THERMODYNAMICS

THERMODYNAMICS: Basic Concepts

Thermodynamics: (from the Greek therme, meaning "heat" and, dynamis, meaning "power") is the study of energy conversion between heat and mechanical work, and subsequently the macroscopic variables such as temperature, volume and pressure

Energy: capacity to do work

Work: motion against an opposing force

System: part of the world which we have interest or being investigated.

Surrounding: is where we make our observations

Definition of Work:

• Work is motion against opposing force.• Work is defined as a force acting through a displacement x, the

displacement being in the direction of the force.

xfw

Consider a Gas Expansion Work:

Initial State Final State

P1, V1, T P2, V2, T

Consider a Gas Expansion Work:

:is pressure, opposing external The ,

)( 12

exP

hmgw

hhmgw

VPw

VVPhAPw

A

mgP

ex

exex

ex

)( 12

or

Consider a Gas Expansion Work:

2

1

1

2

2

1

lnln

,2

1

2

1

P

PnRT

V

VnRTw

V

dVnRTw

V

nRTP

PdVPw

PP

dVPw

v

v

in

in

v

v in

ex

v

v ex

and

gas, the of pressure theis

than greater mallyinfinitesi onlyis instant, every at

volume in increase ssimalinfiniteti against done work a For

in

Consider a Gas Expansion Work: the external pressure is infinitetissimally smaller than the internal pressure at all stages of the expansion - reversible process. A reversible change in thermodynamics is a change that can be reversed by an infinitesimal modification of a variable.

in reversible processes, there is maximum amount of work done that could possible extracted from a process

Gas Expansion Work: Sample ProblemA sample of 4.50 g of methane occupies 12.7 L at 310 K.(A)Calculate the work done when the gas expands isothermally against a constant external pressure of 200 Torr until its volume has increased by 3.3 L.(B) Calculate the work that would be done if the same expansion occurred reversibly.

J 87.97

atm L 1

J101.3 atm L 0.87

atm L 0.87

L) torr760

atm 1 torr

process) ble(irreversi formula the use we constantis pressure opposing since A.

:SOLUTION

:

w

w

w

w

VPw ex

3.3)(200(

Consider a Gas Expansion Work:

J -

KmolK J 1-mol g

g

:expansiongas reversible a for B.

:SOLUTION

1-1-

4.167

7.12

)7.123.3(ln310)3145.8()

16

5.4(

ln

w

L

Lw

V

VnRTw

i

f

Definition of HEAT:

Heat is the transfer of energy between two bodies that are at different temperatures.

Heat appears at the boundary of the system.

heat is transferred from the hotter object to the colder one.

Heat is path dependent.

heat specifics

etemperatur in change T mass,m

Tmsq

THE FIRST LAW OF THERMODYNAMICS:

“Energy can neither be created nor destroyed but can be converted from one form to another.” Law of Conservation of Energy.

ENERGY OF THE UNIVERSE:

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EE

EEE

EEE

therefore

system, given a forenergies inchanges the and

0

If a system undergoes an energy change, the surroundings must also undergo a change equal in magnitude but opposite in direction

Total ENERGY of a system; Internal Energy, U

TOTAL ENERGY:

UE

UPEKEE

total

total

0 are PE and KE rest, at system a for

Internal Energy , U: energy associated with the chemical system depends on the thermodynamic parameters such as T, P, V, composition, etc. consists of translational, rotational, vibrational, electronic energies,

as well as intermolecular interactions

Internal Energy, U:

For a given system, we do not know the exact nature of U, we will be only interested in the changes in U for a particular process that the system undergoes.

12 UUU

Mathematical expression of the first law of thermodynamics:

wqU or for an infinitesimal change:

dwdqdU

The change in the internal energy of a system in a given process is the sum of the heat exchange, q, between the system and its surroundings and the work done on or by the system.

