Thermodynamics

Antoine Lavoisier [1743-

94]

Julius Robert Meyer

James Joule : 1818~1889

Hermann von Helmholtz (1821-1894)

Rudolf Clausius (1822--1888),

THERMOCHEMISTRTHERMOCHEMISTRYY

The study of heat released or required by chemical reactions

Fuel is burnt to produce energy - combustion (e.g. when fossil fuels are burnt)

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) + energy

What is Energy?

EnergyEnergy

Kinetic energy

(EK)

Potential energy

(EP)

Energy due to motion

Energy due to position (stored energy)

Total Energy = Kinetic Energy + Potential Energy

E = EK + EP

Kinetic energy & potential energy are interchangeable

Ball thrown upwards slows &

loses kinetic energy but gains potential energy

The reverse happens as it falls back to the ground

Law of Conservation of EnergyLaw of Conservation of Energy: the total energy of the universe is constant and can

neither be created nor destroyed; it can only be transformed.

The internal energyinternal energy, U, of a sample is the sum of all the kinetic and potential

energies of all the atoms and molecules in a sample

i.e. it is the total energy of all the atoms and molecules in a sample

Systems & Surroundings

In thermodynamics, the world is divided into a system and its surroundings

A system is the part of the world we want to study (e.g. a reaction mixture in a flask)

The surroundings consist of everything else outside the system

SYSTEM

CLOSED

OPEN ISOLATED

Thermochemistry is the study of heat change in chemical reactions.

The system is the specific part of the universe that is of interest in the study.

open

mass & energyExchange:

closed

energy

isolated

nothing

SYSTEMSURROUNDINGS

OPEN SYSTEM: can exchange both matter and energy with the surroundings (e.g. open reaction

flask, rocket engine)

CLOSED SYSTEM: can exchange only energy with the surroundings (matter remains

fixed) e.g. a sealed reaction flask

ISOLATED SYSTEM: can exchange neither energy nor

matter with its surroundings (e.g. a thermos flask)

HEAT and WORK

HEAT is the energy that transfers from one object to another when the two things are at different temperatures and in some kind of contact

e.g. kettle heats on a gas flame

cup of tea cools down (loses energy as heat)

Thermal motion (random molecular motion) is increased by heat energy

i.e. heat stimulates thermal motion

Work is the transfer of energy that takes place when an object is moved against an opposing force

i.e. a system does work when it expands against an external pressure

Car engine: petrol burns & produces gases which push out

pistons in the engine and transfer energy to the wheels

of car

•Work stimulates uniform motion

• Heat and work can be considered as energy in transit

UNITS OF ENERGY

S.I. unit of energy is the joule (J)

Heat and work ( energy in transit) also measured in joules

1 kJ (kilojoule) = 103 J

Calorie (cal): 1 cal is the energy needed to raise the temperature of 1g of water by 1oC

1 cal = 4.184 J

INTERNAL ENERGY (U)INTERNAL ENERGY (U)

Internal energy changes when energy enters or leaves a system

U = Ufinal - Uinitial

U change in the internal energy

Heat and work are 2 equivalent ways of changing the internal energy of a system

+=Change in internal energy

Energy supplied to system as

heat

Energy supplied to system as

work

U = q (heat) + w (work)

q

w

q

wU

U like reserves of a bank: bank

accepts deposits or withdrawals in two currencies (q & w) but stores

them as common fund, U.

First Law of Thermodynamics:

the internal energy of an isolated system is constant

Signs (+/-) will tell you if energy is entering or leaving a system

+ indicates energy enters a system

- indicates energy leaves a system

•An important form of work is EXPANSION EXPANSION WORK WORK i.e. the work done when a system

changes size and pushes against an external force

e.g. the work done by hot gases in an engine as they push back the pistons

WORK

HEAT

In a system that can’t expand, no work is done (w = 0)

U = q + w

when w = 0, U = q (at constant volume)

•A change in internal energy can be identified with the heat supplied at constant volume

ENTHALPY (H)

(comes from Greek for “heat inside”)

• the change in internal energy is not equal to the heat supplied when the system is free to change its

volume

• some of the energy can return to the surroundings as expansion work

U < q

The heat supplied is equal to the change in another thermodynamic property called enthalpy (H)

i.e. H = q

• this relation is only valid at constant pressure

As most reactions in chemistry take place at constant pressure we can say that:

A change in enthalpy = heat supplied

EXOTHERMIC & ENDOTHERMIC REACTIONSEXOTHERMIC & ENDOTHERMIC REACTIONS

Exothermic process: a change (e.g. a chemical reaction) that releases heat.

