chapter 3 thermodynamics

Chemical Thermodynamics

Chapter 3

2

LESSON OUTCOMES At the end of this lecture, students

should be able to:1. Recognize and use thermodynamic

terms: system, surrounding, heat, work, internal energy, Gibbs Free energy, enthalpy and entropy

2. Describe the first and second law of thermodynamics

3. Calculate the change in internal energy


Thermochemistry

SCOPE

Thermochemistry

Forms of Energy and Their InterconversionEnthalpy: Heats of Reaction and Chemical ChangeCalorimetry: Laboratory Measurement of Heats of

ReactionStoichiometry of Thermochemical EquationsHess’s Law of Heat SummationStandard Heats of Reaction

http://www.chem.ed.ac.uk/bunsen_learner/lab_page.html

All changes in matter - accompany by changes in the energy/heat content of the matter.

1. Snow melts - energy is absorbed2. H2O vapor condenses to rain - energy is release

Thermodynamics: the study of heat & its transformation.Thermochemistry: branch of thermodynamics, deals with heat involved in chemical reaction.

3.1 Thermochemistry

Energy

Forms of Energy and Their Inter conversion

All energy (potential or kinetic) - convertible from one to the other

An object has:

(i) potential energy by virtue of its position(ii) kinetic energy by virtue of its motion

The System and Its Surroundings

System must be defined - to make a meaningful observation & measurement of a change in energy.

System: part of universe whose change to be observed.

Surroundings: everything relevant to the change of the system

8

System refers to the particular chemical species being studied

Surroundings are everything else Universe refers to the system and the

surroundings Boundary defined as region across which heat

flows

The System and Its Surroundings

The system and Its Surroundings

Contents of the flask: systemFlask & the laboratory: surroundings

Sum of potential and kinetic energy for all the particles in the system = internal energy, E

Change or difference in internal energy, E = difference between the system’s internal energy after the change (Efinal) and before the change (Einitial)

E = Efinal - Einitial = Eproducts – Ereactants

A reacting chemical system can change its internal energy in either of two ways by:

(i) losing some energy to surroundings: Efinal Einitial

(ii) gaining some energy from surroundings: Efinal > Einitial

Change in internal energy - a transfer of energy from system to surroundings & vice versa

Energy Flow to and from a System

Energy diagrams for the transfer of internal energy (E) between a system and its surroundings

Heat and Work: Two Forms of Energy Transfer

Energy transfer outward from the system or inward from the surroundings can appear in two forms; heat & work

Heat, q (or thermal energy)

energy transferred between a system and its surroundings

as a result of a difference in temperatures between the system and surroundings

Work, wenergy transferred when an object is moved by a force

Total change in a system’s internal energy:

E = q + w (1)

q & w can be either +ve/-ve:

energy into the system: +veenergy out from the system: -ve

Heat and Work: Two Forms of Energy Transfer

Examples of energy transfer as heat onlyA system that does no work but transfers energy

as heat (q), work, w = 0. E = q

Hot water in a beaker - Heat out from a system System: hot water

q = -ve as heat was lost by the system. E = -ve

Ice water in a beaker - Heat into a system System: ice water

q = +ve as heat was gained by the system. E = +ve

Example of energy transfer as work onlyHeat , q = 0. E = w

Push piston down - Work done on a systemIf the external pressure on the piston is increased, the

work is done on the system by the surroundings. System gains energy, w = +ve, E = +ve

Work done on system E = +ve

Work done by a system

- Reaction between Zn & HCl in an insulated

container attached to a piston-cylinder assembly

- System: atoms making up the substances

Zn (s) + 2HCl (aq) H2(g) + Zn2+ (aq) + 2Cl-

(aq)

- As the H2 gas forms, the system used some of its

internal energy on the surroundings and push the

piston outward.

- Energy is lost by the system as work, w = -ve,

E = -ve

3.1.1 Energy Example of energy transfer as work only

q = +ve : system gains heat

q = -ve: system loses energy

w = +ve: work done on system

w = -ve: work done by system

3.1.1 Energy

The Law of Energy Conservation / The First Law of Thermodynamics

Energy

- can be converted from one to another

- cannot simply appear or disappear

- cannot be created or destroyed

First Law of Thermodynamics: The total energy

of the universe is constant

3.1.1 Energy

The Law of Energy Conservation / The First Law of Thermodynamics

Energy of the system + energy of the surroundings remains constant: energy is conserved.