HEAT and WORK sign conventions:

Process Sign

Work done by the system on the surroundings

-

Work done on the system by the surroundings

+

Heat absorbed by the system from the surroundings

+

Heat absorbed by the surroundings from the system

-

Constant volume adiabatic bomb calorimeter

Adiabatic means no heat exchange with the surroundings

Sample inside an oxygen filled container is ignited by an electrical discharge.

Heat released is measured by the increase in the temperature of the water.

v

v

v

qU

VPq

wqU

V

0 VP constant at ,

,

ENTHALPHY

In processes carried out under constant P,

PVUHEnthalpy

PVUPVUq

VVPqUU

orVPwqU

p

p

v

or

,

)()(

),(

,

1122

1212

Total enthalpy of a system can not be measured directly, so the change in enthalpy, ΔH, is a more useful value than H itself.

For a constant P process;

VPdUdH

VPUH

change, simalinfiniteti for

Entalphy, H vs. Internal Energy, U

Consider the following reaction:

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

Heat evolved by the reaction is 367.5 kJ, reaction takes place at constant pressure,

qp= H = -367.5KJ.

Internal Energy U:

The volume of the H2(1 mole) generated by the reaction occupies 24.5 L, therefore –PV = -24.5 L at or -2.5 kJ Finally,

U = -367.5 kJ – 2.5 kJ

U = -370 kJ.

- Difference is due to expansion work.

VPHU

Constant Pressure Calorimeter:

A constant-pressure calorimeter made of two plastic cups. The outer cup helps insulate the reacting mixture from the surroundings. Two solutions of known volume containing the reactants at the same temperature are carefully mixed in the calorimeter. The heat produced or absorbed by the reaction can be determined by the temperature change, the quantities and specific heats of the solutions used, and the heat capacity of the calorimeter.

HEAT CAPACITIES:

Calorimetry: measurement of heat changes in chemical and physical processes.

When heat is added to the system, the corresponding temperature rise will depend ona)The amount of heat delivered,b)The amount of the substance,c)The chemical nature and the physical state of the substance,d)The conditions at which the energy is added to the system.

HEAT CAPACITIES:

Thus for a given amount of a substance, change in temperature is related to the heat added:

capacity heat molaris C mol K J

capacity heatis C K J

1-1-

1-

,

,

,

Tn

qC

n

CT

qC

orTCq

HEAT CAPACITY at constant volume

TCnTTCdTCU

dTCdUT

U

T

qC

vv

T

T v

v

vv

)( 12

2

1

or

q U volume, constant at v

HEAT CAPACITY at constant pressure processes

TCnTTCdTCH

dtCdHT

H

T

qC

pp

T

T p

p

pp

)( 12

2

1

or

q H process, pressure constant at p

for gases, Cp > Cv because of the work to be done to the surroundings in constant pressure processes

for condensed phases, Cp and Cv are identical for most purposes

for ideal gases, RCC vp

THERMOCHEMICAL EQUATIONS:

SAMPLE PROBLEM:

How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is –5,500 kJ/mol C8H18?

C8H18 + 25/2 O2 8CO2 + 9H2O

THERMOCHEMISTRY:

Heat of reaction is the heat change in the transformation of reactants at a given temperature and pressure to products at the same conditions.

For constant pressure processes, the heat of reaction qp is equal to enthalpy change of the reaction, rH.

positive.is H gs.surroundin the from

heatabsorbs system :reactions cEndothermi

negative.is H gs.surroundin

to heat offgives system :reactions rmicExothe

r

r

THERMOCHEMISTRY:

2H2(g) + O2(g) 2H2O (g)

241.8 kJ of heat is given off. The enthalpy change for this process is called standard enthalphy of reaction, rHo

In general, standard enthalpy change of a chemical reaction is the total enthalpy of the products minus the total enthalpy of the reactants

tcoefficien tricstoichiomeis v

enthalpy; molar standard theis H

)reactants(Hv)products(HvHo

ooor

THERMOCHEMISTRY:

)tstanreac(Hv)products(HvH

general; in or

))B(Hb)A(Ha()D(Hd)C(HcH

:is reaction the of enthalpy standard the

dDcCbB Aa

of

of

or

ooooor

formation. of enthalpy molar standard theis H of

- is the enthalpy change when 1 mole of a compound is formed from its constituent elements at 1 bar and 298 K.