A release of heat corresponds to a decrease in enthalpy

Exothermic process: H < 0 (at constant pressure)

Burning fossil fuels is an exothermic

reaction

Endothermic process: a change (e.g. a chemical reaction) that requires (or absorbs) heat.

An input of heat corresponds to an increase in enthalpy

Endothermic process: H > 0 (at constant pressure)

Photosynthesis is an endothermic reaction

(requires energy input from sun)

Forming Na+ and Cl- ions from NaCl is

an endothermic

process

Measuring HeatMeasuring Heat

reaction

reaction

Exothermic reaction, heat Exothermic reaction, heat given off & temperature of given off & temperature of

water riseswater rises

Endothermic reaction, Endothermic reaction, heat taken in & heat taken in &

temperature of water temperature of water dropsdrops

How do we relate change in temp. to the energy transferred?

Heat capacity (J/oC) = heat supplied (J)

temperature (oC)

Heat Capacity = heat required to raise temp. of an object by 1oC

• more heat is required to raise the temp. of a large sample of a substance by 1oC than is needed for a

smaller sample

b. Specific heat is the amount of heat required to raise the temperature of 1 kg of a material by one degree (C or K).

1) C water = 4184 J / kg C

2) C sand = 664 J / kg C

This is why land heats up quickly during the day and

cools quickly at night and why water takes longer.

Why does water have such a high specific heat?

Water molecules form strong bonds with each other; therefore it takes more heat energy to break them. Metals have weak bonds and do not need as much energy to break them.

How to calculate changes in thermal energy

Q = m x T x Cp

Q = change in thermal energy

m = mass of substance

T = change in temperature (Tf – Ti)

Cp = specific heat of substance

c. A calorimeter is used to help measure the specific heat of a substance.

First, mass and temperature of

water are measured

Then heated sample is put

inside and heat flows into

water

T is measured for water to help get its heat gain

This gives the heat lost by

the substance

Knowing its Q value, its mass, and its T, its Cp can be calculated

Specific heat capacity is the quantity of energy required to change the temperature of a 1g sample

of something by 1oC

Specific Heat Capacity (Cs)

Heat capacity

Mass

=

J / oC / g J / oC

g

=

VaporisationVaporisation

Energy has to be supplied to a liquid to enable it to overcome forces that hold molecules together

• endothermic process (H positive)

Melting

Energy is supplied to a solid to enable it to vibrate more vigorously until molecules can move past each other and

flow as a liquid

• endothermic process (H positive)

Freezing

Liquid releases energy and allows molecules to settle into a lower energy state and form a solid

• exothermic process (H negative)

(we remove heat from water when making ice in freezer)

Reaction EnthalpiesReaction Enthalpies

All chemical reactions either release or absorb heat

Exothermic reactions:

Reactants products + energy as heat (H -ve)

Endothermic reactions:

Reactants + energy as heat products (H +ve)

e.g. burning fossil fuels

e.g. photosynthesis

Bond StrengthsBond Strengths

Bond strengths measured by bond enthalpy HB (+ve values)

• bond breaking requires energy (+ve H)

• bond making releases energy (-ve H)

Lattice EnthalpyLattice Enthalpy

A measure of the attraction between ions (the enthalpy change when a solid is broken up into a gas of its ions)

• all lattice enthalpies are positive

• I.e. energy is required o break up solids

Enthalpy of hydration Enthalpy of hydration HHhydhyd

• the enthalpy change accompanying the hydration of gas-phase ions

•Na+ (g) + Cl- (g) Na+ (aq) + Cl- (aq)

• -ve H values (favourable interaction)

WHY DO THINGS DISSOLVE?