A mathematical expression of the first law of thermodynamics

Euniverse = Esystem + Esurroundings = 0

3.1.1 Energy

Units of Energy

SI unit = Joule (J)

1 cal (non SI unit) = 4.184 J

Heat, work, potential energy, kinetic energy are

expressed in Joules

In the case of work:

- Work, w = Force (F) x distance (d)

- where F = mass (m) x acceleration (a) in units of

- w = Jsmkg

mxsmkg

2

2

2

2smkg

3.1.1 Energy

ExampleWhen gasoline burns in a car engine, the

heat released causes the products CO2 and H2O expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system.

If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (E) in J, kJ and kcal.

SolutionDefine the system and surroundings

System - reactants and productsSurroundings - pistons, cooling system & rest of

the carHeat released by the system, q = -ve, work

done by the system to push the piston outward, w = -ve

Calculating E in J:q = -325 J, w = -451 J, E = q + w

= -325 J + (-451J) = -776 J = -0.78 kJ= -185.47 cal = 0.19 kcal (1cal =

4.184 J)

State Functions and the Path Independence of the Energy Change

System’s internal energy, E - state of function

property determined by the current state of the system

not dependent on the path the system took to

reach the state

Changes in state function (such as E, P, V) depend only on the initial and final states of the system

For a given change, E (sum of q and w) is constant, even though the specific values of q and w can vary

Heat, q and work, w are not state function their values depend on the path the system takes in undergoing the energy change

Note: Symbols for state functions (such as E,P,V) are capitalized

State Functions and the Path Independence of the Energy Change

Enthalphy is a thermodynamic property of a

system.

Enthalpy is defined by H = E + PV

Thus, change in enthalphy, H = E + PV

(eq. 1)

Total change in a system’s internal energy

under conditions of constant pressure, E = qp

+ w (eq. 2)

Substituting eq. 2 into eq. 1,

® H = qp + w + PV (eq. 3)

3.1.2 Enthalpy

In Chemistry, we are most often concern with the

reactions/systems that occur in solution under

conditions of constant atmospheric pressure. Hence

the system is free to expand or contract and this

means that it potentially do work. Thus work done by

the system, w = PV (eq. 4)

Substituting eq. 4 in eq. 3,

® H = qp (eq. 5)

® Thus, the heat of reaction at constant

pressure is equal to H (change in

enthalpy)

3.1.2 Enthalpy

Comparing E and H

For many reactions, H is equal / very close to E.

Three cases:

(i) Reactions that do not involve gases

- e.g. 2KOH(aq) + H2SO4(aq) K2SO4(aq) + 2H2O(l)

- Liquids & solids undergo very small volume

changes,

- V 0, PV 0, H E

3.1.2 Enthalpy

(ii) Reactions in which the amount (mol) of gas does not change. Total amount of gaseous reactants = total amount of gaseous products

- e.g. N2(g) + O2(g) 2NO(g), V = 0, PV = 0, H E

(iii) Reactions in which the amount (mol) of gas does change. PV 0, qp is usually much larger than PV

- From eq. 1, H = E + PV, H = qp = E + PV

- E = qp PV = H, E qp H, E H

3.1.2 Enthalpy

Comparing E and H

Exothermic and Endothermic Processes

Enthalphy is a state function

The enthalphy change of reaction, H, is

also called the heat of reaction, Hrxn.

H = Hfinal - Hinitial = Hproducts –

Hreactants

3.1.2 Enthalpy

Exothermic and Endothermic Processes

An exothermic (“heat out”)- process releases heat- results in a decrease in the enthalphy- Exothermic: Hfinal Hinitial, H 0/-ve

An endothermic (“heat in”)- process absorbs heat - results in an increase in the enthalphy- Endothermic: Hfinal Hinitial, H 0/+ve

3.1.2 Enthalpy

Example 1

In each of the following cases, determine the sign of H, state

whether the reaction is exothermic or endothermic(a) H2(g) + 1/2 O2(g) H2O(l) + 285.8 kJ

(b) 40.7 kJ + H2O(l) H2O(g)

