ThermochemistryThermodynamics

Gases

Paul FranklynC305

Consultation: Mondays 8:30-11:30Or email paul.franklyn@wits.ac.za

Lecture notes at:

THERMOCHEMISTRY CHAPTER 5

WHY ENERGY?

• Life requires energy

• From plants – photosynthesis

• To equipment – electricity

• Vehicles – fossil fuels

• To animals - food

ENERGY FROM WHERE?

• Most of our energy as humans– From food– i.e. from chemical reactions

• In the world– From chemical reactions

WHAT IS THERMOCHEMISTRY?

Quantitative study of chemical reactions involving heat changes

Produce energy

Consume energy

Examples:

Food we eat - provides us with energy

Burning coal- to produce electrical energy

Energy from the sun - responsible for chemical reactions in plants which make them grow

RECALL • FORCE (F) – push/pull exerted on an object

• WORK (w) – energy used to cause an object to move against a force (w = Fd)

• HEAT (q) – energy transfer as a result of a temperature difference.

Heat always flows from a hotter to a colder object until they have the same temperature

• ENERGY (E) – capacity to do work or to transfer heat

HOT STUFF?

• Can something transfer cold?

• Does ice make a drink cold?

When you blow hot soup do you make it cold?

SO IN FACT:

• In theory stirring soup long enough and fast enough could make it boil

Heat always flows from a hotter to a colder object until they have the same temperature

BUT I KNOW IT’S COLD!!

Heat from our body is transferred,

making something

feel cold because we

lose heat from the area

touching the cold object

NATURE OF ENERGY

2 types of energy

Kinetic energy Potential energy

Energy due to motion of object

Energy due to position of object relative to others (Stored energy)

KINETIC ENERGY

Note:

Ek increases as velocity increases and Ek

increases as mass increases.

2k mv

2

1E

Atoms and molecules have mass and are in motion, thus they have kinetic energy!

The total thermal energy of the object is the sum of the individual energies of the atoms or molecules.

Thermal energy

Associated with kinetic energy of atoms and molecules.

The higher the temperature, the faster the atoms and molecules move the more kinetic energy molecules and atoms

have the greater the thermal energy of the object.

energy an object possesses because of its temperature.

MISSING: KINETIC ENERGY, PLEASE HELP

• What is it to have no kinetic energy?– Absolute zero!

• Can we detect that?– No!– Why?!– Because in order to detect it we would

have to interact with it which would heat it

POTENTIAL ENERGY

We know Ep = mgh

Forces other than gravity that lead to potential energy

= ELECTROSTATIC forces

Attractions and repulsions dueto oppositely charged objects(e.g. positive and negative ions)

Recall:

Eel = electrostatic energyk = Coulomb constantQ = charged = distance between

charges

UNITS OF ENERGY

SI unit = J (Joule)

1 J = 1 kg.m2.s-2

 (Others 1 cal (calorie) = 4.184 J)

dQkQ

E 21el

Ek=0.5mv2

=(kg)(m.s-1)2

Describes the electrostatic energy between 2 charges.

(The amount of heat energy to raise temp of 1g H2O by 1oC)

A BRAINTEASER

• If everything has kinetic energy

• And chemical potential energy

• Why is everything not moving around constantly?

• Why are solids not felt to vibrate?

SYSTEM AND SURROUNDINGS

Portion singled out for study

Everything else

System Surroundings

System + surroundings = universe

Open system – can exchange both energy and matter with the surroundings

Closed system – exchanges energy but NOT matter with the surroundings

Isolated system – neither energy nor matter can be exchanged with the surroundings

A QUOTE NOT TO LIVE BY

“Thermodynamics is a funny subject. The first time you go through it, you don't understand it at all. The second time you go through it, you think you understand it, except for one or two small points. The third time you go through it, you know you don't understand it, but by that time you are so used to it, it doesn't bother you anymore.”

http://www.eoht.info/page/Thermodynamics+humor

FIRST LAW OF THERMODYNAMICS

Energy is neither created nor destroyed, it is merely converted from one form into another.

Energy is transferred between the system and the surroundings in the form of work and heat.

The total energy of the universe remains constant.

INTERNAL ENERGY (U)

Internal energy (of the system)

= sum of all Ek and Ep of all components

of the system

Symbol for change = Change in internal energy of the system

U

U = UFinal - Uinitial

Number & Unit Magnitude

Sign Direction (in which energy is transferred)

Also given as E

U = UFinal - Uinitial

Usyst > 0 Uf > Ui

GAINED energy from surroundings

Usyst < 0 Uf < Ui

LOST energy to surroundings

RELATING U TO WORK AND HEAT

General ways to change ENERGY of a closed system:

HEATlost/gainedby system

WORKDone by/on

system

Relationship:

U = q + w

We can’t measure U, but we can determine U.

