Chapter 6 Thermochemistry

Prentice Hall © 2003 Chapter 5

Chapter 6Chapter 6ThermochemistryThermochemistry

CHEMISTRY


ThermodynamicsThermodynamics

- is the study of energy and the transformations it undergoes in chemical reactions.

Units: Joules (J)calorie = 4.184 J

calorie – the amount of E needed to raise the temperature of 1 gram of water 1 oC


EnergyEnergy

• Examples of energy:

• Potential E• Kinetic E• Work – E needed to move an object against a force• Heat – E transferred from hot to cold objects

- capacity to do work or transfer heat


Today’s TopicsToday’s Topics

• 4 Thermodynamic Functions• Definition of State Function• Internal Energy• Point of Views – System, Surroundings Universe• Sign Conventions


4 Thermodynamic Functions4 Thermodynamic Functions

• E - Internal Energy• H - Enthalpy• S - Entropy• G - Gibb’s Free Energy


What is a state function?What is a state function?

• A state function is a property that depends on the present condition and not on how the change occurs.

• Derived from calculating the change: = [Final value – initial value]


4 State Functions4 State Functions

• Therefore:•

E = Ef -Ei

H = Hf - Hi

S = Sf - Si

G = Gf - Gi


Internal Energy (E)Internal Energy (E)

• Internal Energy - is the sum of the kinetic and potential energy of a system

• Is the sum of heat (q) and work (w)

E = Efinal – Einitial

E = q + w


Point of ViewsPoint of Views

• System - the reaction we are studying

• Surroundings – anything else besides the reactionFor example: the container, you, etc….

• Universe – system + surroundings


q = heat and w = workq = heat and w = work

• If q is (+), system gaining heat from surroundings (endo)

• If q is (-), system giving up heat to the surroundings (exo)

• If w is (+), system is the recipient of work from the surroundings. (in short, surroundings is doing work on the system.


Thermodynamic FunctionsThermodynamic Functions

• Have value, unit and magnitude.

• The sign of E depends on the magnitude of q and w.

• Knowing the value of E does not tell us which variable is larger, q or w.


Problem 1Problem 1

• Calculate the change in internal energy of the system for a process in which the system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings.


Problem 2Problem 2

• Consider the reaction of hydrogen and oxygen gases to produce water. As the reaction occurs, the system loses 1150 J of heat to the surroundings. The expanding gas does 480 J of work on the surroundings as it pushes against the atmosphere. Calculate the change in the internal energy of the system?


Problem 3Problem 3

• Calculate E and determine whether the process is endothermic or exothermic.

• 1.) q = 1.62 kJ and w = -874 J• 2.) The system releases 113 kJ of heat to the

surroundings and does 39 kJ of work.• 3.) The system absorbs 77.5 kJ of heat while doing

63.5 kJ of work on the surroundings.


Thermodynamic FunctionsThermodynamic Functions

• Have value, unit and magnitude.

• The sign of E depends on the magnitude of q and w.

• Knowing the value of E does not tell us which variable is larger, q or w.


EnthalpyEnthalpy

• Enthalpy – accounts for heat flow in chemical reactions that occur at constant P when nothing other than P-V work are performed

H = Hfinal - Hinitial

• If: H = E + PV• Then:H = E + PV


Ways of Measuring Ways of Measuring HH

• Calorimetry

• Hess’s Law

• Heats of Formation (Hof)


CalorimetryCalorimetry

• - Is the measurement of heat flow

• Specific Heat capacity (C) - the amount of heat needed to raise the temperature of 1 g of substance 1 oC.

• The greater the heat capacity, the greater the heat required to produce a rise in temp.


Calorimetry EquationCalorimetry Equation

• q = mCT

• - qsystem = qsurroundings

• - qsystem = qwater + qcalorimeter

• - qsubstance = (mCTsolution + mCTcalorimeter)

• Simplifies to:

• - qsubstance = (mCTsolution + CTcalorimeter)


Calorimetry ProblemCalorimetry Problem

• A 30.0 gram sample of water at 280 K is mixed with 50.0 grams of water at 330 K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The specific heat capacity of the solution is 4.18 J/g-oC.



• A 46.2 gram sample of copper is heated to 95.4 oC and then placed in a calorimeter containing 75.0 gram of water at 19.6 oC. The final temperature of the metal and water is 21.8 oC. Calculate the specific heat capacity of copper, assuming that all the heat lost by the copper is gained by the water.