Solution(a) Heat is a product (on the right), so H 0 and the

reaction is exothermic(b) Heat is a reactant (on the left), so H 0 and the

reaction is endothermic

Some Important Types of Enthalpy Change

(i) Heat of combustion (Hcomb): when 1 mole of substance reacts with O2 in a combustion reactione.g. C4H10(l) + 13/2 O2(g) 4CO2(g) + 5H2O, H = Hcomb

(ii)Heat of formation (Hf): when 1 mole of

compound is produced from its elemente.g. K(s) + 1/2 Br2 (l) KBr(s), H = Hf

(iii)Heat of fusion (Hfus): when 1 mole of a substance meltse.g. NaCl(s) NaCl(l), H = Hfus

(iv)Heat of vaporization (Hvap): when 1 mole of substance vaporize

Calorimetry: Measurement of Heats of Reaction

Enthalphy of a system in a given state - cannot be measured

Change in enthalphy - can be measured

Specific Heat Capacity

Quantity of heat (q) absorbed by an object is proportional to its temperature change

q T or q = constant x T or q/T = constant

Calorimetry: Measurement of Heats of Reaction

Specific Heat Capacity

Heat capacity, C = q/T = quantity of heat required to change the temperature of substance by 1K. (Unit: J/K)

Specific heat capacity, c = q/m T = quantity of heat required to change the temperature of 1 gram of a substance by 1K (Unit: J/g.K)

Heat absorbed or released, q = c x mass x T

Example

A welded layer of copper on the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 250C to 3000C?

The specific heat capacity (c) of Cu is 0.387 J/gK.

T = Tfinal - Tinitial = 3000C - 250C = 2750C = 275 K

q = c x mass (g) x T = 0.387 J/gK x 125 g x 275 K

= 1.33 x 104 J

Solution

The Practice of Calorimetry

Calorimeter – an equipment used to measure the heat released/absorbed by a physical or chemical process, two common types:

constant-pressure & constant-volume calorimeters

Constant-Pressure CalorimetryA coffee/styroform-cup calorimeterUse to measure the heat (qp) of many processes

that are open to the laboratory atmosphereCommon use - to determine the specific heat

capacity of a solid (does not react/dissolve in water)

Process:solid (system) is

weighed, heated to some known temperature, added to a sample of water (surroundings) of known temperature & mass in the calorimeter

Continual stirring which distribute the released heat

Final water temperature (final temperature of the solid) is measured

Heat lost by the system (-q) is equal in magnitude but opposite in sign to the heat gained (+q) by the surroundings:-qsolid = qH2O, substituting in equation

-(csolid x masssolid x Tsolid) = cH20 x massH20 x TH20

All the quantities are known except csolid

Example

A 25.64 g sample of solid was heated in a test tube to 100.000C in boiling water and carefully added to a coffee-cup calorimeter containing 50.00 g water. The water temperature increased from 25.100C to 28.490C. What is the specific heat capacity of the solid? (Assume all the heat is gained by the water)

Solution Plan -summarize the information

given.Calculating csolid

Kg

J0.387

K)71.51(xg25.64

K3.39xg50.00xKgJ

4.184

ΔTxmass

ΔTxmassxcc

solidsolid

OHOHOHsolid

222

Constant-Volume Calorimetry

Common type: bomb calorimeter

Use to measure very precisely the heat released in combustion reaction

Known mass of the sample and the heat capacity of the calorimeter, the measured T is used to calculate the heat released

Example 4

A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories per serving”. To test the claim, a chemist at the Department of Consumers Affairs places one serving in a bomb calorimeter and burns it in O2. The temperature increases by 4.9370C. Is the manufacturer’s claim correct? (Heat capacity of the calorimeter = 8.151 J/K)

Solution When the dessert burns, the heat

released is gained by the calorimeter: -qsample = qcalorimeter

Heat is found by mutiplying the heat capacity of the calorimeter by T

qcalorimeter = heat capacity x T = 8.151 J/k x

4.937 K = 40.24 J = 9.62

cal (1cal = 4.184 J)

Stoichiometry of Thermochemical Equations

Thermochemical equationsa balanced equation that states the heat of reaction

(Hrxn)Note: Hrxn value refers to the enthalphy change for

the amounts of substances in that specific reaction

Two aspects of enthalphy change:

