Thermodynamics

Honors Unit 5

Energy: Basic Principles Thermodynamics – the study of energy

changes

Energy – the ability to do work or produce heat

Note: Work is force acting over a distance

Energy: Basic Principles

Kinetic Energy – energy of motion

KE =

Potential Energy – energy due to position or composition

2

2

1mv

Law of Conservation of Energy A.k.a. first Law of Thermodynamics

Energy can be converted from one form to another but can’t be created or destroyed

This means the total energy of the universe is CONSTANT!

Heat vs. Temperature

Temperature – measure of the random motion of a substance

Temperature is proportional to kinetic energy (it is a measure of the average kinetic energy in a substance)

Heat (q) – flow of energy due to a temperature difference

Important Aspects of Thermal Energy & Temperature

Heat is NOT the same as temperature!

The more kinetic energy a substance has, the greater the temperature of its atoms and molecules.

The total thermal energy in an object is the sum of its individual energies of all the molecules.

For any given substance, its thermal energy depends not only on its composition but also on the amount of substance

System vs. Surroundings

A system is the part of the universe we are studying.

The surroundings are everything else outside of the system.

Direction of Heat Flow Heat transfer occurs when two objects are

at two different temperatures.

Eventually the two objects reach the same temperature

At this point, we say that the system has reached equilibrium.

Thermal Equilibrium Heat transfer

always occurs with heat flowing from the HOT object to the COLD object.

Thermal Equilibrium

The quantity of heat lost by the hotter object and the quantity of heat gained by the cooler

object are EQUAL.

Exothermic vs. EndothermicExothermic process heat is transferred

from the system to the surroundings Heat is lost from the system

(temperature in system decreases)

Endothermic process heat transferred from the surroundings to the system Heat is added to the system

(temperature in system increases)

Exothermic Process

Endothermic Process

Units of Energy Joule (J) is the SI unit of energy & heat

One kilojoule (kJ) = 1000 joules (J)

calorie (cal) = heat required to raise the temperature of 1.00 g of water by 1 °C

1 calorie = 4.184 J

Units of Energy Food is measured in Calories (also known

as kilocalories) instead of calories

1 Cal = 1 kcal = 1000 calories

Units of Energy3800 cal = __________ Cal = _________ J

Units of EnergyThe label on a cereal box indicates that 1 serving provides 250 Cal. What is the energy in kJ?

Heat TransferDirection and sign of heat flow – MEMORIZE!

ENDOTHERMIC: heat is added to the system & the temperature increases (+q)

EXOTHERMIC: heat is lost from the system (added to the surroundings) & the temperature in the system decreases (-q)

Heat Capacity The quantity of heat required to raise an

object’s temperature by 1 °C (or by 1 Kelvin)

Heat capacity is an extensive property. Which will take more heat to raise the

temperature by 1 °C?

Specific Heat (Specific Heat Capacity) Specific Heat (C) - The quantity of heat

required to raise the temperature of one gram of a substance by 1 °C Intensive property

Units: J/(g°C) or J/(g°K) cal/(g°C) or cal/(g°K)

Examples of Specific HeatAt the beach, which gets hotter, the sand or the water?

Higher specific heat means the substance takes longer to heat up & cool down!

Examples of Specific Heat Specific heat (C)= the heat required to

raise the temperature of 1 gram of a substance by 1 °C

Cwater = 4.184 J/(g°C)

Csand = 0.664 J/(g°C)

Calculating Changes in Thermal E

q = m x C x Δtq = mCΔt

q = heat (cal or J)m = mass (g)C = specific heat capacity, J/(g°C)Δt = change in temperature, Tfinal – Tinitial

(°C or K)

***All units must match up!!!***

ExampleHow much heat in J is given off by a 75.0 g sample of pure aluminum when it cools from 84.0°C to 46.7°C? The specific heat of aluminum is 0.899 J/(g°C).

q = mCΔt

ExampleWhat is the specific heat of benzene if 3450 J of heat are added to a 150.0 g sample of benzene and its temperature increases from 22.5 °C to 35.8 °C?

q = mCΔt

ExampleA 50.0 g sample of water gives off 1.025 kJ as it is cooled. If the initial temperature of the water was 85.0 °C, what was the final temperature of the water? The specific heat of water is 4.18 J/(g°C).

q = mCΔt

Calorimetry Calorimetry: measurement of quantities of

heat A calorimeter is the device in which heat

is measured.