• If dissolves and solution heats up : exothermic

•If dissolves and solution cools down: endothermic

Breaking solid into ions

Ions associating with water

Dissolving+ =

Lattice Enthalpy

+ Enthalpy of Hydration

= Enthalpy of Solution

Substances dissolve because energy and matter tend to disperse (spread out in disorder)

2nd law of Thermodynamics

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 HESS’S LAWHESS’S LAW——

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

USING ENTHALPYUSING ENTHALPY

Hess’s Law Hess’s Law & Energy Level Diagrams& Energy Level Diagrams

Hess’s Law Hess’s Law & Energy Level Diagrams& Energy Level Diagrams

Forming H2O can occur in a single step or in a two steps.

∆Htotal is the same no matter which path is followed.

Hess’s Law Hess’s Law & Energy Level Diagrams& Energy Level Diagrams

Hess’s Law Hess’s Law & Energy Level Diagrams& Energy Level Diagrams

Forming CO2 can occur in a single step or in a two steps.

∆Htotal is the same no

matter which path is followed.

Forming CO2 can occur in a single step or in a two steps.

∆Htotal is the same no

matter which path is followed.

• This equation is valid because This equation is valid because ∆H is a ∆H is a STATE FUNCTIONSTATE FUNCTION

• These depend only on the state These depend only on the state of the system and of the system and notnot on how on how the system got there.the system got there.

• V, T, P, energy — and your bank V, T, P, energy — and your bank account!account!

• Unlike V, T, and P, one cannot Unlike V, T, and P, one cannot measure absolute H. Can only measure absolute H. Can only measure ∆H.measure ∆H.

∆∆H along one path =H along one path =

∆∆H along another pathH along another path

∆∆H along one path =H along one path =

∆∆H along another pathH along another path

Standard Enthalpy Standard Enthalpy ValuesValues

Standard Enthalpy Standard Enthalpy ValuesValues

Most ∆H values are labeled Most ∆H values are labeled ∆H∆Hoo

Measured under Measured under standard conditionsstandard conditions

P = 1 bar = 10P = 1 bar = 1055 Pa = 1 atm /1.01325 Pa = 1 atm /1.01325 Concentration = 1 mol/LConcentration = 1 mol/L

T = usually 25 T = usually 25 ooCC

with all species in standard stateswith all species in standard states

e.g., C = graphite and Oe.g., C = graphite and O22 = gas = gas

Enthalpy ValuesEnthalpy Values Enthalpy ValuesEnthalpy Values

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

∆∆H˚ = -242 kJH˚ = -242 kJ

2 H2 H22(g) + O(g) + O22(g) --> 2 H(g) --> 2 H22O(g)O(g)

∆∆H˚ = -484 kJH˚ = -484 kJ

HH22O(g) ---> HO(g) ---> H22(g) + 1/2 O(g) + 1/2 O22(g) (g)

∆∆H˚ = +242 kJH˚ = +242 kJ

HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(liquid)O(liquid)

∆∆H˚ = -286 kJH˚ = -286 kJ

Depend on Depend on how the reaction is writtenhow the reaction is written and on and on phases of reactants and productsphases of reactants and productsDepend on Depend on how the reaction is writtenhow the reaction is written and on and on phases of reactants and productsphases of reactants and products

Standard Enthalpy Standard Enthalpy ValuesValues

NIST (Nat’l Institute for Standards and Technology) gives NIST (Nat’l Institute for Standards and Technology) gives values ofvalues of

∆ ∆HHffoo = standard molar enthalpy of formation = standard molar enthalpy of formation

— — the enthalpy change when 1 mol of compound is formed the enthalpy change when 1 mol of compound is formed from elements under standard conditions.from elements under standard conditions.