When a system undergoes a chemical/physical change

SYSTEM

SIGN CONVENTION

q>0 Heat transferred TO system

w>0Work done ON system

q<0 Heat

transferred FROM

system

w<0 Work done BY system

The sign of U will depend on the sign and magnitude of q and w since U = q + w.

Example:

Calculate the change in internal energy for a process in which the system absorbs 120 J of heat from the surroundings and does 64 J of work on the surroundings.

System120 J heat

64 J work

U = q + w

SOME DEFINITIONS

Extensive property – dependent on the amount of matter in the system. E.g. mass, volume etc

Intensive Property – NOT dependent on the amount of matter in the system. E.g. density, temperature etc

State function – a function that depends on the state or conditions of the system

and NOT on the details of how it came to be in that state.

A B

C

DEFINITIONS APPLIED TO INTERNAL ENERGY

Recall:

Total internal energy of a system = sum of all Ek and Ep of all components of the system

Thus total internal energy of a system total quantity of matter in system

U is an extensive property

U only depends on Ui and Uf and not how

the change occurred.

NOTE: Heat and work are not state functions!

Internal energy depends on the state or conditions of the system (e.g. pressure, temperature, location)

Does not depend on how it came to be in that state.

state function

e.g. if a gas sample undergoes:

A pressure X heat B, or

A heat Y pressure B

U is the same in both cases.

ENTHALPY (H)

We cannot measure enthalpy (H), we can only measure change in enthalpy (H).

Change in enthalpy (H) is the heat gained or lost by the system when a process occurs under constant pressure.

H = Hfinal - Hinitial = qp

H is a state function and an extensive property

E.g. atm pressure

NOTE:

At constant pressure, most of the energy lost / gained is in the form of heat.

Very little work is done for the expansion / contraction against the atmospheric force, especially if the reaction does not involve gases.

qp>0H>0

qp<0H<0

Exothermic process – system evolves heat

Endothermic process - system absorbs heat

SYSTEM

Example of an exothermic reaction:

Example of an endothermic reaction:

FINDING THE RELATIONSHIP BETWEEN U AND H – CONSIDER

PV WORK

We know that: U = q + w

(e.g. expanding gases in cylinder of a car does PV work on the piston)

Assume we do PV work only

Gas expands and moves piston distance d

Sign: system is doing work on the pistonW = -PV

W = PV

W = PAdBut F = PA

W = Fd

A

FP

U = q + w and w = -PV

U = q - PV

At constant volume, V = 0

At constant pressure:

U = qp - PV

But H = qp

U = qv

U = H - PV

The volume change in many reactions is very small, thus PV is very small and hence the difference between U and H is small.

ENTHALPIES OF REACTION

enthalpy change that accompanies a reaction (heat of reaction)

Hrxn = H(products) – H(reactants)

Thermochemical equation:

2H2(g) + O2(g) 2H2O(g) H = - 483.6 kJ Balanced equation H associated with

the reaction

final initial

exothermic

Coefficients in balanced equation = no. of moles of reactant/product producing associated H (extensive)

2H2(g) + O2(g) 2H2O(g) H = -483.6 kJ

Note:Since enthalpy is an extensive property, H depends on the amount of reactant consumed.

Also, H for a reaction is equal in magnitude, but opposite in sign to H for the reverse reaction.

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

4H2(g) + 2O2(g) 4H2O(g) H =

Calculate the heat needed to convert 25 ml water to steam at atmospheric pressure if the enthalpy change is 44 kJ/mol.

(Assume the density of water is 1.0 kg/l)

Example:

Example:

Hydrogen peroxide can decompose to water and oxygen by the reaction:

2H2O2 (l) 2H2O (l) + O2 (g) H = -196 kJCalculate the heat produced when 2.50 g H2O2

decomposes at constant pressure.

Example – burning money:Ethanol burns in air to give water vapour and carbon dioxide1. Calculate the enthalpy of reaction given enthalpy of

formations of: ethanol (l) = -277.7 kJ/mol water (l) = -285.8 kJ/mol carbon dioxide (g)= -393.5 kJ/mol2. Calculate the minimum amount of water needed to prevent the

paper from burning after be soaked in 1g of ethanol

First: find the balanced chemical equation

Enthalpy of reaction:

Example – burning money:Ethanol burns in air to give water vapour and carbon dioxide2. Calculate the minimum amount of water needed to prevent the

paper from burning after be soaked in 1.00g of ethanol

H2O(l) H2O(g) H = 44 kJ/molRecall:

Find the heat released by burning 1.00 g of ethanol

CALORIMETRY

The experimental determination of heat flow associated with a chemical reaction by measuring the temperature changes it produces.

Units: J.K-1

Extensive property

Units: J.K-1mol-1

Heat Capacity

The amount of heat required to raise the temperature by 1K.

