CHEM 251 Lecture Slides (1st Law Thermo) - Pt 1cheminnerweb.ukzn.ac.za/Files/Chem 251/(2) 1st Law...

7/26/2011

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CHEM 251LECTURE SERIES 2: THE FIRST LAW OF THERMODYNAMICS

LECTURER: Dr. Patrick Ndungu

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!

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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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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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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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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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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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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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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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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More Work... P ≠ Constant10







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Reversible Work Integral...13



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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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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� 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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� 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



� 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



� 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

∂

∂=



� 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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� 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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...



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



& 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!



� 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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� 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

−=∆

−=



� 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 += ,,



� 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



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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Summary of Various Enthalpy Changes29

Ref: Atkins’ Physical Chemistry 8th

Edition. Peter Atkins & Julio de Paula



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



Sample Calculations31

Hess’s Law in action!

αααα

� αααα



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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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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Enthalpy & Temp...34

dTCdH p=

dTCHHp

T

TTT ∫+=

2

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dTCHHP

T

TrTrTr

ΟΟΟ

∫ ∆+∆=∆2

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∑∑ ΟΟΟ−=∆

tsac

mP

oducts

mPPr vCvCCtanRe

,

Pr

,



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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CHEM 251 Lecture Slides (1st Law Thermo) - Pt 1cheminnerweb.ukzn.ac.za/Files/Chem 251/(2) 1st Law...

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