Annual 2012CH1101 Section ACH1101 Physical Chemistry Tutorial 1.Prof. Mike Lyons.

Internal Energy (U) : total kinetic & potential energy of system.e.g. Gas in container with (n,P,V,T). Gas molecules translate,rotate & vibrate so U = Etrans + Erot + Evib

Units: Joules JInternal Energy

First law Thermodynamics relates change in internal energy of system ∆Uto energy changes arising from transfers of heat q and work w.

U q W∆ = +

First Law of Thermodynamics: Internal energy U or E.

Change in internal energy (∆U or ∆E) : energy change as system goes from an initial state withenergy Ui to a final state with energy Uf. Hence ∆U = Uf – Ui.We suppose that a closed system undergoes a process by which it passes from a state A to a state B. Then the change in internal energy ∆U = UB – UA is given by ∆U = q +W.

q+ W+U∆+

Kotz, section 5.4. pp.222-224

AB

Initial stateFinal State Remember: analogy to fund transfer &

how Bank A/c balance is effected.

System

Surroundings U q W∆ = +

Increase in internalenergy

Heat transfer to system

Work done on systemWe area led therefore to the importantconclusion that the energy of an isolatedsystem is constant.In an isolated system there is no transferof heat or work and so q = 0 and W = 0.Hence ∆U = 0 hence U = constant.

This is a mathematical statement of theFirst Law of Thermodynamics.

Spontaneous processes and entropy

A process is said to be spontaneous if it occurs without outside intervention.Spontaneous processes may be fast or slow.Thermodynamics can tell us the direction in which a process will occur butcan say nothing about the speed or the rate of the process. The latter isthe domain of chemical kinetics.There appears to be a natural direction for all physical and chemical processes.

• A ball rolls down a hill but never spontaneously rolls back up a hill.• Steel rusts spontaneously if exposed to air and moisture. The iron

Kotz, Ch.19, pp.862-868.Discussion on energy dispersalVery good.

• Steel rusts spontaneously if exposed to air and moisture. The ironoxide in rust never spontaneously changes back to iron metal and oxygen gas.

• A gas fills its container uniformly. It never spontaneously collects at one endof the container.

• Heat flow always occurs from a hot object to a cooler one. The reverse process never occurs spontaneously.

• Wood burns spontaneously in an exothermic reaction to form CO2 and H2O,but wood is never formed when CO2 and H2O are heated together.

• At temperatures below 0°C water spontaneously freezes and at temperaturesabove 0°C ice spontaneously melts.

The First Law of thermodynamics led to the introduction of theinternal energy, U. The internal energy is a state function that lets us assess whether a change is permissible: only those changes may occur for which the internal energy of an isolated system remains constant.

The law that is used to identify the signpost of spontaneous change, the Second Law of thermodynamics, may also be expressed in terms of another state function, the entropy, S.

The entropy (which is a measure of the energy dispersed The entropy (which is a measure of the energy dispersed in a process) lets us assess whether one state is accessible from another by a spontaneous change.

The First Law uses the internal energy to identify permissible changes; the Second Law uses the entropy to identify the spontaneous changesamong those permissible changes.

The characteristic common to all spontaneously occurring processes is an increase in a property called entropy (S). Entropy is a state function.This idea form the basis of the Second Law of Thermodynamics.The change in the entropy of the universe for a given process is a measureof the driving force behind that process.

In simple terms the second law of thermodynamics says that energy of all kinds in the material world disperses or spreads out if it is not hindered from doing so. In a spontaneous process energy goes from being more concentrated to being more dispersed.Entropy change measures the dispersal of energy: how much energy is Entropy change measures the dispersal of energy: how much energy is spread out in a particular process or how widely spread out it becomes at a specific temperature.

Second law of Thermodynamics

The second law of thermodynamics states that a spontaneous processis one that results in an increase in the entropy of the universe, ∆Suniverse> 0, which corresponds to energy being dispersed in the process.

gssurroundinsystemtotal SSS ∆+∆=∆

Entropy measures the spontaneous dispersal of energy :How much energy is spread out in a process, or how widely spread out it becomes – at a specific temperature.