• A 15.0 gram sample of nickel metal is heated to 100.0 oC and dropped into 55.0 grams of water, initially at 23 oC. Assuming that no heat is lost to the calorimeter, calculate the final temperature of the nickel and water. The specific heat of nickel is 0.444 J/g-oC. The specific heat of water is 4.18 J/g-oC.


Problem 4Problem 4

• The specific heat of water is 4.18 J/g-K.• How much heat is needed to warm 250 g of water

from 22 oC to 98 oC?

• What is the molar heat capacity of water?

• Molar heat Capacity = C x molar mass


Problem 5Problem 5

• Large beds of rocks are used in solar heated homes to store heat. Assume that the specific heat of rocks is 0.82 J/g-K.

• A. Calculate the amount of heat absorbed by 50 kg. of rocks if their temperature rose by 12.0 oC.

• B. What temperature change would these rocks undergo if they emitted 450 kJ of heat?


• Work – E needed to move an object against a force

• When P is constant, P-V work is given byw = - PV


EnthalpyEnthalpy

• Enthalpy of a reaction or heat of Reaction: H = Hproducts - Hreactants

• 1. sign of H depends on the amount of reactant consumed

• 2. H sign is opposite for backwards reaction• 3. Hrxn depends on the physical state of the

reactants and products.


Problem 1Problem 1

• Given the reaction:

• 2H2 (g) + O2 (g) 2 H2O (g) H = -483 kJ

• Calculate the H value for:

• 2 H2O (g) 2H2 (g) + O2 (g)


Problem 2Problem 2

• Given the reaction:

• 2H2 (g) + O2 (g) 2 H2O (g) H = -483 kJ

• How much heat is released when 10.5 grams of H2 is burned in a constant-pressure system?


Ways of Measuring Ways of Measuring HH

• Calorimetry

• Hess’s Law

• Heats of Formation (Hof)


Hess’ LawHess’ Law

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


Problem 1Problem 1

• Given:

• C (s) + O2 (g) CO2 (g) H = -393.5 kJ

• CO(g) + ½ O2 (g) CO2 (g) H = -283.5 kJ

• Calculate the enthalpy of combustion for:C(s) + ½ O2 (g) CO (g)


Problem 2Problem 2

• Given:

• C(graphite) + O2 (g) CO2 (g) H = -393.5 kJ

• C(diamond) + O2 (g) CO2 (g) H = -395.4 kJ

• Calculate the enthalpy of combustion for:C(graphite) C (diamond)


Problem 3Problem 3

• Given:• C2H2 (g) + 5/2 O2 (g) 2CO2 (g) + H2O (l) H = -1299.6 kJ

• C (s) + O2 (g) CO2 (g) H = -395.4 kJ

• H2 (g) + 1/2 O2 (g) H2O (l) H = -285.8 kJ

• Calculate the enthalpy of combustion for:2C (s) + H2 (g) C2H2 (g)


Enthalpies of FormationEnthalpies of Formation

• - known as heat of formation • - gives the energy needed for a compound to form

• Standard enthalpy - is the enthalpy change (H) when the reactants and products are in their standard state, usually 1 atm and 25 oC - denoted by Ho ( ex. Ho

f)


Standard Enthalpy of Standard Enthalpy of Formation, (Formation, (HHoo

ff))

• By definition:

• The standard enthalpy of formation ( ex. Hof) of the

most stable form of any element is ZERO because there is no formation reaction needed when the element is in its standard state.

• - important for diatomic molecules• - need knowledge of standard states of compounds


• If there is more than one state for a substance under standard conditions, the more stable one is used.

• Standard enthalpy of formation of the most stable form of an element is zero.

Using Enthalpies of Formation of Calculate Enthalpies of Reaction

• We use Hess’ Law to calculate enthalpies of a reaction from enthalpies of formation.






• For a reaction


reactantsproductsrxn ff HmHnH


Next TopicNext Topic

S = Entropy = disorder


Thermodynamic QuestionThermodynamic Question

• Can a process occur?• Spontaneous or Non-spontaneous?

– Forward vs. Reverse reactions

– Example: Gas Expansion

• Reversible or irreversible?


EntropyEntropy

• Entropy, S, is a measure of the disorder of a system.• Spontaneous reactions proceed to lower energy or

higher entropy.• In ice, the molecules are very well ordered because of

the H-bonds.• Therefore, ice has a low entropy.


• As ice melts, the intermolecular forces are broken (requires energy), but the order is interrupted (so entropy increases).

• Water is more random than ice, so ice spontaneously melts at room temperature.

• Conclusion: The higher the entropy the more spontaneous the reaction.


