TOPIC 8: THERMOCHEMISTRY

TOPIC 8: THERMOCHEMISTRY

Thermochemistry is the branch of chemistry dealing with determining quantities of heat by measurement and calculation. Some of these calculations will allow us to establish indirectly, a quantity of heat that would be difficult to measure directly.

Some Main Concepts in Thermochemistry

Opensystem

Closedsystem

Isolated system

System: The part of the universe we choose to study.Surroundings: The parts of the universe with which the system interacts. Systems are classified into 3 groups:Open system: Free transfer of energy and matter between the system and its surroundings.Closed system: Only free transfer of energy between the system and its surroundingsIsolated system: No possible transfer of energy and or matter between the system and its surroundings.

Some main Concepts in ThermochemistryEnergy transfer can occur as heat (q) or in several other forms, known as work (w). Heat: The form of energy which is transfered from a system with higher temperature to a system with low temperature due to the temperature difference.Work: The effect that causes the change in position of a body by exerting a force on it. Energy: The capacity of a system to perform a work.Energy transfers occuring as heat and work affect the total amount of energy contained within a system, its internal energy (E). The components of internal energy of special interest to us are thermal energy and chemical energy. Thermal energy: Energy associated with random molecular motion.Chemical energy: Energy associated with chemical bonds and intermolecular forces.

EnergyKinetic energy (KE): The energy of a moving object. m: mass, v: velocity

2

21 vmKE

Energy Unit:

2

2

)(smkgJulJ

Kinetic energy

Potential energy: The stored energy or «energy of position» associated with forces of attraction or repulsion between objects. The stored energy arises from the kinetic energy of molecules and atoms.

hgmPE Potential Energy:

g: acceleration of gravity, 9,81 m/s2

h: height of the matter, mm: mass of the matter,kg

Heat transfer between the system and its surroundings occurs as a result of temperature difference.•Heat moves from the hot environment to the cold environment.

– Temperature is variable.– Change of the state may occur

The heat flux occuring at constant temperature is called ISOTHERMAL PROCESS.

Heat

Units of Heat

• Calorie (cal)– The quantity of heat required to change the

temperature of one gram of water by one degree Celcius.

• Joule (J)– SI unit of work and energy

1 cal = 4.184 J

The quantity of heat energy ,q , depends upon:

How much the temperature is to be changed

The quantity of substance

The nature of substance (type of atoms or molecules)

Heat Capacity

• The quantity of heat required to change the temperature of a system by one degree .– Molar heat capacity:

• The System is one mole of substance– Specific heat (capacity), c.

• The system is 1 g of substance– Heat capacity, C

• Mass x specific heat .

q = mcT

q = CT

Conservation of Energy• In interactions between a system and its

surroundings, the total energy remains constant- energy is neither created nor destroyed.

• Applied to the exchange of heat, this means :

qsystem + qsurroundings = 0

qsystem = -qsurroundings

Thus, heat lost by a system is gained by its surroundings, and vice versa.

Experimental Determination of Specific Heats

Lead

The transfer of energy, as heat, from the lead to the cooler water causes the temperature of the lead decrease and that of the water increase, to the point where both the lead and water are at the same temperature. If we consider lead to be the system and water as the surroundings, we can write :

qlead = -qwater

Experimental Determination of Specific Heats

qlead = -qwater

q water = mcT = (50.0 g water)(4.184 J/g water °C)(28.8 - 22.0)°Cq water = 1.4x103 Jq lead = -1.4x103 J = mcT = (150.0 g lead)(c)(28.8 - 100.0)°C

c lead = 0.13 Jg-1°C-1

Heats of Reaction and Calorimetry

• Chemical Energy. – Type of energy related to the internal energy of the

system.– The energy which occurs as a result of chemical

reactions. The batteries and accumulators are the vehicles which convert the chemical energy into electrical energy. The accumulation of electrical energy in batteries is performed by chemical methods. Chemical energy can be also converted into the mechanical, heat and light energy.

• Heat of reaction, qrxn.– The quantity of heat exchanged between a system and its

surroundings when a chemical reaction occurs within a system at constant temperature and pressure.

Heat of Reaction

• Exothermic reaction.– Gives off heat to the surroundings, qrxn < 0.

– Temperature increase in the system (a)

• Endothermic reaction.– Gain of heat from the surroundings, qrxn > 0.