Mathematically we can define entropy as follows : entropy change = energy dispersed/temperature.In chemistry the energy that entropy measuresas dispersing is ‘motional energy’, the translational,vibrational and rotational energy of molecules,and the enthalpy change associated withphase changes.

T

HS

T

qS

changephasesystem

revsystem

∆=∆

=∆

Energy transferred as heatUnder reversible conditions.

TSsystem =∆

Entropy units : J mol-1 K-1

Note that adding heat energy reversibly means that it is added very slowly so that at any stage the temperature difference between the systemand the surroundings is infinitesimally small and so is always close to thermal equilibrium.

Entropy is not disorder. Entropy is not a measure of disorder orchaos. Entropy is not a driving force. The diffusion, dissipation or dispersion of energy in a final stateas compared with an initial state is the driving force in chemistry.Entropy is the index of that dispersal within a system and betweenthe system and its surroundings.In short entropy change measures energy’s dispersion at a stated temperature.Energy dispersal is not limited to thermal energy transfer between system and surroundings (‘how much’ situation). It also includes redistribution of the same amount of energy in a system

What entropy is not and what it is.

It also includes redistribution of the same amount of energy in a system (‘how far’ situation) such as when a gas is allowed to expand adiabatically (q = 0) into a vacuum container resulting in the total energy being redistributed over a larger final total volume.

Entropy measures the dispersal of energy among molecules in microstates.An entropy increase in a system involves energy dispersal among more microstates in the system’s final state than in its initial state.

Possible ways of distributingtwo packets of energy between four atoms.Initially one atom has 2 quantaand three with zero quanta.There are 10 different ways (10 microstates) to distribute this quantity ofenergy

Microstate

21 ways (21 microstates) to distribute 2 quanta of energy among 6 atoms.

WkS B ln=

{ }−=

−=∆ initialfinal

WWk

SSS

lnln

Entropy measures the dispersal of energy among molecules in microstates.An entropy increase in a system involves energy dispersal among more microstates in the system’s final state than in its initial state.

Entropy in the context of Molecular Thermodynamics.

{ }

=

−=

initial

finalB

initialfinalB

W

Wk

WWk

ln

lnln

W = number of accessibleMicrostateskB = Boltzmann constant

= 1.38 x10-23 J K-1

Microstates with greatest energy dispersionare most probable.

A 100 W electric heater is used to heat gas in a cylinder for 10 min. The gas expands from 500 cm3

to 21.4 dm3 against a pressure of 1.10 atm. What is the change in internal energy of the gas?

StrategyCalculate the energy transferred as heat from the power of the heater and the time for which it operates. Then determine the energy transferred as work as the gas expands using w = -Pext ∆V. Finally, use ∆U = q + W, to determine the change in the internal energy as the sum of the energy transferred as heatand as work. It will be necessary to pay close attention to the units and signs ofthe various quantities.

SolutionPower P = dE/dt = rate of supply of energy.P = 100 W = 100 Js-1

Time t = 10 min = 600 s.q = + 60 kJ

∆U = + 57.6 kJ

Time t = 10 min = 600 s.Energy transferred as heat: q = (100 Js-1) x (600 s) = 60 x 103J = 60 kJ.When gas expands work done by system isnegative, W = - Pext∆V.Pext = 1.10 atm = (1.10 atm)x(101325 Pa atm-1)= 111,458 Pa (or Nm2).Change in volume ∆V = Vfinal – Vinitial = 21.4x10-3m3 - 500x10-6 m3

≈ 0.0209 m3.

Hence, work done by gas:W = -(111,458 Nm-2) x(0.021 m3) = - 2400 Nm = - 2400 J = - 2.4 kJChange in internal energy ∆U = q + W = + 60 kJ – 2.4 kJ = + 57.6 kJ

W = - 2.4 kJ

Calculate the entropy change when 1.0 mol of water at 373 K vaporizes to steam.You may assume that the enthalpy change ∆vapH

0 for water is 40.7 kJ mol-1.