Calorimetry Assumptions:

Heat lost = -heat gained by the system

In a simple calorimeter, no heat is lost to the surroundings

Coffee Cup Calorimetry

***Use styrofoam instead of a beaker to keep heat in

Steps:1. Add hot solid metal to cool water2. Water will heat up (T rises) as metal cools3. Eventually, water & metal are at same T.

qmetal + qwater = 0

qmetal = -qwater

Heat lost by metal = -heat gained by the water

Calorimetryqmetal = -qwater

Heat lost by metal = -heat gained by water

Since q = mCΔT,

mmCmΔTm = -[mH2OCH2OΔTH2O]

Sample ProblemA 358.11 g piece of lead was heated in water to

94.1 °C. It was removed from the water and

placed into 100. mL of water in a Styrofoam

cup. The initial temperature of the water was

18.7 °C and the final temperature of the lead

and water was 26.1 °C. What is the specific

heat of lead according to this data?

Bomb Calorimeter Constant volume “bomb”

calorimeter Burn sample in O2

Some heat from reaction warms water qwater = mCH2OΔT

Some heat from reaction warms bomb qbomb = CbombΔT

qrxn + qH2O + qbomb = 0

Energy & Changes of State All changes of state involve energy

changes (more in Unit 9) Note that fusion = melting

State Functions A property where the change from initial

to final state does not depend on the path taken Ex.) The change in elevation from the top to

bottom of a ski slope is independent of the path taken to go down from the slope

Enthalpy Changes for Chemical Rxns. Heat of reaction

The heat absorbed or given off when a chemical reaction occurs at constant T (temp.) and P (pressure)

Enthalpy Enthalpy (H)

The heat content of a reaction (chemical energy)

ΔH = change in enthalpy The amount of energy absorbed by or

lost from a system as heat during a chemical process at constant P

ΔH = ΔHfinal - ΔHinitial

Properties of Enthalpy Enthalpy is an extensive property

It does depend on quantity

Enthalpy is a state function Depends only on the final & initial values

Every reaction has a unique enthalpy value since ΔH = Hproducts - Hreactants

Representation of Enthalpy as a Graph

Two Ways to Designate Thermochemical Equations

Endothermic:

a) H2 (g) + I2 (s) 2 HI (g) ΔH = 53.0 kJ

b) H2 (g) + I2 (s) + 53.0 kJ 2 HI (g)

Two Ways to Designate Thermochemical Equations

Exothermic:

a) ½ CH4 (g) + O2 (g) ½ CO2 (g) + H2O (l) ΔH = -445.2 kJ

b) ½ CH4(g) + O2(g) ½ CO2(g) + H2O(l) + 445 .2 kJ

Two Ways to Designate Thermochemical Equations

Note the meaning of the sign in ΔH in the equations above!!

Endothermic: ΔH = +Exothermic: ΔH = -

Two Ways to Designate Thermochemical Equations

Note the important of designating the physical state or phase of matter. Why??

Because this will change the heat of reaction! (ΔH)

Thermochemical EquationsWhat do the coefficients stand for? How can they differ from the ones we have used before?

Coefficients = the number of moles (as before)

BUTWe can use fractional coefficients now!

Thermochemical EquationsWhat is the standard state? How do we designate conditions of temperature and pressure that are not at standard state?

Standard state = 1 atm pressure & 25 °C

ΔH° = ΔH at standard state

Must show conditions over arrow if not at standard state!

Thermochemical Equations

How can we find the enthalpy of reaction when we reverse it?

Reverse the reaction, reverse the sign of ΔH!

Example:CO (g) + ½ O2 (g) CO2 (g) ΔH = -283 kJ

CO2 (g) CO (g) + ½ O2 (g) ΔH = +283 kJ

ExampleGiven Rxn. #1, find the ΔH for Rxns. 2 & 3

Reaction #12 SO2 (g) + O2 (g) 2 SO3 (g) ΔH = +197.8 kJ

Reaction #2SO2 (g) + ½ O2 (g) SO3 (g) ΔH =

Reaction #34 SO3 (g) 4 SO2 (g) + 2 O2(g) ΔH =

ΔH as a Stoichiometric Quantity

Given the reaction below, how much heat is produced when 15.0 g of NO2 are produced?

2 NO (g) + O2 (g) 2 NO2 (g) ΔH = -114.1 kJ

ΔH as a Stoichiometric Quantity

Given: ΔH = -283 kJCO (g) + ½ O2 (g) CO2 (g)

(a) Calculate the enthalpy of the above reaction when 3.00 g of product are formed

ΔH as a Stoichiometric Quantity

Given: ΔH = -283 kJCO (g) + ½ O2 (g) CO2 (g)

(b) If only 10.0 grams of oxygen and an unlimited supply of CO are available to run this reaction, how much heat will be given off?