Recall: a change of 1K = a change of 1oC

Molar Heat Capacity The heat capacity of 1 mol of substance.

Specific Heat Capacity (C) heat capacity of 1 g of substance

Units: J.K-1g-1

determined by measuring the change in temperature that a known mass of substance undergoes when it gains/loses a specific quantity of heat.

Tm

qC

Specific Heats for Some Subst’s at 298KH2O (l) 4.18 J.K-1g-1

N2 (g) 1.04 J.K-1g-1

Al (s) 0.90 J.K-1g-1

Fe (s) 0.45 J.K-1g-1

Hg (l) 0.14 J.K-1g-1

Example:

a) How much heat is needed to warm 1.00 L of water from 23.0oC to 98.0oC?

(Assume the density of water is 1.00 kg/L).

b) What is the molar heat capacity of water?

a) How much heat is needed to warm 1.00 L of water from 23.0oC to 98.0oC? (Assume the density of water is 1.00 kg/L).

b) What is the molar heat capacity of water?

CONSTANT PRESSURE CALORIMETRY

Any calorimeter at atmospheric pressure has a constant pressure, e.g. “coffee-cup” calorimeter.

Can calculate qsoln = CmT,

but we want qrxn

Since heat given off by the reaction is absorbed by the solution:

qrxn = -qsoln

The heat produced by a reaction is entirely absorbed by the solution at constant pressure.i.e. heat does not escape the calorimeter

Recall: H = qp

Example:

When 4.25 g solid ammonium nitrate dissolves in 60.0 g water in a coffee-cup calorimeter, the temperature drops from 22.0oC to 16.9oC.

Calculate H (in kJ/mol NH4NO3) for the dissolution process.

Assume the specific heat of solution is the same as that for pure water.

NH4NO3(s) NH4+(aq) + NO3

-(aq)

When 4.25 g solid ammonium nitrate dissolves in 60.0 g water in a coffee-cup calorimeter, the temperature drops from 22.0oC to 16.9oC. Calculate H (in kJ/mol NH4NO3) for the dissolution process.

When 4.25 g solid ammonium nitrate dissolves in 60.0 g water in a coffee-cup calorimeter, the temperature drops from 22.0oC to 16.9oC. Calculate H (in kJ/mol NH4NO3) for the dissolution process.

CONSTANT VOLUME CALORIMETRY – BOMB CALORIMETER

At constant volume – measure U rather than H, but for most U H

A bomb calorimeter is used to study combustion reactions etc.We calculate the heat of combustion from the measured change in temperature.

We need to know the heat capacity of the calorimeter (Ccal).

Recall: U = qv

qrxn = -Ccal x T

NB: This equation cannot be applied blindly!!!

Always check units for Ccal.Always include units in calculations to ensure they cancel.

Bomb Calorimeter

Example:

A 2.200 g sample of quinone, C6H4O2, is burned in a bomb calorimeter whose total heat capacity is 7.854 kJ/oC. The temperature of the calorimeter increases from 23.44oC to 30.57oC.

What is the heat of combustion per gram of quinone? per mole of quinone?

A 2.200 g sample of quinone, C6H4O2, is burned in a bomb calorimeter whose total heat capacity is 7.854 kJ/oC. The temperature of the calorimeter increases from 23.44oC to 30.57oC.

What is the heat of combustion per gram of quinone? per mole of quinone?

Example:

50.0 g of water at 62.5oC was poured into a calorimeter containing 50.0 g water at 18.7oC. The final temperature was 35.0oC. How much heat was lost to the surroundings during this process? (Heat capacity of water = 4.184 J.oC-1.g-1)

Ans: -2.34 kJ

50.0 g of water at 62.5oC was poured into a calorimeter containing 50.0 g water at 18.7oC. The final temperature was 35.0oC. How much heat was lost to the surroundings during this process? (Heat capacity of water = 4.184 J.oC-1.g-1)

HESS’S LAW

If a reaction is carried out in a series of steps, H for the reaction will be equal to the sum of the enthalpy changes for the individual steps.

A + B C + D H1+ H2

A + B X + Y H1

X + Y C + D H2

H for a reaction is calculated from H data of other reactions.

Recall that enthalpy is a state function H is independent of the path followed.

It is useful to use Hess’s Law in cases where H cannot be measured.

Example:

Carbon occurs in two forms, graphite and diamond. The enthalpy of combustion of graphite is –393.5 kJ/mol and that of diamond is –395.4 kJ/mol.

Calculate H for the conversion of graphite to diamond.

C(s,graph) + O2(g) CO2(g) H = –393.5 kJ/mol

C(s,diam) + O2(g) CO2(g) H = –395.4 kJ/mol

C(s,graph) C(s,diam) H = ?