(i) SignSign of H depends on the reaction of the changeSign of a forward reaction is opposite that of the

reverse reaction e.g. Decomposition of 2 mol water (endothermic): 2H2O(l) 2H2(g) + O2(g) Hrxn = 572 kJ

Formation of 2 mol water (exothermic): 2H2(g) + O2(g) 2H2O(l) Hrxn = -572 kJ

(ii) MagnitudeMagnitude of H is proportional to the amount of

substances in the reaction e.g. Formation of 1 mol water: H2(g) + 1/2 02(g) H20(l) Hrxn =

-286 kJ

Formation of 2 mol water 2H2(g) + O2(g) 2H2O Hrxn = -

572 kJ

Hess’s Law of Heat SummationHess’s law of heat summation: the enthalpy

change of an overall process is the sum of the enthalpy changes of its individual steps

Hess Laws - the difference between the enthalpies of the reactant and the product has the same value even though the reaction takes place differently

Calculating an unknown H involves three steps:(i) identify the target equation, note the number of moles of reactants and products.

(ii) manipulate the equations of known H - the target numbers of moles of reactants and products are on the correct sideschange the sign of H when reversing an

equationmultiply number of moles and H by the

same factor

(iii) Add the manipulated equations to obtain the target equation. (cancel terms that appear on both sides of equations). Add their H values to obtain the unknown H

Example 5Application of Hess’s Law:

Oxidation of sulfur to sulfur trioxideEq. 1: S(s) + O2(g) SO2(g) H1 = -296.8 kJ

Eq. 2: 2SO2(g) + O2(g) 2SO3(g) H2 = -198.4 kJ

Eq. 3: S(s) + 3/2 O2(g) SO3(g) H3 = ?

Manipulate equation 1 and/or 2 so that they add up to equation 3identify eq. 3 as a target, carefully note the

number of moles of reactants and productseq. 1 and 3 contain the same amount of S, so

leave eq. 1 unchanged

Eq. 2 has twice as much SO3 as Equation 3, so multiply it by 1/2, multiply H2 by 1/2 as well

Add eq. 1 to the halved eq. 2, cancel terms that appear on both sides:

Eq. 1: S(s) + O2(g) SO2(g) H1 = -296.8 kJ

1/2(Eq. 2): SO2(g) + 1/2 O2(g) SO3(g) 1/2(H2) =

-99.2 kJ

Eq. 3: S(s) + O2(g) + SO2(g) + 1/2 O2(g) SO2(g) + SO3(g)

S(s) + 3/2 O2(g) SO3(g)

H3 = H1 + 1/2 (H2) = -296.8 kJ + (-99.2 kJ) = -396.0 kJ

Formation Equations and Their Standard Enthalpy

In a formation equation, 1 mol of compound forms from its elements

Standard heat of formation (H0f) - enthalpy

change accompanying the formation equation when all the substances are in their standard states.

e.g. C(graphite) + 2H2(g) CH4(g) H0

f = -74.9 kJ

Example 6

Write balanced equations for the formation of 1 mole of the following compounds from their elements in their standard states and include H0

f (To obtain values, refer to Table 6.5 - page 244, Chemistry: the molecular nature of matter and change, 2nd Edition, Silberberg)(i) Silver chloride, AgCl, a solid at standard conditions

(ii) Calcium carbonate, CaCO3, a solid at standard conditions(iii) Hydrogen cyanide, HCN, a gas at standard conditions

Solution

(i) Ag(s) + 1/2 Cl2(g) AgCl(s) H0f = -127.1 kJ

(ii) Ca(s) + C(graphite) + 3/2 O2(g) CaCO3 H0f = -1206.9 kJ

(iii) 1/2 H2(g) + C(graphite) + 1/2 N2(g) HCN (g) H0f = 135.1 kJ

Standard Heats of Reaction (H0

rxn)

Standard states (a set of specified conditions and concentration) are used to compare heats of reaction and other thermodynamic data:(i) Gas: standard state - 1 atm(ii) Substance in aqueous solution: standard state - 1 M(iii) Pure substance (element or compound): standard state: usually the most standard at 1 atm and the temperature of interest (250C)

A right superscript zero indicates when thermodynamics variable has been determined with all substances in their standard states.e.g. Standard heat reaction, H0

rxn, is the H0rxn

measured with all substances in their standard states.