See Table 6.2See Table 6.2

∆∆HHffoo, standard molar , standard molar

enthalpy of formationenthalpy of formationEnthalpy change when 1 mol of compound is Enthalpy change when 1 mol of compound is

formed from the corresponding elements under formed from the corresponding elements under standard conditionsstandard conditions

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

∆∆HHffoo (H (H22O, g)= -241.8 kJ/molO, g)= -241.8 kJ/mol

By definition, By definition,

∆∆HHffoo = 0 for elements in their standard = 0 for elements in their standard

states.states.

Using Standard Enthalpy Using Standard Enthalpy ValuesValues

Using Standard Enthalpy Using Standard Enthalpy ValuesValues

Use ∆H˚’s to calculate Use ∆H˚’s to calculate enthalpy changeenthalpy change for for

HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g)

(product is called “(product is called “water gaswater gas”)”)

Using Standard Enthalpy Using Standard Enthalpy ValuesValues

Using Standard Enthalpy Using Standard Enthalpy ValuesValues

HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g)

From reference books we findFrom reference books we find

• HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g) ∆HO(g) ∆Hff˚ = - 242 ˚ = - 242

kJ/molkJ/mol

• C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g) ∆H ∆Hff˚ = - 111 ˚ = - 111

kJ/molkJ/mol

Using Standard Enthalpy Using Standard Enthalpy ValuesValues

Using Standard Enthalpy Using Standard Enthalpy ValuesValues

HH22O(g) --> HO(g) --> H22(g) + 1/2 O(g) + 1/2 O22(g) ∆H(g) ∆Hoo = +242 kJ = +242 kJ

C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g) ∆H∆Hoo = -111 kJ = -111 kJ

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

To convert 1 mol of water to 1 mol each of To convert 1 mol of water to 1 mol each of

HH22 and CO requires 131 kJ of energy. and CO requires 131 kJ of energy.

The “water gas” reaction is ENDOthermic.The “water gas” reaction is ENDOthermic.

HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g)

∆∆HHoonetnet = +131 kJ = +131 kJ

HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g)

∆∆HHoonetnet = +131 kJ = +131 kJ

Using Standard Enthalpy Using Standard Enthalpy ValuesValues

Using Standard Enthalpy Using Standard Enthalpy ValuesValues

In general, when In general, when ALLALL enthalpies of formation enthalpies of formation are known: are known:

Calculate ∆H of Calculate ∆H of reaction?reaction?

∆∆HHoorxnrxn = = ∆H ∆Hff

oo (products) - (products) - ∆H ∆Hff

oo (reactants)(reactants)∆∆HHoo

rxnrxn = = ∆H ∆Hffoo

(products) - (products) - ∆H ∆Hffoo

(reactants)(reactants)

Remember that ∆ always = final – initial

Using Standard Enthalpy Using Standard Enthalpy ValuesValues

Using Standard Enthalpy Using Standard Enthalpy ValuesValues

Calculate the heat of combustion of Calculate the heat of combustion of methanol, i.e., ∆Hmethanol, i.e., ∆Hoo

rxnrxn for for

CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) --> CO(g) --> CO22(g) + 2 H(g) + 2 H22O(g)O(g)

∆∆HHoorxnrxn = = ∆H ∆Hff

oo (prod) - (prod) - ∆H ∆Hff

oo (react)(react)

Using Standard Enthalpy Using Standard Enthalpy ValuesValues

Using Standard Enthalpy Using Standard Enthalpy ValuesValues

∆∆HHoorxnrxn = ∆H = ∆Hff

oo (CO(CO22) + 2 ∆H) + 2 ∆Hff

oo (H(H22O) O)

- {3/2 ∆H- {3/2 ∆Hffoo (O(O22) + ∆H) + ∆Hff

oo (CH(CH33OH)} OH)}

= (-393.5 kJ) + 2 (-241.8 kJ) = (-393.5 kJ) + 2 (-241.8 kJ)

- {0 + (-201.5 kJ)}- {0 + (-201.5 kJ)}

∆∆HHoorxnrxn = -675.6 kJ per mol of methanol = -675.6 kJ per mol of methanol

CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) --> CO(g) --> CO22(g) + 2 H(g) + 2 H22O(g)O(g)

∆∆HHoorxnrxn = = ∆H ∆Hff

oo (prod) - (prod) - ∆H ∆Hff

oo (react)(react)

Measuring Heats of Reaction

CALORIMETRYCALORIMETRYCALORIMETRYCALORIMETRY

Constant Volume “Bomb” Calorimeter

• Burn combustible sample.

