occur under
conditions of constant PRESSURE (not Volume)ENTHALPY: Heat of reaction at constant pressure!
PqH
Vpbut
VpUH
pVUH
- w
Use a “coffee-cup” calorimeter to measure it
PP
PP
qTCH
T
HC
Heat capacity
Exercise: When 50mL of 1M HCl is mixed with 50mL of 1M NaOH in a coffee-cup 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?
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Perfect gas enthalpy
RTUH
pVUH
mm
mmm
• Use intensive property of molar enthalpy and internal energy
• At 25°C, RT = 2.5 kJ/mol
Thermicity-RevisionEndothermic 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
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Temperature variation of enthalpy
Come to the lecture to see what is on this slide!
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Relation between heat capacities
RCC
RT
U
T
H
TRUH
RTUH
RTUH
mVmp
mm
mm
mm
mm
,,
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ThermochemistryChemists 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
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H°
Substance Freezing point, Tf/K fusHo/(kJ mol 1) Boiling point, Tb/K vapHo/(kJ mol 1)
Ammonia, NH3 195.3 5.65 239.7 23.4
Argon, Ar 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
Helium, He 3.5 0.02 4.22 0.08
Mercury, Hg 234.3 2.292 629.7 59.30
Methane, CH4 90.7 0.94 111.7 8.2
Methanol, CH3OH 175.5 3.16 337.2 35.3
Propanone, CH3COCH3 177.8 5.72 329.4 29.1
Water, H2O 273.15 6.01 373.2 40.7
* For values at 298.15 K, use the information in the Data section.
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Sublimation 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
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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
25 000
Na Mg Al Si P S Cl Ar
494 738 577 786 1060 1000 1260 1520
4560 1451 1 820
7740 2 740
11 600
ionH°(T)= Ionisation energy(0) + (5/2)RT (see Atkins & de Paula, Table 3.2)
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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?
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Bond enthalpies (kJ/mol) H C N O F Cl Br I S P Si
H 436
C 412 348 (1)
612 (2)
838 (3)
518 (a)
N 388 305 (1) 163 (1)
613 (2) 409 (2)
890 (3) 945 (3)
O 463 360 (1) 157 146 (1)
743 (2) 497 (2)
F 565 484 270 185 155
Cl 431 338 200 203 254 242
Br 366 276 219 193
I 299 238 210 178 151
S 338 259 496 250 212 264
P 322 200
Si 318 374 466 226
Values are for single bonds except where otherwise stated (in parentheses). (a) Denotes aromatic.
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Problem
Estimate the standard reaction enthalpy for the formation of liquid methanol from its elements as 25°C
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Enthalpies of combustionEnthalpies (heats) of combustion: complete reaction of compounds with oxygen.
Measure using a bomb calorimeter.
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
RTnUH
P
RTn
P
nRTV
VPUH
g
g
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Heats of formationIf 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.
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Hess’s LawTo 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 = fHom(products) - fHo
m(reactants)
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Variation of rH° with T
rH°(T2) = rH°(T1) + rCp°(T2-T1)
If heat capacity is temperature dependent, we need to integrate
over the temperature range
2
1
d)()( 12
T
T
opr
or
or TCTHTH
rCp° = Cp,m°(products) - Cp,m°(reactants)
Kirchoff’s Law
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Thermodynamics: the 2nd lawDeals 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!
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EntropyThe 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 system2nd Law: The Entropy
of an isolated system increases in
the course of spontaneous change
Stot>0
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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)
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Heat Engines
Come to the lecture to see what is on this slide!
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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 TAlso: Because S is a state function, get the same
value for an irreversible expansion
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Heating Entropy
capacityheat constant for
lndCd
Cdd
ddd
d T in change malinfinitesifor or
T in change malinfinitesifor d
d
i
fT
T
T
T
rev
T
TC
T
TC
T
TS
T
TS
TCqT
qC
T
qC
T
qS
f
i
f
i
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Entropy of phase transition
Entropy of fusion
Entropy of vapourisation
f
ffusfus T
THS
)(
b
bvapvap T
THS
)(
Trouton’s rule
The entropy of vapourisation is approximately the same (85
J/K.mol) for all non-polar liquids
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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).
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Entropy changes in the surroundings
T
qS
T
qS
T
qS
sur
sursur
revsursur
,
Stot = 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
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Spontaneity of water freezing
Come to the lecture to see what is on this slide!
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Problem
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
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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.
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Absolute and standard molar entropies (S and S0
m)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)
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Standard molar entropies
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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:
Note: absolute entropies, S, and standard molar entropies, S0
m, are discussed in section 4.7 of the textbook
rxnSo = Som(products) - So
m(reactants)
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Spontaneity of reactions
Come to the lecture to see what is on this slide!
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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 – TS (constant T) So that G = – TStot (constant T and P)
G = H – TSIn a spontaneous change at constant temperature and pressure, the Gibbs
energy decreases
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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?
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Problem solved
Come to the lecture to see what is on this slide!
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Summary 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