THERMOCHEMISTRY:

0))g(O(H ;0))g(H(H

0H :etemperatur particular a at

forms allotropic stable most their inelements for

:H

2o

f2o

f

of

of

H2(g) + 1/2O2(g) H2O(g)

kJ/mole 8.241H of

FORMATION REACTIONS:

formation of 1 mole of a compound from its constituent elements at 1 bar and 298 K.

Example: Formation reaction of NaOH:

Na(s) + ½ O2(g) + ½ H2(g) NaOH(s)

MEASUREMENTS OF of H

DIRECT METHOD: measure fHo of direct synthesis from their elements:

2H2(g) + O2(g) H2O(g)

8.241H

))g(O(H))g(H(H2))g(OH(H2H

)tstanreac(Hv)products(HvH

or

2o

f2o

f2o

fo

r

of

of

or

MEASUREMENTS OF of H

INDIRECT METHOD: HESS’S LAW: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

of H for carbon monoxide:

C(graphite) + 1/2O2(g) CO(g)

C(graphite) + O2(g) CO2(g)

CO + 1/2O2(g) CO2(g)

kJ/mole 5.393H or

kJ/mole 0.283H or

MEASUREMENTS OF of H

of H for carbon monoxide:

C(graphite) + O2(g) CO2(g)

CO2(g) 1/2O2(g) + CO(g)

kJ/mole 5.393H or

kJ/mole 0.283H or

C(graphite) + 1/2O2(g) CO(g) kJ/mole 5.110H or

MEASUREMENTS OF of H

C(graphite) + O2(g) CO2(g) + 393.5kJ

kJ/mole 5.393 or HC(graphite) + O2(g) CO2(g)

• Enthalpy changes are additive. • the of the reverse reaction will have the opposite sign• multiplying a reaction by a factor, you also multiply the by the same factor.

or H

or H

Hess’s Law can be rephrased as: ‘ The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which the reaction might be divided.’

SAMPLE PROBLEM:

Calculate the standard molar enthalpy of formation of acetylene (C2H2) from its elements.

kJ/mole 8.2598H or

kJ/mole 5.393H or

2C(graphite) + H2(g) C2H2(g)

The equations for combustion and the corresponding enthalpy changes are:

(1) C(graphite) + 1/2O2(g) CO2(g)

(2) H2(g) + 1/2O2(g) H2O(l)

(3) 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

kJ/mole 8.285H or

SAMPLE PROBLEM:o

f H for carbon monoxide:

C(graphite) + O2(g) CO2(g)

CO2(g) 1/2O2(g) + CO(g)

kJ/mole 5.393H or

kJ/mole 0.283H or

C(graphite) + 1/2O2(g) CO(g) kJ/mole 5.110H or

SAMPLE PROBLEM:

TEMPERATURE DEPENDENCE OF ENTHALPY CHANGE

KIRCHHOFFS’s LAW: The difference in enthalpies of a reaction at two different temperatures, T1 and T2, is just the difference in the enthalpies of heating the products and reactants from T1 to T2.

)( 1212 TTCHH pr r

Where Cp is the difference in molar heat capacities between the products and reactants

Bond Dissociation Enthalpy:

Bond Enthalpy: change in enthalpy when bonds form or break in diatomic molecule

Bond Dissociation Enthalpy: average change in enthalpy in polyatomic molecules when individual bonds form or break.