• Temperature decrease in the system (b)

• Calorimeter– Device for measuring quantities of heat

CaO(s) + H2O(l) → Ca(OH)2 (aq)

Ba(OH)2·8H2O + 2NH4Cl(s)→BaCl2(s) + 2 NH3(aq) + 8 H2O(l)

Bomb Calorimetry

qrxn = -qcalqcal = qbomb + qwater + qwire +…

heat

qcal = miciT = CcalT

Heat capacity of calorimeter

The type of calorimeter shown in the figure is called bomb calorimetry. It is ideally suited for measuring the heat evolved in combustion reaction. It is an isolated system from its surroundings

When the combustion reaction occurs, chemical energy is converted to thermal energy and temperature of the system rises.The heat of reaction qrxn is the quantity of heat that the system would have to lose to its surroundings to be restored to its initial temperature and pressure.

Determination of heat of reaction by the bomb calorimeter

The combustion of 1,010 g sucrose, C12 H22 O11 , in a bomb calorimeter causes the temperature rise from 24,92˚ to 28,33˚C. The heat capacity of the calorimeter assembly is 4,90 kJ/ ˚C

(a) What is the heat of combustion of sucrose (kJ/mol) ?

(b) Verify the claim of sugar producers that one teaspoon of sugar (4,8 g) contains only 19 calories .

Example

Example 7-3

Calculate qcal qcal = CT = (4,90 kJ/°C)(28,33-24,92)°C = (4,90)(3,41) kJ

= 16,7 kJ

Calculate qrxn

qrxn = -qcal = -16,7 kJ

in 1,01 g sample of sugar

Solution

Example 7-3

Conversion of the unit ofqrxn as kJ/mol :

qrxn = -qcal = -16,7 kJ1,010 g

= -16,5 kJ/g

343,3 g1,00 mol

= -16,5 kJ/g

= -5,65 x 103 kJ/mol

qrxn

(a)

For a teaspoon

4,8 g1 ts

= (-16,5 kJ/g)(qrxn (b))( )= -19 cal/teaspoon1,00 kal4,184 J

Solution

(b) Verification of the claim that a teaspon of sugar(4,8 g) has 19 Calories.

The Coffee-Cup Calorimeter

• Easy to handle .– An isolated system composed of styrofoam

cup.– Reactants are mixed up in the cup– Temperature difference is measured at the end

of reaction.– System is at constant pressure.

qrxn = -qcal

qrxn = - Ccal T

WORK

• Apart from transfer of heat, some chemical processes may do work.(the expansion or compression of gases)

• The gas pushes the atmosphere.• The volume changes.

• Pressure-volume work

In thermodynamics work means the transfer of energy between the system and its surroundings due to an external makroscopic force.

Pressure-Volume Work

xFw

VPw Unit of work

JatmL 3,101

cedisforceWork tan

extFF

APFA

FP

ext

ext

xFw ext

xAPw iVVfV

Since work is done by the system , it has a negative sign - .

If the volume increases, work has - sign.

Example 7-3

We assume that He behaves as an ideal gas:

Vi = nRT/Pi

= (0,100 mol)(0,08201 L atm mol-1 K-1)(298K)/(2,40 atm)

= 1,02 L

Vf = nRT/Pf= 1,88 L

Example

Calculation of Pressure-Volume Work.

What is the work done in joules, when 0,100 mol He(g) at 298K expands from an initial pressure of Pi= 2,4 atm to a final pressure Pf= 1,3 atm?

V = 1.88-1.02 L = 0.86 L

Example 7-3

Calculation of work done by the system

w = -PV

= -(1.30 atm)(0.86 L)(

= -1.1 x 102 J

Example

) 101 J1 L atm

Conversion factor:8.3145 J/mol K ≡ 0.082057 L atm/mol K

1 ≡ 101.33 J/L atm

Pressure-Volume WorkExample: What is the quantity of work, in joules, done by the gas in the figure next if it expands against a constant pressure of 0,980 atm and the change in Volume(ΔV) is 25 L .

kJwatmL

JatmLw

LatmwVPw

48,21

3,1015,24

25980,0

First Law of Thermodynamics

• Internal Energy, U.– The TOTAL energy of the system (potential and

kinetic). •Translational kinetic energy.

•Rotations.

•Bond vibrations.

•Intermolecular attractions

•Chemical bonds.

•Electrons.

The First Law of Thermodynamics

• A system keeps the energy only in form of internal energy.– A system does not contain energy in the form of

heat or work.– Heat and work are the means by which the

system exchanges energy with its surroundings.– Heat and work only exist during a change.

• Law of Conservation of energy– The energy of an isolated system is constant.