• Generally, when an increase in entropy in one process is associated with a decrease in entropy in another, the increase in entropy dominates.

• Entropy is a state function. • For a system, S = Sfinal - Sinitial.• If S > 0 the randomness increases, if S < 0 the

order increases.


Entropy• Suppose a system changes reversibly between state 1

and state 2. Then, the change in entropy is given by

– at constant T where qrev is the amount of heat added reversibly to the system. (Example: a phase change occurs at constant T with the reversible addition of heat.)

Entropy and the Second Entropy and the Second Law of ThermodynamicsLaw of Thermodynamics

)(constant revsys T

Tq

S


The Second Law of Thermodynamics• Spontaneous processes have a direction.• In any spontaneous process, the entropy of the

universe increases. Suniv = Ssys + Ssurr: the change in entropy of the

universe is the sum of the change in entropy of the system and the change in entropy of the surroundings.

• Entropy is not conserved: Suniv is increasing.



The Second Law of Thermodynamics

• Reversible process: Suniv = 0.

• Spontaneous process (i.e. irreversible): Suniv > 0.

Ssys for a spontaneous process can be less than 0 as long as Ssurr > 0. This would make Suniv still (+).

• For an isolated system, Ssys = 0 for a reversible process and Ssys > 0 for a spontaneous process.



Third Law of Third Law of ThermodynamicsThermodynamics

• The entropy of a perfect crystal at 0 K is zero.

• Entropy changes dramatically at a phase change.

• As we heat a substance from absolute zero, the entropy must increase.



• Absolute entropy can be determined from complicated measurements.

• Standard molar entropy, S: entropy of a substance in its standard state. Similar in concept to H.

• Units: J/mol-K. Note units of H: kJ/mol.• Standard molar entropies of elements (S ) are not 0.• For a chemical reaction which produces n moles of

products from m moles of reactants:

Entropy Changes in Entropy Changes in Chemical ReactionsChemical Reactions

reactantsproducts mSnSS



For a spontaneous reaction the entropy of the universe must increase.

• Reactions with large negative H values are spontaneous.

• How do we correlate S and H to predict whether a reaction is spontaneous?

• Gibbs free energy, G, of a state is

• For a process occurring at constant temperature

Gibbs Free EnergyGibbs Free Energy

TSHG

STHG


Gibb’s Free EnergyGibb’s Free Energy

• Is the capacity to do maximum useful work.

• Heat decreases the amount of useful work done.


WORKWORK

• At constant P, wsys = - PV

• At constant temperature, wsys = - TS

• If G = H – TS, for maximum useful work, H must be = 0.


Three important conditions:– If G < 0 then the forward reaction is spontaneous.

– If G = 0 then reaction is at equilibrium and no net reaction will occur.

– If G > 0 then the forward reaction is not spontaneous. If G > 0, work must be supplied from the surroundings to drive the reaction.

• For a reaction the free energy of the reactants decreases to a minimum (equilibrium) and then increases to the free energy of the products.



ProblemProblem

• Correlate S and H to predict whether a reaction is non-spontaneous or spontaneous. If spontaneous, determine whether the reaction will be spontaneous at all temperature, at high temperature, or at low temperature.

• A. If S is (-) and H is (+).• B. If S is (-) and H is (-).• C. If S is (+) and H is (+).• D. If S is (+) and H is (-).


• Consider the formation of ammonia from N2 and H2.

• Initially ammonia will be produced spontaneously (Q < Keq).

• After some time, the ammonia will spontaneously react to form N2 and H2 (Q > Keq).

• At equilibrium, ∆G = 0 and Q = Keq.


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


Standard Free-Energy Changes Gf• Standard states are: pure solid, pure liquid, 1 atm (gas), 1

M concentration (solution), and G = 0 for elements.

G for a process is given by

• The quantity G for a reaction tells us whether a mixture of substances will spontaneously react to produce more reactants (G > 0) or products (G < 0).


reactantsproducts ff GmGnG


Free Energy and Free Energy and TemperatureTemperature


• Recall that G and K (equilibrium constant) apply to standard conditions.

• Recall that G and Q (equilibrium quotient) apply to any conditions.

• It is useful to determine whether substances under any conditions will react:

Free Energy and The Free Energy and The Equilibrium ConstantEquilibrium Constant

QRTGG ln


• At equilibrium, Q = K and G = 0, so

• From the above we can conclude:– If G < 0, then K > 1.

– If G = 0, then K = 1.

– If G > 0, then K < 1.

