Thermodynamics

Heat

Work

Heat is a form of energy

• Mechanical work done on a system produces a rise in temperature like heat added to the system.

Joule’s ExperimentJoule showed that mechanical energy could beconverted into heat energy.

F

M

xH2O

T

W = Fx

• Thermodynamics Thermodynamics is concerned with interconversions of different forms of energy.

• It was developed as a mathematical tool for studying phenomena such as the way in which heat energy can be converted into mechanical energy in heat engines

• Thermodynamics provides a means for deciding whether a process will occur spontaneously

• Spontaneous” in this context implies nothing about how fast it will take place - e.g. combustion of a diamond

• Thermodynamics is linked at a fundamental level to the nature of the universe

• The Industrial Revolution depended on heat engines, most of which (like these steam engines) were of very low efficiency. The development of physical theories, and mathematical tools, to analyse these systems led to rapid improvements in technology.

• Energy can be divided into two categories - kinetic energy and potential energy

• Kinetic energy includes all forms of energy that result from movement - either linear motion or rotation.

• Heat, which is molecular motion.• Radiant energy - the kinetic energy of photons of light and other

electromagnetic radiation

• Mechanical energy

• Electrical energy as currents of moving electrons or charged particles

• Potential energy includes all forms of energy that are stored.

– Energy stored in chemical bonds

– Energy stored in concentration gradients

– Energy stored as electrical potential (separation of charges)– Energy stored in the nuclei of atoms

• The basic unit of energy is the Joule (J)

Thermodynamic Systems

• A thermodynamic system is a collection of matter which has distinct boundaries.

OR• A real or imaginary portion

of universe whish has distinct boundaries is called system.

OR• A thermodynamic system is

that part of universe which is under thermodynamic study.

Thermodynamic SystemA quantity of matter or a region in space chosen for study.

SurroundingsEverything external to the system.

BoundarySurface that separates the system from the surrounding. It may be fixed or movable

Systems

Closed system (Control mass)A fixed amount of mass chosen for study (no mass can cross its boundary). Heat and work can cross the boundary, volume may also change.e.g. piston cylinder.

Open system (Control volume)A selected region chosen for study. Mass, heat and work can cross its boundary, volume may also change.e.g. water heater, car radiator, turbine, nozzle.

Isolated systemA system closed to mass, heat and work flows. It is not affected by the surroundings.

Closed and Open Systems

Open Systems

Properties of a system

Thermodynamic PropertyA measurable quantity that defines the condition of a systeme.g. temperature T pressure P mass m volume V density ρ

Extensive and Intensive properties

• Properties are of 2 types:

• Intensive properties Independent of mass. e.g. P, T, v, ρ.

• Extensive properties Change with mass. e.g. m, V, Energy

each

consider a quantity of substance part same intensive property

divide into 2 parts will 1/2 extensive property

have

Temperature and 0th law of thermodynamicsTemperature: Degree of hotness of coldness

0th law of thermodynamicsWhen 2 bodies have equality of temperature with a 3rd body, then they have equality of temperature with each other.

TA TB

TC

if &

then A C B C

A B

T T T T

T T

Absolute scale of temperature :

273.15 note that in in o oK C T K T C

Internal Energy (E)

• Definition – The total of the kinetic and potential energy in a system.

• E = Kinetic Energy + Potential Energy

Internal Energy• Internal Energy: total energy of a system.

• Involves translational, rotational, vibrational motions.

• Cannot measure absolute internal energy.

• Change in internal energy,

The First Law of ThermodynamicsThe First Law of Thermodynamics

initialfinal EEE

First Law of Thermodynamics

• The change in the internal energy of a system E, is equal to the heat input Q minus the work done by the system.

• ΔE = Q−W

• The internal energy is the energy stored in the system.

• For an ideal gas the internal energy is the kinetic energy of the gas

Relating E to Heat(q) and Work(w)• Energy cannot be created or destroyed.• Energy of (system + surroundings) is constant.• Any energy transferred from a system must be transferred

to the surroundings (and vice versa).

• From the first law of thermodynamics:

The First Law of ThermodynamicsThe First Law of Thermodynamics

wqE

•Work is a form of energy. It’s the energy involved in moving something. If nothing moves, no work is done.

•Work in chemical terms is usually done with pressure and volume changes.

The First Law of The First Law of ThermodynamicsThermodynamics

QUIZ……

• Calculate the energy change for a system undergoing a process in which 15.4 kJ of heat flows and where 6.3 kJ of work is done on the system.