Use entropy change = ∆vapS = qrev/T = ∆vapH/Tb

Now T = Tb = 373 K and ∆vapH0 = + 40 .7 kJ mol-1 = + 40,700 J mol-1.Hence, ∆vapS0 = ∆vapH0/Tb

= + 40,700 J mol-1/373 K = + 109 JK-1mol-1.

Annual 2012 CH1101 Section A

Chemical Equilibrium• Many reactions do not proceed to

completion. Instead a definite quantity of reactants and products are present in reaction vessel after a long reaction time has elapsed (the state of equilibrium).

• Countless experiments with chemical systems have shown that in a state of equilibrium, the concentrations of reactants and products no longer change with time.

• This apparent cessation of activity occurs because under such conditions, all reactions are microscopically reversible.

[ ][ ] ↓

↑

t

t

ON

NO

42

2

Kinetic

[ ][ ] eq

eq

ON

NO

42

2

Concentrations varywith time

Concentrations timeinvariant

Equilibrium constant enables one toCalculate the composition of a reactionMixture (amount of reactants remaining& products formed) at equilibrium.

reactions are microscopically reversible.

• We look at the dinitrogen tetraoxide/

nitrogen oxide equilibrium which

occurs in the gas phase.

Kinetic

regimeNO2

N2O4

timeco

ncen

trat

ion

N2O4 (g) 2 NO2 (g)

colourless brown

Kinetic analysis.

[ ][ ]2

2

42

NOkR

ONkR

′=

=�

�

Equilibrium:

[ ] [ ][ ][ ] K

k

k

ON

NO

NOkONk

RR

eq

eqeq

=′

=

′=

=

42

22

2242

��

∞→t

∞→t

Equilibriumconstant

Rate equations

THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANT

For any type of chemical equilibrium of the typeFor any type of chemical equilibrium of the type

a a AA + b + b BB �������� p p PP + q + q QQ

the following is a CONSTANT (at a given T)the following is a CONSTANT (at a given T)

[ ] [ ][ ] [ ]

p q

eq eq

a b

P QK =

Equilibriumconstant

Product concentrations

The size of the equilibrium constant indicates whether the reactants or the products are favoured .

If K is known, then we can predict concentrations of products or reactants in the reaction mixture at equilibrium and hence the yield of the reaction.

[ ] [ ]a b

eq eq

KA B

=

Reactant concentrations

Reactants favouredwhen Kc is small

Reactants and productsare in almost equalabundance when Kc

near unity

Products favouredwhen Kc is large

Relationship between Gibbs Energy and Equilibrium Constant.

We now derive an expression which relates the change in Gibbs energy for areaction as a function of the composition of the reaction mixture at any stagein the reaction.

qQpPbBaA +→+

Hence after some algebra and simplification the change in Gibbs energy for reaction can be computed.

( ) ( )bGaGqGpGG +−+=∆

We can define Gibbs energy in terms ofthe activity ak of the species k. kkk aRTGG ln0 +=

If the reaction is allowed to proceed

Activity = generalised concentration

( ) ( )BAQPr bGaGqGpGG +−+=∆

( ) ( )( ) ( )BBAA

QQPPr

aRTGbaRTGa

aRTGqaRTGpG

lnln

lnln00

00

+−+−

+++=∆

QRTGaa

aaRTGG rb

BaA

qQ

pP

rr ln.

.ln 00 +∆=

+∆=∆

Reaction quotient Q( ) ( )00000BAQPr bGaGqGpGG +−+=∆

If the reaction is allowed to proceedTo equilibrium then we replace Q byThe equilibrium constant K and set∆rG = 0 by definition.

KRTG

KRTG

r

r

ln

0ln0

0

−=∆

=+∆

a A + b B c C + d D

[ ] [ ][ ] [ ]QRTG

BA

DCRTGG ba

dc

ln

ln

0

0

+∆=

+∆=∆

• Under non-equilibrium conditionsGibbs energy change is :

Q = reaction quotient [ ] [ ]dc DCQ =

KQ < KQ >Gibbs energy change for reaction mixture

Expression showshow ∆G varieswith compositionof reaction mixture.