Free Energy and The Free Energy and The Equilibrium ConstantEquilibrium Constant

eq

eq

KRTG

KRTG

QRTGG

ln

ln0

ln


Calculating Calculating H, H, S, S, GG

• Use Hess’ Law• Standard Heats of Formation (Ho, So, Go )• Equations


• Next Chapter: ELECTROCHEMISTRY


• Galvanic Cell – spontaneous

• Electrochemical Cell – non-spontaneous





Hess’ LawHess’ Law

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



Kinetic Energy and Potential Energy• Kinetic energy is the energy of motion:

• Potential energy is the energy an object possesses by virtue of its position.

• Potential energy can be converted into kinetic energy. Example: a bicyclist at the top of a hill.

The Nature of EnergyThe Nature of Energy

2

21

mvEk


Systems and Surroundings• System: part of the universe we are interested in.• Surroundings: the rest of the universe.

We sometimes use the calorie instead of the joule:1 cal = 4.184 J (exactly)

A nutritional Calorie:1 Cal = 1000 cal = 1 kcal



Transferring Energy: Work and Heat• Force is a push or pull on an object.• Work is the product of force applied to an object over a

distance:

• Energy is the work done to move an object against a force.

• Heat is the transfer of energy between two objects.• Energy is the capacity to do work or transfer heat.


dFw


Internal Energy• Internal Energy: total energy of a system.• Cannot measure absolute internal energy.• Change in internal energy,

The First Law of The First Law of ThermodynamicsThermodynamics

initialfinal EEE


Relating E to Heat and Work• 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:

when a system undergoes a physical or chemical change, the change in internal energy is given by the heat added to or absorbed by the system plus the work done on or by the system:


wqE



Exothermic and Endothermic Processes• Endothermic: absorbs heat from the surroundings.• Exothermic: transfers heat to the surroundings.• An endothermic reaction feels cold.• An exothermic reaction feels hot.

• State function: depends only on the initial and final states of system, not on how the internal energy is used.



• Chemical reactions can absorb or release heat.• However, they also have the ability to do work.• For example, when a gas is produced, then the gas

produced can be used to push a piston, thus doing work.

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

• The work performed by the above reaction is called pressure-volume work.

• When the pressure is constant,

EnthalpyEnthalpy

VPw


• Enthalpy, H: Heat transferred between the system and surroundings carried out under constant pressure.

• Enthalpy is a state function.• If the process occurs at constant pressure,

EnthalpyEnthalpy

PVEH

VPE

PVEH


• Since we know that

• We can write

• When H, is positive, the system gains heat from the surroundings.

• When H, is negative, the surroundings gain heat from the system.

EnthalpyEnthalpy

VPw

wq

VPEH

P


EnthalpyEnthalpy


• For a reaction:

• Enthalpy is an extensive property (magnitude H is

directly proportional to amount):

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(g) H = 1604 kJ

Enthalpies of ReactionEnthalpies of Reaction

reactantsproducts

initialfinal

HH

HHH


• When we reverse a reaction, we change the sign of H:

CO2(g) + 2H2O(g) CH4(g) + 2O2(g) H = +802 kJ

• Change in enthalpy depends on state:

H2O(g) H2O(l) H = -88 kJ

Enthalpies of ReactionEnthalpies of Reaction


Heat Capacity and Specific Heat• Calorimetry = measurement of heat flow.• Calorimeter = apparatus that measures heat flow.• Heat capacity = the amount of energy required to raise

the temperature of an object (by one degree).• Molar heat capacity = heat capacity of 1 mol of a

substance.• Specific heat = specific heat capacity = heat capacity of 1

g of a substance.


Tq substance of gramsheat specific


Constant Pressure Calorimetry• Atmospheric pressure is constant!


T

qq

qH P

solution of grams

solution ofheat specificsolnrxn


• Hess’s law: if a reaction is carried out in a number of steps, H for the overall reaction is the sum of H for each individual step.

• For example:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

2H2O(g) 2H2O(l) H = -88 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ

Hess’s LawHess’s Law


• If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation, Ho

f .

• Standard conditions (standard state): 1 atm and 25 oC (298 K).

• Standard enthalpy, Ho, is the enthalpy measured when everything is in its standard state.

• Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states.



• If there is more than one state for a substance under standard conditions, the more stable one is used.

• Standard enthalpy of formation of the most stable form of an element is zero.


• We use Hess’ Law to calculate enthalpies of a reaction from enthalpies of formation.






• For a reaction


reactantsproductsrxn ff HmHnH

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Chapter 6 Thermochemistry