U = q + w

Uisolated = 0

The First Law of Thermodynamics

System

Any energy entering the system carries + sign.Thus,if heat is absorbed by the system, q>0 .If work is done on the system ,w>0.

Any energy leaving the system carries a – sign.Thus if heat is given off by the system , q<0. If work is done by the system, w<0.

If, on balance more energy enters the system than leaves, ΔU is positive.If more energy leaves than enters, ΔU is negative.

Functions of State

• Any property having a unique value when the state of the system is defined is called function of state.

• Example:• The state of water at 293,15 K and 1,00 atm is specific • In this state d = 0.99820 g/mL • Density just depends on the state of the system.• It does not depend on how it is reached.

Functions of State

• U is a function of state.– Can not be measured.– We do not need to know

the real value.

• The U between U2 and U1 have a unique value.– It can be measured easily.

The value of the internal energy is the value of heat energy given off from the surroundings to the system to reach from the state U1 to U2

State 2

State 1

In

tern

al e

nerg

y

tot

• Heat and work are not functions of state!• The values of heat and work depend upon the path

to be followed for a change in the system.

• 0,1 mol He 298 K, 2,40 atm (State 1) (1,02 L) A↓

298 K, 1,30 atm (State 2) (1,88 L)

Functions depending on the path to be followed

wBC = (-1,80 atm)(1,36-1,02)L – (1,30 atm)(1,88-1,36)L

= -0,61 L atm – 0,68 L atm = -1,3 L atm

= -1,3 x 102 J in contrast;

wA = -1,1 x 102 J

298 K, 1,80 atm (1,36 L)

B

C

Heats of Reaction and Entalpy Change: U and H

Reactants → ProductsUi Uf

U = Uf- Ui

U = qrxn + wIn a system at a constant volume(Bomb

calorimeter):

U = qrxn + 0 = qrxn = qv

However, lots of chemical reactions occur in the earth under constant pressure!

What is the relationship between qp and qv ?

w = -PV= 0

Heat of Reactions

In

tern

al e

nerg

y

First state

Final state Final state

First state

qV = qP + w

Heat of reactions

w = - PV and U = qv :qV = qP + w

U = qP - PV qP = U + PV

P,V,U are functions of stateH = U + PV

H = Hf – Hi = U + PV

At constant pressure and temperature:

H = U + PV = qP

Enthalpy

Comparison of heat of reactions

qP = -566 kJ/mol

= H

PV = P(Vf – Vi)

= RT(nf – ni)

= -2,5 kJ

U = H - PV

= -563,5 kJ/mol

= qV

Constant volume

Constantpressure

heat

heat

2 CO(g) + O2(g) → 2CO2(g)

Example• The heat of combustion at constant volume of

CH4 (g) is measured in a bomb calorimeter at 25˚C and is found to be – 885,389 J/mol. What is the enthalpy change,ΔH ?

• Solution: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

ΔU= - 885,389 JΔn= Σnproducts - Σnreactants 1-(2+1)= - 2 (Note that, the

mole number of H2O is not included in the calculation!!!Since volume change of liquids and solids are too small, it is here neglected.

Example

• ΔH= ΔU + PΔV• ΔH= ΔU + ΔnRT• ΔH= - 885,389 -2x 8,314 J/molK x 298,15K • ΔH=- 885,389 kJ- 4,957 kJ• ΔH=- 890,346 kJNote that the value measured by the bomb

calorimeter is equal to ΔU!!!

Example

• For the reaction ;B2H6(g) + 3 O2(g) B2O3(s) + 3 H2O(l)

ΔU is – 2143,2 kJa) Calculate ΔH for the reaction at 25 ˚Cb) Determine the value of standard enthalpy of

formation of B2H6(g).

For B2O3 (s) ΔHf˚=-1264,0 kJ/mol and for H2O(l) ΔHf˚=-285,9 kJ/mol

Solve the problem by yourself!

Enthalpy during the phase transitions

H2O (l) → H2O(g) H = 44,0 kJ ; 298 K

Molar enthalpy of vaporization:

Molar enthalpy of melting :

H2O (s) → H2O(l) H = 6,01 kJ ; 273,15 K

qp= n*Hphchange

Heat energy during the phase transition:

Example 7-3Example

Think about the solution in two steps: First step: Increase in the temperature of the water. Second step: The vaporization

Enthalpy change during phase transition

Calculate the molar heat of vaporization of 50,0 g sample of the water whose temperature is risen from 25,0°C to 100°C . cwater= 4,184 J/g °C, Hvap = 44,0 kJ/mol

= (50,0 g)(4,184 J/g °C)(100-25,0)°C + 50,0 g18,0 g/molx 44,0 kJ/mol

Solution:qP = mcH2OT + nHvap

= 15,69 kJ + 122 kJ = 137,69 kJ

Hvap = 44,0 kJ/mol

Standard Enthalpies of Formation

• Standard state is the pure substance at 1 atm pressure at the temperature of interest.