ANSWER

• ∆E = q + w

• q = - 15.40 J

• w = + 06.30 J

• ∆E = - 15.40 J + 06.30 J = - 09.10 J

Exothermic and Endothermic Processes

• Endothermic: absorbs heat from the surroundings.• An endothermic reaction feels cold.

• Exothermic: transfers heat to the surroundings.• An exothermic reaction feels hot.

The First Law of ThermodynamicsThe First Law of Thermodynamics

Pressure-volume work

• P = f / A • Work done by the

system (gas):• w = – f Δx

• w = –P ΔV

• Example 2

• A gas expands by 0.50 L against a constant pressure of 0.50 atm at 25 °C. What is the work in erg done by the system? (1.0 atm = 1.013 x 106 dyne/cm2)

• Solution:• W = - PΔV• = - (0.50 x 1.013 x 106 dyne/cm2) x 500 cm3

• = - 2.50 x 108 dyne/cm• = - 2.50 x 108 erg

Reversible work

• Isothermal work expansion against variable pressure.

• n = number of moles, R = gas constant (=8.314 JK-1mol-

1, = 1.987 cal K-1mol-1, = 0.0802 L.atm. K-1mol-1)

T = absolute temperature

vpwmaxnRTpv

v

nRTp

1

2

1

2 log303.2ln2

1 v

vnRT

v

vnRT

v

dvnRT

v

dvnRT

v

v

• Example:

• What is the maximum work done in the isothermal reversible expansion of 2 moles of an ideal gas from 1 to 5 litres at 25 °C?

• Solution:

» = 7976.43 J

1

2max log303.2

v

vnRTw

L

LKxKJmolmolxx

1

5log298314.82303.2 11

Special Forms of the 1st law

• ∆E = q - w• Case 1: Isothermal process• ∆E = 0, hence, q = w• Case 2: Isochoic process• w = 0, ∆E = qv

• Case 3: Adiabatic process• q = 0, ∆E = w• Case 4: Isobaric • ∆E = q - w

Energy Changes in ChemicalReactions

• How are energy changes measured?

• One Answer – Calorimetry, q = CcalΔT

• What thermodynamic quantities do we get?

• Constant Volume: qv = ΔE.

• Constant Pressure: qP = ΔH

• In most calorimetry, ΔT is very small, initial

•and final states are at nearly constant T

•

ChemicalChemical Reactivity ReactivityChemicalChemical Reactivity ReactivityWhat drives chemical reactions? How do they What drives chemical reactions? How do they

occur?occur?

The first is answered by The first is answered by THERMODYNAMICSTHERMODYNAMICS and the second by and the second by KINETICSKINETICS..

Have already seen a number of Have already seen a number of ““driving forcesdriving forces”” for reactions that are for reactions that are PRODUCT-FAVOREDPRODUCT-FAVORED..

•• formation of a precipitateformation of a precipitate

•• gas formationgas formation

•• HH22O formation (acid-base reaction)O formation (acid-base reaction)

•• electron transfer in a batteryelectron transfer in a battery

Chemical ReactivityChemical ReactivityBut energy transfer also allows us to predict But energy transfer also allows us to predict

reactivity.reactivity.In general, reactions that transfer In general, reactions that transfer

energy to their surroundings are energy to their surroundings are product-favored.product-favored.

So, let us consider heat transfer in chemical processes.So, let us consider heat transfer in chemical processes.

Enthalpy of Reaction from Heat of Formation

• Standard Heat of Formation (ΔHof):

The amount of energy gained or lost when1 mole of the substance is formed from its elements under standard conditions(25°C, 1 atm = 101.3 kPa)

)tan()( tsreacprorcucts ffreaction

Example 2

Standard enthalpies of formation are: C2H5OH(l) -228, CO2 (g) -394, and H2O(l) -286 kJ/mol. Calculate the enthalpy of the reaction,

• C2H5OH (l) + 3O2 (g) = 2 CO2 (g) + 3 H2O (l)

• Solution:

• ∆HR = [3x∆Hf(H2O (l)) + 2x∆Hf(CO2(g))] – [1x∆Hf C2H5OH(l) + 3x∆HfO2(g)]

= [3molX-286 kJ/mol + 2molx-394 kJ/mol]-[1molx-228 kJ/mol + 3molx0.0 kJ/mol]

= [- 858 + (-788)] – [-228] = - 1546 + 228 = - 1418 kJ

)tan()( tsreacprorcucts ffreaction

Heat of Reactions from Bond Energies

•When bonds are formed, energy is released.• In order to break bonds, energy must be absorbed

• Exothermic: Products have stronger bonds than the reactants.