Determining H0rxn from H0

f values of Reactants and Products

By applying Hess’s Law, H0f values can be used to

determine H0rxn for any reaction:

H0rxn = nH0

f (products) - mH0f (reactants)

Suppose we want H0rxn for:

TiCl4(l) + 2H2O(g) TiO2(s) + 4HCl(g)

(Products) (Reactants)

H0rxn = {H0

f [TiO2(s)] + 4H0f [HCl(g)]} - {H0

f [TiCl4(l) + 2H0f

H2O(g)]}

Example 7

Nitric acid, with an annual production of about 8 billion kg, is used to make many products, including fertilizer, dyes and explosive.

The first step in its industrial production is the oxidation of ammonia:

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Calculate H0rxn from H0

f values

SolutionH0

rxn = mH0f (products) - mH0

f (reactants)

= {4H0f [NO(g)] + 6H0

f [H2O(g)]} – {4H0

f [NH3(g)] + 5H0f [O2(g)]}

= [4 mol (90. 3 kJ/mol) + 6 mol (-241.8 kJ/mol)] –

[4 mol (-45.9 kJ/mol) + 5 mol (0 kJ/mol)]

= 361 kJ - 1451 kJ + 184 kJ - 0 kJ

= -906 kJ

Thermodynamics - Chapter Outline

Spontaneous ProcessesEntropy and the Second Law of

ThermodynamicsThe Molecular Interpretation of EntropyCalculation of Entropy ChangesGibbs Free EnergyFree Energy and TemperatureFree Energy and the Equilibrium Constant


Area of chemistry that explores energy relationship

Limitations of First Law Thermodynamics

only tells the amount of heat released/work done by the system & the amount of heat gained/work done on the system

sign of H - not enough to predict whether the reaction will proceed

H - not the only factor governs reactants or products favoured

Reactants or products favoured In addition to H, change in randomness or disorder in chemical reaction is considered

New aspect in discussion of thermodynamics spontaneous processrelated to disorder in chemical reaction


Spontaneous ProcessSpontaneous process: Chemical or physical

change occurs by itself

requires no continuing outside agency / occurs without any outside intervention

A rock at the top of a hill rolls down by itself - occur spontaneously /

naturallyrolls up - not natural process / non

spontaneous: require work

Temperature: have an impact on the spontaneitye.g. endothermic process of melting ice, at T

00C:ice melts spontaneouslyreverse process (liquid water ice), non-

spontaneous

But at T 00C:liquid water converts into ice spontaneouslyconversion of ice liquid water, non

spontaneous

Spontaneous Process

Entropy & the Second Law of Thermodynamics

Entropy, S a thermodynamic quantity - a measure of the

randomness/disorder of a system. SI unit: Joules per Kelvin (J/K) a state function

quantity of S depends only on variables (such as T & P) that determine the state of substance

(i) Solids: have much more regular structure than liquids. Liquids > disordered than solids

(ii) Particles in a gaseous state are in random motion. Gases > disordered than liquids

(iii) Any process that increases the amount of particles, disorder

Consider the melting ice:Ice: an ordered crystalline structureIce melts liquid, crystalline structure breaks down,

resultant a less ordered liquid structureIn ice, H2O molecules occupy regular, fixed

positions; ice has a relatively low SIn liquid water, molecules moves about freely,

disordered structure; liquid water has greater entropy than ice

Entropy change, S

S = Sfinal – Sinitial

S of a system to the disorder of the systemSsystem 0, implies the system becomes more

disorderedSsystem 0, implies the system becomes less

disordered

Molecular Interpretation of Entropy

Structure & behaviour of molecules - affect the entropy of the system

e.g. Entropy of the system increase (S > 0) when:(i) gas molecules spread out in a larger

volume(ii) phase changes: solid liquid, liquid gas

e.g. Entropy of the system decrease (S < 0) when:(i) condensing a gas, (ii) freezing a liquid

A reaction leads to a decrease in the number of

gaseous moleculesgenerally leads to a decrease in entropy

e.g. 2NO(g) + O2(g) 2NO2(g)

entropy change = -ve because 3 molecules of gas react to form 2 molecules of gas

formation of new N-O imposes more order, atoms are more “tied up” in the products than in the reactants

leads to a decrease in the entropy


Formation of the new bonds - decreases the number of degrees of freedom/forms of motion atoms are less free to move in random fashion