ΔH as a Stoichiometric Quantity

Given: ΔH = -283 kJCO (g) + ½ O2 (g) CO2 (g)

(c) How many grams of carbon monoxide are necessary (assuming oxygen is unlimited) to produce 500 kJ of energy in this reaction?

ΔH as a Stoichiometric Quantity

Given: ΔH = -283 kJCO (g) + ½ O2 (g) CO2 (g)

(d) Calculate the heat of decomposition of two moles of carbon dioxide.

Hess’s Law The heat of a reaction (ΔH) is constant,

whether the reaction is carried out directly in one step or indirectly through a number of steps.

The heat of a reaction (ΔH) can be determined as the sum of heats of reaction of several steps.

Hess’s Law: ExampleConsider the formation of water:

H2(g) + ½ O2(g) H2O(g) + 241.8 kJ

(Exothermic Rxn ΔH = -241.8 kJ)

Hess’s Law

Hess’s Law

Hess’s LawΣ ΔH along one path =

Σ ΔH along another path

Since ΔH is a state function!!

Hess’s LawGiven:C(s) + O2(g) CO2(g) ΔH = -393.5 kJ

2 CO(g) + O2(g) 2 CO2(g) ΔH = -577.0 kJ

Determine the heat of reaction for:

C(s) + ½ O2(g) CO(g)

Hess’s LawGiven:C(s) + O2(g) CO2(g) ΔH = -393.5 kJ

C2H4(g)+3 O2(g) 2 CO2(g)+2 H2O(l) ΔH= -1410.9 kJ

H2(g) + ½ O2(g) H2O(l) ΔH = -285.8 kJ

Determine the heat of reaction for:

2 C(s) + 2 H2(g) C2H4(g)

Standard Enthalpies of FormationNIST (National Institute for Standards and Technology) gives values for

ΔHf° = standard molar heat of formation

Definition:The heat content or enthalpy change

when one mole of a compound is formed at 1.0 atm pressure and 25 °C from its elements under the same conditions.

Examples of Formation EquationsH2(g) + ½ O2(g) H2O(g)

ΔHf°(H2O, g) = -241.8 kj/mol

C(s) + ½ O2(g) CO(g)

ΔHf°(CO, g) = -111 kj/mol

***Elements/reactants 1 mol of compoundNotice units are per mole

Standard Enthalpy of Formation Values Can look up values of in reference

book or textbook

By definition, ΔHf° = 0 for elements in their standard states Example: Cl2 (g)

H2 (g)

Ca (s)

Summation Equation

In general, when all enthalpies of formation are known:

ΔHrxn° = ΣΔHf°(products) - ΣΔHf°(reactants)

Must multiply all Hf values by coefficient from balanced equation!!!

Summation Equation ExampleUse the summation equation to determine the enthalpy of the following reaction:

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

ΔHreaction° = ΣΔHf°(products) - ΣΔHf°(reactants)

Unit 5: Part II Thermodynamics: Spontaneity, Entropy and Free Energy

Spontaneous Change

What is a spontaneous process?

A process that occurs by itself without an outside force helping it.

[ A spark to start a process is OK though]

Spontaneous Change

Which of the following are spontaneous processes?

1. Snowman melting in the sun

2. Assembling a jigsaw puzzle

3. Rusting of an iron object in humid air

4. Recharging of a camera battery

Spontaneous Reactions and Energy

Many spontaneous reactions are exothermic, but not all!

Example:H2O (s) H2O (l)

is spontaneous and an ENDOTHERMIC reaction!

(∆ H = + 6.0 kJ)

What other Factor Influences Spontaneity?

The Randomness Factor!

Nature tends to move spontaneously to a more random state.

Entropy: Disorder and Spontaneity

What is entropy?

A measure of the randomness (disorder) of

a system—a STATE property (state function)!

Reaction of K Reaction of K with waterwith water

The Second Law of The Second Law of ThermodynamicsThermodynamics

The Second Law of Thermodynamics states:

In a spontaneous process, there is a net increase of entropy (taking into account system and surroundings).

EXAMPLE:

H2O (s) H2O (l)

Water molecules are more disordered as a liquid than as a solid.

Spontaneous Processes result in more random states (more disorder).