• Standard molar enthalpy of formation(molar heat of formation), Hf– The difference in enthalpy between one mole of a

compound in its standard state and its elements in their most stable forms and standard states.

We will use enthalpies of formation to perform a variety of calculations. The first thing we have to do is to write a chemical calculation and then to sketch an enthalpy diagram

Temperature values must be given with H° !!

Enthalpy Diagrams

En

thal

py

Products

Products

Enth

alpy

Reactants

Reactants

Endothermicreactants Exothermic

reactants

Indirect Determination of H :Hess’s Law

H is an Extensive Property .– Enthalpy change is directly proportional to the amounts

of substances in a system.

N2(g) + O2(g) → 2 NO(g) H = +180.50 kJ

½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ (we divide all coeeficients and H value by two)

H changes sign when a process is reversed.NO(g) → ½N2(g) + ½O2(g) H = -90.25 kJ

Hess’s Law

• Hess’s Law of Constant Heat Summation:– If a process occurs in stages or steps (even if only

hypothetically) the enthalpy change for the overall(net) process (Hnet )is the sum of the enthalpy changes for the individual steps (H).

½N2(g) + O2(g) → NO2(g) H = +33.18 kJ

½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ

NO(g) + ½O2(g) → NO2(g) H = -57.07 kJ

Hess’s Law

½N2(g) + ½O2(g) → NO(g)

NO(g) + ½O2(g) → NO2(g)

½N2(g) + O2(g) → NO2(g)

Enth

alpy

• We assign enthalpies of zero to the elements in their most stable forms when in the standard state.

• Most stable forms of the elements are the ones indicated below

• Na(s), H2(g), N2(g), O2(g), C(graphite), Br2(l)

Hf °

Standard Enthalpy of Formation

Hess’s Law

• Hess’s Law of Constant Heat Summation:– If a process occurs in stages or steps (even if only

hypothetically) the enthalpy change for the overall(net) process (Hnet )is the sum of the enthalpy changes for the individual steps (H).

½N2(g) + O2(g) → NO2(g) H = +33.18 kJ

½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ

NO(g) + ½O2(g) → NO2(g) H = -57.07 kJ

Hess’s Law

½N2(g) + ½O2(g) → NO(g)

NO(g) + ½O2(g) → NO2(g)

½N2(g) + O2(g) → NO2(g)

Enth

alpy

• We assign enthalpies of zero to the elements in their most stable forms when in the standard state.

• Most stable forms of the elements are the ones indicated below

• Na(s), H2(g), N2(g), O2(g), C(graphite), Br2(l)

Hf °

Standard Enthalpy of Formation

Standard Enthalpies of Formation

F

orm

atio

n En

thal

py

Enthalpy of Formation

(formaldehyde)

Standard Enthalpies of Formation

Pozitive Enthalpies of Formation

Standard Enthalpiesof Formation

Negative Enthalpiesof Formation

Standard Enthalpies of Reaction

• If the reactants and products of a reaction are all in their standard states, we call the enthalpy change for a reaction standard enthalpy of reaction.

• ΔHrxn0

• Instead of standard enthalpy of reaction the term is also frequently used:– Standard heat of reaction

Standard Enthalpy of Reaction

Standard Enthalpy of Formation

Htot = -2Hf °NaHCO3

+ Hf °Na2CO3

+ Hf

°CO2

+ Hf °H2O

Enth

alpy

Standard Enthalpy of Formation

• Since enthalpy is a state of function , it is independent on the pathway!!

• The overall(net) enthalpy change of the reaction is equal to the sum of all enthalpy changes of the steps in the whole reaction.