In general - the greater the no. of freedom of system, the greater its entropy

The degrees of freedom of molecules - associate with three types of motion for the molecule:

(i) Translational motion(ii) Vibrational motion(iii) Rotational motion


In general - entropy increases with increasing temperature

Entropy of the phases of a substance:Ssolid < Sliquid < Sgas

In general, the entropy is expected to increase when:

(i) Liquids or solutions are formed from solids

(ii) Gases are formed from either solids or liquids

(iii) No. of molecules of gas increases during a chemical reaction


Calculation of Entropy Change

Standard molar entropies, S0 - molar entropy values of substances in their standard states

Entropy change in a chemical reaction, S0:

= “sum of”n = amounts (mol) of the productsm= amounts (mol) of the reactants

reactantsS Σm-productsnSΣ=SΔ ooo

Calculate S0 for the synthesis of ammonia from N2(g) and H2(g):

N2(g) + 3H2(g) 2NH3(g)

SolutionS0 = 2S0(NH3) - [S0(N2) + 3S0(H2)]

Substituting the appropriate S0 values

KJ

198.3-

K molJ

130.6mol 3K mol

J 191.5mol 1

K molJ

1925mol 2 ΔS

Example 8

Choose the substance that has greater entropy in each pair, and explain your choice:

(a) 1 mol of HCl(s) or 1 mol of HCl(g) at 250C(b) 2 mol of HCl(g) or 1 mol of HCl(g) at 250C(c) 1 mol of N2(s) at 24 K or 1 mol of N2(g) at 298 K

Solution(d) Gaseous HCl has the higher entropy because

gases are more disordered than solids(e) The sample containing 2 mol of HCl has twice the

entropy of the sample containing 1 mol(f) The gaseous N2 sample has the higher entropy

because gases are more disordered than solids

Example 9

Second Law of Thermodynamics

Second Law of Thermodynamics: total energy of a system and its surroundings always increases for a spontaneous process

Note: the difference between the entropy & energyenergy - cannot be created or destroyed during

chemical changeentropy-created during a spontaneous or natural

process

• For a spontaneous process carried out at a given temperature

The second law can be restated in a form that refers only to the system

Consider the changes in entropy that occur in the systemoprocess takes place entropy is created, at the same time heat flows into/out of the system

oentropy accompanies that heat flow


When heat flow into the system entropy flows into the system

The change in entropy, S, of the system at a given temperature:

quantity of entropy created during spontaneous process cannot be directly measured

quantity of entropy created = +ve, by deleting it from the right side of equation:S > q/T for a spontaneous process

Tq

createdentropy S Δ


The restatement of the second law: for a spontaneous process at a given temperature, the change in entropy of the system is greater than the heat divided by the absolute temperature,


Tq

Entropy Change for a Phase Transition

Certain processes occur closely at equilibrium. Under equilibrium conditions, no significant amount

of entropy is created. The S results only from the absorption of heat.

Thus, the change in entropy:

(equilibrium process, Eq.1)

Examples of phase changes under equilibrium: vaporization of a liquid fusion of a solid

Tq

SΔ

Eq.1 - can be used to obtain the entropy change for a phase changee.g. Consider the melting of ice, Hfus 6.0 kJ (1 mol

of ice)heat absorbed = heat of fusion, Hfus

entropy change ,

where T = absolute temp. of the phase transition, 273K

Note: S - usually express in joules per Kelvin

Tq

=ΔS

THΔ

=S fusΔ

Entropy Change for a Phase Transition

The heat vaporization, Hvap, of carbon tetrachloride, CCl4, at 250C is 43.0 kJ/mol.

CCl4(l) CCl4(g); Hvap = 43.0 kJ/mol

If 1 mol of liquid CCl4 at 250C has an entropy of 214 J/K, what is the entropy of 1 mol of the vapor in equilibrium with the liquid at this temperature?

Example 10

Solution

1 mol of CCl4 increases in entropy by 144 J/K when it vaporizes.

Thus, entropy of 1 mol of CCl4 at equilibrium = (214 J/K) + 144 J/K = 358 J/mol.K

K mol

J 144=

K 298mol

J10×43.0

=T

ΔH=ΔS

3

vap

How thermodynamics is applied to the question of whether a reaction is spontaneous?Example:assume a reaction occurs at constant T and P:2NH3(g) + CO2(g) NH2CONH2(aq) + H2O(l)

Is the reaction spontaneous? Does it go forward as written?