• Endothermic: Products have weaker bonds than the reactants.

• O2(g) = 2O(g) ∆H° = 490.4 kJ

• H2(g) = 2H(g) ∆H° = 431.2 kJ

• H2O(g)= 2H(g) + O(g) ∆H° = 915.6 kJ

• 2H2(g) + O2(g) = 2H2O(g)    ∆H° = ?

• ∆H° reaction = bonds broken Energy (absorbed) ─ bonds formed Energy

(released)• =[2 H-H + O=O] – [ 4 O-H]

»= [2 x 431.2 + 1 x 490.4] – [4 x 457.8]»= [862.4 + 490.4] – [1831.2]»= 1532.8 – 1831.2»= - 478.4 kJ

Heat of Combustion

IS the Heat Evolved when One mole of a Fuel is Burned in Enough Oxygen

• EXample:

• How much heat is evolved when 54.0 g glucose (C6H12O6) is burned according to this equation?

• [ΔHcomb = -2808 kJ/mol, C = 12, O = 16, H = 1]

C6H12O6(s) + 6O2(g) 6H2O(l) + 6CO2(g)

• Answer:

• Mass of glucose = 54.0 g C6H12O6

• ΔHcomb = -2808 kJ

• Energy (q) = ?

• Convert: grams of C6H12O6 to moles of C6H12O6

• 54.0 g / 180 g mol-1 = 0.300 mol C6H12O6

• ∆ H = 0.300 mol x - 2808 kJ/mol = - 842 kJ

Heat at constant VOLUME

• If we consider a system at constant volume. ∆V is zero. If ∆V is zero, then the work is zero.

• At constant volume the heat is equal to ∆U.

• We can measure heat at constant volume in a Bomb Calorimeter.

• The calculations are the same for bomb calorimetry as for coffee-cup calorimetry.

• The heat calculated is a measure of the INTERNAL ENERGY change instead of Enthalpy.

Finding ∆H from ∆E

• We can find ∆E from bomb calorimetry. But that is at constant volume.

• How do we find ∆H?

• Here ∆n is the change in number of moles of GAS in the balanced chemical equation.

ΔE = qp + w = ΔH − PΔV = ΔH − ΔnRT

• ∆nRT• This is an amount of energy that would

be represented as work if work could be done.

• This can be zero if there is no change in number of moles of gas.

Application

• For the reaction

•Mg (s) + ½ O2 (g) = MgO (s)

the enthalpy change is -601.2 kJ. What is the internal energy change for this reaction? How many g of magnesium must react to effect an internal energy change of -22.4 kcal?

HessHess’’s Laws Law

The Enthalpy of a Reaction is the Same if it takes place in One or More than One step.

HessHess’’s Law s Law & Energy Level & Energy Level

DiagramsDiagramsForming H2O can

occur in a single step or in a two

steps .

∆Htotal is the same no matter which path is

followed.

Making HMaking H22O from HO from H22 involves two steps. involves two steps.

HH22(g) + 1/2 O(g) + 1/2 O22(g) ---> H(g) ---> H22O(g) + 242 kJO(g) + 242 kJ

HH22O(g) ---> HO(g) ---> H22O(liq) + 44 kJ O(liq) + 44 kJ

-----------------------------------------------------------------

HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(liq) + 286 kJO(liq) + 286 kJ

Example of Example of HESSHESS’’S LAWS LAW——

If a rxn. is the sum of 2 or more others, the If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆Hnet ∆H is the sum of the ∆H’’s of the other s of the other rxns.rxns.

USING ENTHALPYUSING ENTHALPY

QUIZs

• The standard heat of formation, ΔHof, for sulfur dioxide

(SO2) is -297 kJ/mol. How many kJ of energy are given off when 25.0 g of SO2 (g) is produced from its elements? 

• The heat of reaction for the combustion of 1 mol of ethyl alcohol is -9.50 × 102 kJ:C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (l) + 9.5 × 102 kJ. How much heat is produced when 11.5 g of alcohol is burned?

• ΔH for the complete combustion of 1 mol of propane is -2.22 × 103 kJC3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (l)Calculate the heat of reaction for the combustion of 33.0 g of propane.