Q = reaction quotient [ ] [ ][ ] [ ]ba BA

DCQ =

• Q defines reactant and product concentration ratio (i.e. reaction composition) at any stage inchemical transformation.

• When ∆G = 0 at constant T and P wehave equilibrium . Hence Q = Kc.

[ ] [ ][ ] [ ]

KRTG

BA

DCRTG b

eqa

eq

deq

ceq

ln0

ln0

0

0

+∆=

+∆= KRTG ln0 −=∆

∆−=RT

GK

0

exp

0

0

ln

lnr r

r

G G RT Q

G RT K

∆ = ∆ +

∆ = −

ln ln

(ln ln )rG RT K RT Q

RT Q K

∆ = − += −

lnQ

G RT ∆ =

This is the mostuseful form of the

lnr

QG RT

K ∆ =

If Q/K <1 then Q < K and ∆rG is negative, thereaction tends to proceed in forward directionand the reaction is said to be product favoured. The reaction is said to be spontaneouswhen ∆rG is negative.If Q/K > 1 then Q > K and ∆rG is positive. Here thereaction does not proceed spontaneously in theforward direction and is said to be reactant favoured.If Q = K then since ln 1 = 0 ∆rG = 0 and the reactionis at equilibrium.

useful form of theequation for interpretation.

Reaction Gibbs energy

Tp

r

GG

,

∂∂=∆

ξextent of

reaction = ξ

1<<K 1>>K1=K

0=ξ 1=ξ

∆rG is the slope of the G versus ξ graph at any degree of advancement ξ of the chemical reaction.

0=∆ Gr

0=∆ Gr0=∆ Gr

The thermodynamic equilibrium constant K is 6.46 at 373 K for the reaction N2O4(g) � 2NO2(g).1 mol of N2O4 is introduced into a sealed container at 373K.The final pressure in the container is 1.5 bar when the system comesto equilibrium.Find the composition of the mixture at equilibrium. Note that 1 bar = 1 x 105 Pa.

Strategy.We use the stoichiometry of the reaction to derive expressions for the mole fraction,And hence partial pressure of each component in terms of the initial amount of N2O4

And the proportion that has reacted. We use these expressions for partial pressures to generate an expression that relates the equilibrium constant to the proportion ofN2O4 which has reacted. The latter expression is then rearranged and solved for proportion of N2O4 reacted.

If 1 mol N2O4 is present initially and a fraction α reacts then amount of N2O4

present at equilibrium can be determined.

N2O4(g) � 2NO2(g)

2 41 molN On α= −

For every molecule of N2O4 that reacts 2 molecules of NO2 are formed.

22 molNOn α=

Standard pressurep0 = 1 bar

P = final pressure

present at equilibrium can be determined. 2 41 molN On α= −

22 molNOn α=

( )

( )

2 4 2

2 4

2 4

2

2

2 4 2 4

2 2

2 2

2 4 2 4

20 2

0 0

1 2 1

1

1

2

1

1

1

2

1

1

total N O NO

N ON O

total

NONO

total

N O N O

NO NO

NO NO

N O N O

n n n

nX

n

nX

n

p X p p

p X p p

p p pK

p p p p

α α α

αα

αα

αα

αα

= + = − + = +

−= =+

= =+

− = = +

= = +

= =

( ){ }( ) ( ){ } ( ) ( )

( )

2 4

2

22 2

0 0 2 0

2 2 0

2 2 0

2

0 0

0

2

2 1 1 4 4

1 1 11 1

1 4

4

14 4

1

1.5 6.46 1

0.518 0.518 0.719 0.72

1 1 0.72 0.28

2 2 0.72 1.44

N O

NO

p p pK

p p pp

K p p

K K p p

Kp p

Kp K p

p bar K p bar

n mol

n mo

α α α αα α αα α

α α

α α

α

α αα

α

+ = = = − + −− +

− =

− =

= =+ +

= = =

= = = ≅= − = − =

= = × = lEquilibriumComposition