If H & S for the reaction are known second law in the form can be used to answer the above question.

The second law for a spontaneous reaction at constant P:

Tq

=ΔS

TH

> S

TH

=T

q>S p

ΔΔ

ΔΔ

(spontaneous reaction, constant T and P)prediction: reaction is spontaneous left to right as

written

prediction: reaction is non-spontaneous left to right as writtenreaction is spontaneous in the opposite direction

reaction is at equilibrium

(-ve) 0SΔT-HΔ

0 SΔ-T

HΔ

ve)( 0SΔT-HΔ

0=ST-H ΔΔ

Gibbs Free Energy X J Willard Gibbs (1839 -1903) - proposed a way to

use H & S to predict whether a given reaction will be spontaneous

Gibbs proposed a new state function Gibbs free energy (or just free energy)

Gibbs free energy, G = H - TS, T = absolute temp.

At constant T, the change in free energy of the system, G = H - TS

If T & P are constant, the relationship between the sign of G and the spontaneity of a reaction:

(i) If G = -ve, reaction is spontaneous in the forward reaction

(ii) If G = +ve, reaction in the forward reaction is non

spontaneous, work must be supplied from the surroundings to make it occur

reverse reaction will be spontaneous

Gibbs Free Energy

Standard Free Energy Change Standard free energies of formation, Go

f ,are useful in calculating the standard free energy change for chemical process

= “sum of”, m = amounts (mol) of reactants

n = amounts (mol) of products

Quantity of G0 tells whether a mixture of reactants and products (under standard conditions):(i) would spontaneously react in the forward

direction to produce more products G0 < 0(ii) or in the reverse direction to form more

reactants G0 > 0

)(reactantsΔG Σm(products)ΔG ΣnΔG of

of

of

(a) Calculate the standard free-energy change for the following reaction at 298 K:N2(g) + 3H2(g) 2NH3(g)

(b) What is G0 for the reverse of the above reaction?

Solution(a)

(b) 2NH3(g) N2(g) + 3H2(g), G0 = +33.32 kJ

kJ 33.32-

molkJ

0 mol 3molkJ

0 mol 1molkJ

16.66- mol 2

HG mol 3NΔG mol 1NHΔG mol 2 ΔG 2of2

of3

of

o

Δ

Example 11

Free Energy and TemperatureConsider equation :

G = H (enthalpy term) - TS (entropy term)

Generally H & S change very little with T. However, the value of T directly affects the magnitude of -TS

As T increases, the magnitude of the term -TS increases. It will become relatively more important in determining the sign & magnitude of G.

Example: melting of ice liquid water (P = 1 atm) H2O(s) H2O(l), H > 0, S > 0

endothermic process, H = +veentropy increases during process, S = +ve, -TS =

-ve

At T < 00C, magnitude of H > magnitude of S+ve enthalpy term dominates, leading to a +ve value of G.Melting of ice is not spontaneous at T < 00C, reverse process (liquid water to ice) is spontaneous at T < 00C

When T > 00C, magnitude of -TS > magnitude of H-ve entropy term dominates, leading to a negative value of GMelting of ice is spontaneous at T > 00C

At normal melting point of water, T = 00C, & the two phases are in equilibrium.At T = 0, H & -TS are equal in magnitude. G = 0

Under standard conditions (at 250C) G0 = H0 - TS0

The Haber process for the production of ammonia involves the following equilibrium: N2(g) + 3H2(g) → 2NH3(g)

Assume that H0 and S0 for this reaction do not change with temperature

(a) Predict the direction in which G0 for this reaction changes with increasing temperature

(b) Calculate G0 for the reaction at 5000C

Example 12

Solution( )a G0 become less negative (or more positive)

with increasing temperature. Thus the driving force for the production of NH3 becomes smaller with increasing temperature

( )b G0 = H0 - T S0

T = 500 + 273 = 773 K

kJ 61

kJ 153 kJ 92.38-

kJ 10kJ 1

KJ

198.3- K 773 - kJ 92.38G 3o

Δ

The End

chapter 3 thermodynamics

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