From the following enthalpies of reaction:

H2(g) + F2(g) 2HF(g) H = -537 kJ C(s) + 2F2(g) CF4(g) H = -680 kJ 2C(s) + 2H2(g) C2H4(g) H = 52.3 kJ

Calculate H for the reaction of ethylene with F2:

C2H4(g) + 6F2(g) 2CF4(g) + 4HF(g)

Example:

H2(g) + F2(g) 2HF(g) H = -537 kJ C(s) + 2F2(g) CF4(g) H = -680 kJ 2C(s) + 2H2(g) C2H4(g) H = 52.3 kJ

C2H4(g) + 6F2(g) 2CF4(g) + 4HF(g)

STANDARD STATES

The magnitude of H depends on conditions of temperature, pressure, and state of products and reactants.In order to compare enthalpies, need same set of conditions.

Standard state of a substance = pure form at 1 atm and at the temperature of interest(usually 298 K (25oC))

Standard enthalpy (Ho) when all products and reactants are in their standard states

STANDARD ENTHALPIES OF FORMATION

Hfo change in enthalpy for the reaction

that forms 1 mol of a substance from its elements in their standard states.Units: kJ/mol

Note: If there is more than one form of the element present under standard conditions, use the most stable form.

Hfo = 0 kJ/mol for the most stable form of

any element

Carbon(s) can exist as graphite or diamond at 1 atm and 298 K

Examples of most stable form:

Oxygen(g) can exist as O2 or O3 (ozone) at 1 atm and 298 K

Most stable form = O2

Most stable form = graphite

Hfo(O2(g)) = 0 kJ/mol

Hfo(C(s,graphite)) = 0 kJ/mol

Tabulated data of enthalpies of formation can be used to calculate enthalpies of reaction as follows:

)tstanreac(Hn)products(HmH of

of

orxn

Stoichiometric coefficients

Table of standard enthalpies of formation

General KnowledgeGeneral Knowledge

Gray tin and white tin are two solid forms of tin. The denser white metallic form is the most stable phase above 13oC, and the powdery gray form is more stable below 13oC.

The formation of gray tin is said to have contributed to Napoleon’s defeat at Moscow, when his soldier’s buttons fell off their clothes at the low temperatures they encountered there.

Example:Nitroglycerine is a powerful explosive, giving four

different gases when detonated:

2C3H5(NO3)3 (l) 3N2(g) + 1/2O2(g) + 6CO2(g) + 5H2O(g)

Given that the enthalpy of formation of nitroglycerine is Hf

o = -364 kJ.mol-1, calculate the enthalpy change when 10.0 g of nitroglycerine is detonated.

2C3H5(NO3)3 (l) 3N2(g) + 1/2O2(g) + 6CO2(g) + 5H2O(g)Calculate the enthalpy change when 10.0 g of

nitroglycerine is detonated.

2C3H5(NO3)3 (l) 3N2(g) + 1/2O2(g) + 6CO2(g) + 5H2O(g)Calculate the enthalpy change when 10.0 g of

nitroglycerine is detonated.

Example: For you to do!

Calculate Ho for the decomposition of limestone:

CaCO3(s) CaO(s) + CO2(g)Hence calculate the heat required to

decompose 1 kg of limestone.

BOND ENTHALPIES

Bond enthalpy = enthalpy change for breaking a particular bond in a mole of gaseous substance.

The stability of a molecule is related to the strengths of the covalent bonds it contains.

The strength of a covalent bond between two atoms is determined by the energy required to break that bond.

Can determine approximate value for Horxn if

we have the values of average bond enthalpies.

Horxn is calculated by determining the energy

required to break all bonds minus the energy evolved to form all bonds.

)formedbonds(BE)brokenbonds(BEHorxn

Reactants

initial

Products

final

This is only an approximation because the average bond enthalpies used as the bond enthalpies are not exactly the same in all molecules.

e.g. C-H bond energy in CH4 is slightly different to that in C2H6 and so on.

H - C - H

H

H

H - C – C - H

H

H

H

H

O2

N2

CO

CO2

CN

H2CO

You must know these:

O=O

NN

CO

O=C=O

CN

C=OH

H

Example:

Hydroiodic acid reacts with chlorine as follows:

2HI(aq) + Cl2(g) 2HCl(aq) + I2(s)

Approximate Ho for the reaction using tabulated bond energies.

Bonds Broken

Bonds Formed

2HI(aq) + Cl2(g) 2HCl(aq) + I2(s)

Bond Energy / kJ.mol-1

)formedbonds(BE)brokenbonds(BEHorxn

Example:

Estimate H using bond enthalpies for the following reaction:

C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O(g)

Bonds Broken

Bonds Formed

Bond Energy / kJ.mol-1

)formedbonds(BE)brokenbonds(BEHorxn

C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O(g)