chem 251 lecture slides (1st law thermo) - pt 1cheminnerweb.ukzn.ac.za/files/chem 251/(2) 1st law...
TRANSCRIPT
7/26/2011
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CHEM 251LECTURE SERIES 2: THE FIRST LAW OF THERMODYNAMICS
LECTURER: Dr. Patrick Ndungu
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
Some Fundamental Concepts
� Universe = System + Surroundings
� The System: What we look @
� Surroundings: Region outside the system
� Can Exchange energy and/or matter�Open
� Closed
� Isolated
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
More Basics...
� Work: Motion Against an Opposing Force (Pushing Weights)
� Energy: Capacity of the System to do work
� When work is done on the system by the surroundings, energy of the system increases
� If the system does work, its energy decreases
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
And More Basics....
� Heat: Exchange of thermal energy
� Exothermic – Heat release by System
� Endothermic – Heat absorbed by the System
� Reversible: Process is slow enough to be reversed
� Irreversible: Process can not be reversed
� Isobaric – Process done at Constant Pressure
� Isothermal – Process done at Constant Temperature
� Isochoric – Process done at Constant Volume
� Adiabatic – Process were no loss or gain of heat by the system (q = 0)
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
State Functions
� State = System exists under set conditions determined by variables we can measure
� Any property that does not depend on its history/past (or how it got that way doesn’t matter!): “Independent of the pathway leading to the current state”
� Change of a state function only depends on the initial and final state
� Examples include mass, pressure, temperature, volume, energy...
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
Internal Energy
� Total energy of a system
� It’s a State Function
� Extensive Property (depends on amount of substance present: e.g. Mass, volume)� Intensive Property = Independent of amount of substance present; e.g. Temp, density, pressure
� Can be changed by altering state variables (T, P, V, n)
if UUU −=∆
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
1st Law “The Energy of the Universe Remains Constant”
� Internal energy of an isolated system is constant
�q = energy transferred as heat to system
�w = work done on the system
�Add heat only to a system: ∆U = q
�Do work only on the System: ∆U = w
wqU +=∆
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
The 1st Law...
� “Heat & work are equivalent ways of changing a systems internal energy”
� No Such thing as perpetual motion machine!
� Energy is Conserved!
�w>0 work done on the system (free lunch!)
� q> 0 energy in
�w<0 Work done by the system
� q < 0 energy out
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
What is Work?
� Get the def’ from classical mechanics (or Physics)! Fdzdw −=
l
Volume Decrease
-∆V = Al
Area of cross section = A
Pressure Applied= constant
Gas at constant pressure P
� Recall P = F/A
� Thus F = P · A
� Wrev = F · l = P · A · l
� Notice A · l = volume pushed in by the piston thus:
� Wrev = - P∆V
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
More Work... P ≠ Constant10
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
More Work... P ≠ Constant11
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
More Work... P ≠ Constant12
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
Reversible Work Integral...13
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
What is Reversibility & Irreversibility?
� If a system is at equilibrium with its surroundings; and a very small (infinitesimal) change (perturbation) on the conditions of the system are made and results in opposite changes in states
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
Reversible & Irreversible Work Done
� Fig (a) = reversible Work
� Fig (b) irreversible work
� N.B. ‘The work done by the system in a reversible expansion from A to B represents the maximum work that the system can perform in changing from A to B’
∫−=
f
i
V
V
revV
dVnRTW
i
f
revV
VnRTW ln−=
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
� Suppose that water at its boiling point is maintained in a cylinder that has a frictionless piston. For equilibrium to be established, the pressure that must be applied to the piston is 1 atm. Suppose that we now reduce the external pressure by an infinitesimal amount in order to have a reversible expansion. If the pistons sweeps out a volume of 2.00 dm3, what is the work done by the system?
� Answer:� External Pressure is constant: Thus –Wrev = P∆V
� = 101 325 Pa x 2.00 dm3 = 202.65 Pa m3
� In terms of units; Pa = Kg m-1 s-2
� Thus you get kg m2 s-2; which = J
� Thus work done = 202.65 J
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E.g. 1
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
� dU = dq + dw
� If there is no change in volume, there is no work being done thus;
� dU = dq
� i.e. The change in internal energy of a system at constant volume is equal to the heat supplied
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Work at Constant Volume
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
� Def: “The amount of heat required to raise the temperature of any substance by 1 K (or 1 °C)”
� Symbol C
� Units (SI) J K-1
� Specific Heat Capacity: “ The amount of heat required to raise the temperature of unit mass of a material by 1 K
� If mass is in Kg; J K-1 Kg-1
� For molar: J K-1 mol-1
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Heat Capacity
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
� Not a state function, thus must specify the how
� Constant volume (isochoric) Cv
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About (C) Heat Capacity...
dT
dqC V
V ≡V
VT
UC
∂
∂=
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
� Heat capacity relates internal energy to change in temperature
� Infinitesimal changes in temperature result in infinitesimal changes in internal energy, and the constant of proportionality is Cv
� Since at constant Volume, heat supplied ~ change in internal energy we get an expression to directly measure Cv using changes in temperature and energy transferred
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More on Heat Capacity (Cv)...
dTCdU V=
TCU V ∆=∆
TCq VV ∆=
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
� Large heat Capacity means will see small changes in temperature & vice versa
� An infinite heat capacity means no change in temperature no matter how much heat added
� Phase transitions do not get changes in temperature suggesting...
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...
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
Enthalpy22
� Consider any chemical process occurring in an open vessel; pressure ~ Const.