Hrxn = H°dec+ H°

f

TABLE 8.1 Some Standard Enthalpies of Formation at 298 K

SUBSTANCE ΔHf˚,298 K

kJ/molSUBSTANCE ΔHf

˚,298K kJ/mol

CO(g) -110,5 HF(g) -271,1CO2(g) -393,5 HI(g) 26,48

CH4(g) -74,81 H2O(g) -241,8

C2H2(g) 226,7 H2O(l) -285,8

C2H4(g) 52,26 H2S(g) -20,63

C2H6(g) -84,68 NH3(g) -46,11

C3H8(g) -103,8 NO(g) 90,25

C4H10(g) -125,7 N2O(g) 82,05

CH3OH(l) -238,7 NO2(g) 33,18

C2H5OH(l) -277,7 N2O4(g) 9,16

HBr(g) -36,40 SO2(g) -296,8

HCl(g) -92,31 SO3(g) -395,7

Formation Decomposition

Overall

Exotermic reaction

Ent

halp

y

Reactant

Products

Formation Decomposition

Overall

Endothermic reaction

Ent

halp

y

Reactants

Product

Standard Enthalpy of Formation

Hrxn = ΣυpHf °

products- Συr Hf°reactants

°

Determination of the enthalpy change of reaction

The enthalpy change of the combustion reaction of C6H6 is -6535 kJ. Determine the molar enthalpy of formation of C6H6? (ΔHCO2(g)=-393,5 kJ/mol ; ΔHH2O(l)= -285,8 kJ/mol

Ionic Reactions in Solutions

• Many chemical reactions in aqueous solution are best thought of as reactions between ions and represented by net ionic equations.

• To calculate the net enthalpy change of reaction,we need to know the the enthalpy of formation data for individual ions.

• However, we can not create ions of a single type in a chemical reaction.

• We must choose a particular ion to which we assign an enthalpy of formation of zero in aqueous solutions. We then compare the enthalpies of formation of other ions to this reference ion.

• ΔH0(H+) (aq) = 0

Table 7.3 Enthalpies of Formation of Ions in Aqueous Solutions

Example• Calculating the enthalpy change in an ionic reaction• Given that ΔHf

˚(BaSO4 (s))= 1473 kJ/mol,what is the enthalpy change for the precipitation of barium sulfate?• Solution: • Ba2+ (aq) + SO4

2- (aq) BaSO4(s)

• The enthalpy of formation of BaSO4(s) is given and those of Ba2+(aq) and SO42-(aq) are found in Table 7.3

• ΔH˚=ΔHf˚(BaSO4 (s))- ΔHf

˚(Ba2+ (aq))-ΔHf˚(SO4

2- (aq))

ENTROPY and FREE ENERGY• THE MEANING OF SPONTANEOUS CHANGE• Spontaneity: A spontaneous process is a process that

occurs in a system left to itself• A nonspontaneous process will not occur unless some

external action is continiously applied.• Examples: a) the reaction of NaOH(aq) and HCl(aq) is • a neutralization process and spontaneous• b) the electrolysis of liquid water is• a nonspontaneous process, since electric current is

required to decompose liquid water into its elements• c)The melting of an ice cube is• spontaneous above the melting point(0˚C) ,• but it is nonspontaneous below the melting point

ENTROPY and FREE ENERGY• Entropy: The thermodynamic property related

to the degree of disorder in a system, designated by the symbol S.

• The greater the degree of randomness in a system , the greater its entropy.

• The entropy change, ΔS, is the difference in entropy between two states and also has a unique value.

• A(g) + B(g) mixture of A(g) and B(g)

• ΔS=Smix of gases-(SA(g)+ SB(g)) > 0

ENTROPY AND FREE ENERGY• The increase in disorder (ΔS>0) outweighs the fact that

heat must be absorbed(ΔH>0) and the process is spontaneous.

• Entropy increases are expected when• Pure liquids or liquid solutions are formed from solids• Gases are formed, from either solids or liquids• The number of molecules of gases increases as a result of

chemical reaction• The temperature of a substance increases.(Increased

temperature means increased molecular motion (either vibrational motion of atoms or ions in a solid, or translational motion of molecules in a liquid or gas))

ENTROPY and FREE ENERGY

• Ice melting in a warm room is a common way of increasing entropy

ENTROPY and FREE ENERGY

ENTROPY and FREE ENERGY

ENTROPY and FREE ENERGY

CASE ΔH ΔS ΔG RESULT EXAMPLE1 - + - Spontaneous at all temperature 2 N2O(g) 2

N2(g)+ O2(g)

2 - - -+

Spontaneous at low temperatureNonspontaneous at high temperature

H2O(l) H2O(s)

3 + + +-

Nonspontaneous at low temperature Spontaneous at high temperature

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

4 + - + Nonspontaneous at all temperature 3 O2(g) 2O3(g)

Evaluating Entropy and Entropy Changes