N2(g) 2N(g)

H2O(g) H(g) + OH(g) rHo = 502 kJ/molOH(g) H(g) + O(g) rHo = 427 kJ/mol

SAMPLE PROBLEM:

ENTHALPY CHANGE and BOND ENERGIES:

The Enthalpy of a Reaction can be estimated by the enthalpies of the total number of bonds broken and formed in the reaction.

released energy total- input energy total

(products))(reactantsr

BEBEH o

SAMPLE PROBLEM:

Estimate the enthalpy of combustion of methane:

CH4(g) + O2(g) CO2(g) + H2O(g)

Using bond enthalpies in Table 4.4. Compare your result with that calculated from the enthalpies of formation of products and reactants.

C-H 410 kJ/moleO=O 494 kJ/moleC=O 563 kJ/moleH-O 460 kJ/mole

The Second Law of Thermodynamics

LIMITATION OF THE FIRST LAW:

Does not address whether a particular process is spontaneous or not.

Deals only with changes in energy.

Consider this examples:

Drop a rock from waist-high height, rock will fall spontaneously

Plunger of a spray is presses , gas comes out spontaneously

Metallic sodium is placed in a jar with chlorine gas, reaction occurs

Spontaneous Processes:

Why does the color spread when placing a drop of dye in a glass of clean water?

SPONTANEOUS PROCESSES:

Spontaneous processes are those that can proceed without any outside intervention.

The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously go back to B.

SPONTANEOUS PROCESSES:

Processes that are spontaneous in one direction are not spontaneous in the reverse direction.

SPONTANEOUS PROCESSES

Processes that are spontaneous at one temperature may be not spontaneous at other temperatures.Above 0C it is spontaneous for ice to melt.Below 0C the reverse process is spontaneous

SPONTANEOUS PROCESSES:

Changes in the extent of disorder.

When we spill a bowl of sugar, why do the grains go everywhere and cause such a mess?

natures’s way to seek disorder. It is easy to create disorder; difficult to create order.

ENTROPY:

Entropy can be thought of as a measure of the randomness or disorder of a system.

It is related to the various modes of motion in molecules.

The Concept of Entropy (S)

Entropy refers to the state of order.

A change in order is a change in the number of ways of arranging the particles and dispersing their energy of motion, and it is a key factor in determining the direction of a spontaneous process.

solid liquid gasmore order less order

crystal + liquid ions in solutionmore order less order

more order less order

crystal + crystal gases + ions in solution

PROCESSES THAT RESULT IN PREDICTABLE ENTROPY CHANGES:

1. Phase Changes

1. Temperature Changes

1. Volume Changes

1. Mixing of substances

1. Increase in number of particles

1. Changes in the number of moles of gaseous components

7. Atomic size or molecular complexity

Spontaneity and the Sign of S

Why does a room with a fragrance bottle at the other end of the room is suddenly filled with the aroma?

Limonene (l) Limonene (g)

S(process) = S (final state) - S(initial state)

A spontaneous process is accompanied by S of positive sign.

Reaction Spontaneity by Inspection

Why do damp clothes become dry when hung outside?

S(g) >> S(l) > S(s)

By inspection alone, decide whether the sublimation of solid carbon dioxide is spontaneous or not. How about the condensation of water?

THERMODYNAMIC DEFINITION OF ENTROPY:

ST

q

V

VlnnR

T

q

V

VnRTln q

process reversible a for

-wq

process, isothermal In

rev

1

2rev

1

2rev

Although for the expansion of gases, they are

applicable to all types of processes at constant

temperature

The Second Law of Thermodynamics:

0SSS

,Thus

0)1V/2Vln(nR

SSS

processes leirreversib for

0T

)V/Vln(nRT

T

)V/Vln(nRTT

q

T

q

SSS

processes; reversible for

:universe the of entropy in Change

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The entropy of an isolated system increases in an irreversible processes and remain unchanged in a reversible process.