� Thus dU = dqP + PdV...(Infinitesimal Change?!!!)
� So the heat absorbed: dqP = dU + PdV.
� Critical criteria is that only PV work is done!
� Thus we can integrate the expression:
∫ ∫+=
2
1
2
1
U
U
V
V
P PdVdUq
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
& Enthalpy...23
� P is constant! (So we can move it to the other side):
∫ ∫+=
2
1
2
1
U
U
V
V
P dVPdUq
� & we then get: =(U2 – U1) + P(V2 – V1)
� or = (U2 + PV2) – (U1 + PV1)
� U, P, & V are state functions
� This particular combination of state functions happens all over thermochemistry i.e. ENTHALPY.... Also a state function!
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
� Since H = U + PV
� We get: qP = H2 – H1 = ∆H
� So if all work done is PV: qP = ∆H
� When a process releases heat; qP and ∆H are negative – Exothermic
� When a process absorbs heat; qP and ∆H are positive – Endothermic
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H = U + PV (Enthalpy)
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
� Lets look at that heat capacity thing again at const Pressure...
� Constant pressure (isobaric) Cp
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Now We Know (qP = ∆H)...
dT
dqC
p
P ≡P
PT
HC
∂
∂=
∫=
2
1
,,
T
T
mPmP dTCq)(
)(
12,
12,,
TTCH
TTCq
mPm
mPmP
−=∆
−=
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
� For solids and liquids Cv,m & Cp,m are the same since
∆Um & ∆Hm are very close
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N.B...(more on Cv,m & Cp,m)
mmm PVUH +=
RTPV
n
VV
nRTPV
m
m
=
=
=
RTUH mm +=
dT
RTd
dT
dU
dT
dH mm )(+=
RCC mVmP += ,,
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
� Study of enthalpy changes in chemical processes
� When a process releases heat; ∆H is negative –Exothermic
� When a process absorbs heat; ∆H is positive –Endothermic
� Standard State of a Substance: The pure form of a substance at a specified temperature at 1 bar
� Conventional Temp = 298.15 K = 25.0 C
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Thermochemistry
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
Change in H, when initial & final substances are in their Std States
enthalpy change/mol when a pure liquid at 1 bar vaporizes to a gas
enthalpy change/mol when solid changes to a liquid
When one mole of a compound is formed from its constituent pure elements in their standard states
When one mole of a compound undergoes complete combustion in excess oxygen under standard conditions
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Change in Enthalpy...
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
Summary of Various Enthalpy Changes29
Ref: Atkins’ Physical Chemistry 8th
Edition. Peter Atkins & Julio de Paula
ChemChemChemChem 251: Phys 251: Phys 251: Phys 251: Phys ChemChemChemChem for for for for ChemChemChemChem EngEngEngEng
1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
Measuring Enthalpy Changes
� H is a state function
� 1st Law (Conservation of Energy)
� Thus:
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Direct Calorimetry Hess’s Law (Indirect Calorimetry)
� Usually used on chemical processes that occur to completion without any side reactions� Strong acid & Strong Base
� Combustion of organic compounds in excess oxygen
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
Sample Calculations31
Hess’s Law in action!
αααα
� αααα
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
Using Hess’s Law32
Solution
1. C6H12O6(s) + 6O2(g) � 6CO2(g) + 6H2O(l) ∆∆∆∆cHO = - 2809.1 kJ mol-1
2. C12H22O11(s) + 12O2(g) � 12CO2(g) + 11H2O(l) ∆∆∆∆cHO = - 5645.5 kJ mol-1
C6H12O6(s) � C12H22O11(s) + 1/2H2O(l)
3. 6CO2(g) + 11/2H2O(l) � 1/2C12H22O11(s) + 6O2(g) ∆∆∆∆cHO = 2822.8 kJ mol-1
[1] C6H12O6(s) + 6O2(g) � 6CO2(g) + 6H2O(l) ∆∆∆∆cHO = - 2809.1 kJ mol-1
[3] 6CO2(g) + 11/2H2O(l)� 1/2C12H22O11(s) + 6O2(g) ∆∆∆∆cHO = 2822.8 kJ mol-1
C6H12O6(s) � C12H22O11(s) + 1/2H2O(l) : ∆∆∆∆HO = 13.7 kJ mol-1
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
But You Could...
∑ ∑ ΟΟΟ∆−∆=∆
oducts tsac
ffr HvHvHPr tanRe
C3H8(l) + 5O2(g) � 3CO2(g) + 4H2O(l)
� C3H8(l) ∆∆∆∆fHO = - 103.8 kJ mol-1
� CO2(g) ∆∆∆∆fHO = - 393.5 kJ mol-1
� H2O(l) ∆∆∆∆fHO = - 285.8 kJ mol-1
∆∆∆∆cHO = {3(-393.5) + 4(-285.8)} – {-103.8 + 5( )}
= -2219.9 kJ mol-1
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
Enthalpy & Temp...34
dTCdH p=
dTCHHp
T
TTT ∫+=
2
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dTCHHP
T
TrTrTr
ΟΟΟ
∫ ∆+∆=∆2
112
∑∑ ΟΟΟ−=∆
tsac
mP
oducts
mPPr vCvCCtanRe
,
Pr
,
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1111stststst Law of Thermo!Law of Thermo!Law of Thermo!Law of Thermo!
Exact Differentials
� Exact differential:Integrated, gives a result that is independent of the path taken.
� Inexact differential:Integrated result depends on the path between the initial and final states.
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