• Measure heat evolved in a reaction.

• Derive ∆E for

reaction.

Constant Volume “Bomb” Calorimeter

• Burn combustible sample.

• Measure heat evolved in a reaction.

• Derive ∆E for

reaction.

CalorimetryCalorimetryCalorimetryCalorimetry

Some heat from reaction warms waterqwater = (sp. ht.)(water mass)(∆T)

Some heat from reaction warms “bomb”qbomb = (heat capacity, J/K)(∆T)

Total heat evolved = qtotal = qwater + qbomb

Calculate heat of combustion of octane. Calculate heat of combustion of octane. CC88HH1818 + 25/2 O + 25/2 O22 --> 8 CO --> 8 CO22 + 9 H + 9 H22OO

•• Burn 1.00 g of octaneBurn 1.00 g of octane• Temp rises from 25.00 to 33.20 Temp rises from 25.00 to 33.20 ooCC• Calorimeter contains 1200 g waterCalorimeter contains 1200 g water• Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K

Measuring Heats of ReactionMeasuring Heats of ReactionCALORIMETRYCALORIMETRY

Measuring Heats of ReactionMeasuring Heats of ReactionCALORIMETRYCALORIMETRY

Step 1Step 1 Calc. heat transferred from reaction to water.Calc. heat transferred from reaction to water.q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 Jq = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J

Step 2Step 2 Calc. heat transferred from reaction to bomb.Calc. heat transferred from reaction to bomb.q = (bomb heat capacity)(∆T)q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J= (837 J/K)(8.20 K) = 6860 J

Step 3Step 3 Total heat evolvedTotal heat evolved 41,170 J + 6860 J = 48,030 J41,170 J + 6860 J = 48,030 JHeat of combustion of 1.00 g of octane = - 48.0 kJHeat of combustion of 1.00 g of octane = - 48.0 kJ

Measuring Heats of ReactionMeasuring Heats of ReactionCALORIMETRYCALORIMETRY

Measuring Heats of ReactionMeasuring Heats of ReactionCALORIMETRYCALORIMETRY

Second Law of Thermodynamics:

the disorder (or entropy) of a system tends to increase

ENTROPY (S)

•Entropy is a measure of disorder

• Low entropy (S) = low disorder

•High entropy (S) = greater disorder

• hot metal block tends to cool

• gas spreads out as much as possible

Total entropy change

entropy change of system

entropy change of

surroundings

+=

Dissolving

disorder of solution

disorder of surroundings

• must be an overall increase in disorder for dissolving to occur

1. If we freeze water, disorder of the water molecules decreases , entropy decreases

( -ve S , -ve H)

2. If we boil water, disorder of the water molecules increases , entropy increases

(vapour is highly disordered state)

( +ve S , +ve H)

A spontaneous change is a change that has a tendency to occur without been driven by an

external influence

e.g. the cooling of a hot metal block to the temperature of its surroundings

A non-spontaneous change is a change that occurs only when driven

e.g. forcing electric current through a metal block to heat it

•A chemical reaction is spontaneous if it is accompanied by an increase in the total entropy of the

system and the surroundings

• Spontaneous exothermic reactions are common (e.g. hot metal block spontaneously cooling) because they release heat that increases the entropy of the

surroundings.

•Endothermic reactions are spontaneous only when the entropy of the system increases enough to

overcome the decrease in entropy of the surroundings

System in Dynamic EquilibriumSystem in Dynamic Equilibrium

A + B C + D

Dynamic (coming and going), equilibrium (no net change)

• no overall change in disorder

S 0 (zero entropy change)

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