ENTROPY CHANGES PHASE CHANGE:

T

HS

T

HS

Hq so,process pressure constant

OHOH

:Consider

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fusfus

fusrev

)l(2atm1,C0

)s(2

o

SAMPLE PROBLEM

The enthalpy of vaporization of methanol is 35.27 kJ/mol at its normal boiling point of 64.1oC. Calculate (a) the entropy of vaporization of methanol at this temperature

SAMPLE PROBLEM:

Calculate the change in entropy when 1.0 mole of ice at -10oC is heated until it is a liquid at 25oC.

mol-J/K 6.33O(g)H C

mol-J/K 3.75O(l)H C

mol-J/K 7.37O(s)H C

mol-J/K 4.109S

molK/J 0.22S

2p

2p

2p

vap

fus

THIRD LAW OF THERMODYNAMICS:

01lnkS

1W K,0 at WlnkS

dTT

CS

zero; to etemperatur the lower we Suppose

b

b

T

0

p

Every substance has a finite positive entropy, but at absolute zero the entropy maybe come zero, and it does in case of pure perfect crystalline substance.

THIRD LAW OF THERMODYNAMICS:

The entropy of a perfect crystal at 0 K is zero.

It is impossible to reach a temperature of absolute zero

It is impossible to have a (Carnot) efficiency equal to 100% (this would imply Tc = 0).

THIRD LAW OF THERMODYNAMICS:

T = 0, S = 0 T > 0, S > 0

ENTROPY OF CHEMICAL REACTIONS:

)tstanreac(Sv)products(SvS

(B)Sb-(A)Sa-(D)Sd(C)ScS

:by givenis reaction of entropy eth

dDcCbBaA

reaction, alhypothetic a For

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SAMPLE PROBLEM:

Calculate the value of the standard molar entropy changes for the following reactions at 298 K.

210.6 :NO 205, :O 191.5, :N

69.9 :OH 205, :O 130.6, :H

213.6 :CO 39.8, :CaO 92.9, :CaCO

mol) (J/K :1bar) K,(298 values S

2NO(g) (g)O (g)N c)

O(l)2H (g)O (g)2H b)

(g)CO CaO(s) (s)CaCO a)

22

222

23

o

22

222

23

The GIBBS Energy:

ST-HG :EnergyGibbs

TS-HG :function a define now we

0TdS-dH

T by multiply

0T

dqdS

0dSdSdS

not? ors spontaneouIs

l)O(gH (g)O(g)H

reaction, the Consider

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22

THE GIBBS ENERGY: Factors affecting

H S G

+ + Positive at low Temp; negative at high Temp; Reaction is spontaneous at forward at high T and spontaneous in reverse direction at low temperature.

+ - Positive at all temperatures, Reaction is spontaneous in the reverse reaction at ll temperatures

- + Negative at all temperatures. Reaction is spontaneous in the forward direction at all T.

- - Negative at low temperatures; positive at high temperatures. Reaction is spontaneous at Low temperatures. Tends to reverse at high temperatures.

The HELMHOLTZ Energy:

0A

:mequilibriu and yspontaneit for criteria

processes. volume and etemperatur constant for

Energy Helmholtz

TS-UA

sys

STANDARD MOLAR GIBBS ENERGY OF FORMATION

)tstanreac(Gv)products(GvG

(B)Gb-(A)Ga-(D)Gd(C)GcG

:by givenis reaction of EnergyGibbs molar standard the

dDcCbBaA

reaction, alhypothetic a For

ooor

ooooor

GIBBS FREE ENERGY, GGIBBS FREE ENERGY, G

∆ ∆GGoo = ∆H = ∆Hoo - T∆S - T∆Soo

Two methods of calculating ∆GTwo methods of calculating ∆Goo

A. Determine ∆HA. Determine ∆Hoorxnrxn and ∆S and ∆Soo

rxnrxn and use GIbbs equation. and use GIbbs equation.