Sample Problem

Predict which of the following processes have a positive change in entropy:

(an increase in the randomness or disorder)

a. Taking dry ice from a freezer and allowing it to warm from -80oC to room temperature

b. dissolving blue food coloring in water

c. freezing water into ice cubes

EntropyEntropy is used to quantify

randomness or disorder.

Like enthalpy, entropy is also a state function.

The Third Law of Thermodynamics

The Third Law of Thermodynamics states:

A completely ordered pure crystalline solid has an entropy of

zero at 0 K.

Standard Molar Entropies: ΔSo (1 mole, standard conditions): Tells you entropy at 25oC and 1 atm

(standard state conditions)

Units: J/mol K

Note: Elements DO NOT have ΔSo = 0!(like they did with ΔHo)

Standard Molar Entropies

1. Phase change occurs from s l g

2. # moles of gas increase from reactants to products

3. T increases (KE increases)

For a substance, Entropy generally increases as:

1. Reactants (solids or liquids) Products (gases)

2. Total # moles of products > Total # moles of reactants

3. Total # moles of gaseous products > Total # moles of gaseous reactants

4. T is increasing.

For a reaction, entropy generally increases as:

Sign of ΔSo for a reaction means:

+∆S Entropy increases; S prod > S react

-∆S Entropy decreases; S react > S prod

Predict the sign of ΔS in each of the following reaction, and explain your prediction.

NH3 (g) + HCl (g) NH4Cl (s)

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

CO (g) + H2O (g) CO2 (g) + H2 (g)

Example

Calculation is similar to ∆Ho (from Part I)

Note units are JOULES not kJ as before!

Calculating ∆S for a Reaction

∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)

Example: Calculate ΔSo for the following reaction using the tables in your reference book and the summation equation.

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

Gibbs Free Energy and Free Energy ChangeThe Gibbs (also known as Gibbs-Helmholtz) Equation

shows relationship between Energy, Entropy and Spontaneity:

ΔG = ΔH - T ΔS

Change in Free Energy = Change in Enthalpy – (Temp. x Change in Entropy)

What is free energy?

Free energy =AVAILABLE energy

The Relationship between ΔGreaction

and Spontaneity

1. If ΔG = positive, reaction is NONSPONTANEOUS.

2. If ΔG = zero, reaction is at equilibrium (balanced).

3. If ΔG = negative, reaction is SPONTANEOUS.

Gibbs Free Energy, G

Spontaneous Processes Must Have

a Negative Free Energy!

J. Willard GibbsJ. Willard Gibbs1839-19031839-1903

How are these factors and spontaneity related?

EXOTHERMIC reactions with Increasing Entropy are EXOTHERMIC reactions with Increasing Entropy are

Always spontaneous!Always spontaneous!

EXOTHERMIC reactions with Increasing Entropy are EXOTHERMIC reactions with Increasing Entropy are

Always spontaneous!Always spontaneous!

Case #

ΔH ΔS ΔG Result

1 - + - spontaneous at all T

2 --

--

-+

spontaneous toward low T HOWEVERnonspontaneous toward high T

3 ++

++

+-

nonspontaneous toward low T HOWEVERspontaneous toward high T

4 + - + nonspontaneous at all T

Predict if the reaction will be spontaneous or not. Use ΔH as given and your estimate of the sign of ΔS.

a. C6H12O6 (s) + 6 O2 (g) 6 CO2 (g) + 6 H2O (g)

ΔH = -2540 kJ

b. Cl2 (g) 2 Cl (g) ΔH is positive

Example

Two methods of calculating ∆Go

∆Go = ∆Ho - T∆So

a) Determine ∆Horxn and ∆So

rxn and use Gibbs

equation.

b) Use tabulated values of free energies

of formation, ∆Gfo.

(we will not do this calculation, since it is similar to the ∆Ho one we did in Part I)

Standard Free Energy Change, ∆Go

∆Go = ∆Ho - T∆So

Note:

The units for ΔHo are generally in kJ The units for ΔSo generally are in J

You must convert FIRST before beginning the problem!

T is in K (oC + 273)

Calculate ΔGo for the reaction below, and predict whether the reaction is spontaneous at 25oC.

C (s)+ 2H2 (g) CH4 (g)

ΔSo = -80.8 J/mol K ΔHo = -74.8 kJ/mol T= 298 K

Example: Gibbs Equation

Standard Free Energy of Formation

ΔGorxn. = ΣGo

f(products) - ΣGof(reactants)

SAME SUMMATION EQUATION as ΔHo !!!!!

(Use reference book for values)