B. Use tabulated values of B. Use tabulated values of free energies of formation, ∆Gfree energies of formation, ∆G ffoo..

∆∆GGoorxnrxn = = ∆G ∆Gff

oo (products) - (products) - ∆G ∆Gffoo (reactants) (reactants)∆∆GGoo

rxnrxn = = ∆G ∆Gffoo (products) - (products) - ∆G ∆Gff

oo (reactants) (reactants)

FREE ENERGIES OF FORMATIONFREE ENERGIES OF FORMATION

Note that ∆G˚Note that ∆G˚ff for an element = 0 for an element = 0

SAMPLE CALCULATION,SAMPLE CALCULATION, ∆G ∆Goorxnrxn

For the combustion of acetyleneFor the combustion of acetylene

CC22HH22(g) + 5/2 O(g) + 5/2 O22(g) --> 2 CO(g) --> 2 CO22(g) + H(g) + H22O(g)O(g)

a)a) by inspection is the reaction spontaneous or not?by inspection is the reaction spontaneous or not?

b)b) Calculate the Calculate the ∆G∆Goorxnrxn using standard molar enthalpies and entropies. using standard molar enthalpies and entropies.

c) Is the reaction spontaneous or not? Is it entropy or enthalpy driven?c) Is the reaction spontaneous or not? Is it entropy or enthalpy driven?

∆∆H (acetylene) 52.26 s = 219.56 g = 68.15H (acetylene) 52.26 s = 219.56 g = 68.15

O g = 205.14O g = 205.14

Co2 h=-393.51 s = 213.74 g = -394.36Co2 h=-393.51 s = 213.74 g = -394.36

H20 h = -241.82 s 188.83 g = -228.57H20 h = -241.82 s 188.83 g = -228.57

FREE ENERGY AND TEMPERATUREFREE ENERGY AND TEMPERATUREIron metal can be produced by reducing its ore (Iron(III) oxide with Iron metal can be produced by reducing its ore (Iron(III) oxide with

graphite:graphite:

2 Fe2 Fe22OO33(s) + 3 C(s) ---> 4 Fe(s) + 3 CO(s) + 3 C(s) ---> 4 Fe(s) + 3 CO22(g)(g)

∆∆HHoorxnrxn = +467.9 kJ = +467.9 kJ ∆S∆Soo

rxnrxn = +560.3 J/K = +560.3 J/K

∆∆GGoorxnrxn = +300.8 kJ = +300.8 kJ

A) Is the reaction spontaneous or not?A) Is the reaction spontaneous or not?

B) At what temperature will the reaction become spontaneous?B) At what temperature will the reaction become spontaneous?At what T does ∆GAt what T does ∆Goo

rxnrxn just change from being (+) to being (-)? just change from being (+) to being (-)?

When ∆GWhen ∆Goorxnrxn = 0 = ∆H = 0 = ∆Hoo

rxnrxn - T∆S - T∆Soorxnrxn

FACT: ∆GFACT: ∆Goorxnrxn is the change in free energy when pure is the change in free energy when pure

reactants convert COMPLETELY to pure products.reactants convert COMPLETELY to pure products. FACT: Product-favored systems have FACT: Product-favored systems have

KKeqeq > 1. > 1.

Therefore, both ∆G˚Therefore, both ∆G˚rxnrxn and K and Keqeq are related to reaction are related to reaction

favorability.favorability.

Thermodynamics and KThermodynamics and Keqeq

KKeqeq is related to reaction favorability and so to ∆G is related to reaction favorability and so to ∆Goorxnrxn..

The larger the value of K the more negative the value The larger the value of K the more negative the value of ∆Gof ∆Goo

rxnrxn

∆ ∆GGoorxnrxn = - RT lnK = - RT lnK

where R = 8.31 J/K•molwhere R = 8.31 J/K•mol

THERMODYNAMICS AND KTHERMODYNAMICS AND Keqeq