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Chapter 6Chapter 6

EnergyEnergyThermochemistryThermochemistry

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The Nature of EnergyThe Nature of Energy The concept of energy is quite familiar, energy itself is The concept of energy is quite familiar, energy itself is

rather difficult to define precisely.rather difficult to define precisely. We will define We will define energyenergy as the ability to do work or to as the ability to do work or to

produce heat.produce heat. One of the most important characteristics of energy is One of the most important characteristics of energy is

that it is conserved.that it is conserved. Law of conservation of energyLaw of conservation of energy state that: energy can be state that: energy can be

convert from one form to another but can be neither convert from one form to another but can be neither created nor destroyed.created nor destroyed.

Energy can be classified as:Energy can be classified as:- potential energy.- potential energy.- kinetic energy.- kinetic energy.

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Potential energy (P.E.): Potential energy (P.E.): energy due to position or energy due to position or composition. composition. e.g.: e.g.: water behind dam has P.E. that can be convert to water behind dam has P.E. that can be convert to work when water flows down through turbines, creating work when water flows down through turbines, creating electricity.electricity.e.g.: e.g.: the energy released when gasoline is burned result the energy released when gasoline is burned result from differences in attractive forces between the nuclei from differences in attractive forces between the nuclei and electrons in the reactants and products.and electrons in the reactants and products.

Kinetic energy (K.E.): Kinetic energy (K.E.): due to motion of the object. due to motion of the object. Depend on the mass of the object (m) and its velocity(v)Depend on the mass of the object (m) and its velocity(v)

K.E. = ½ (mvK.E. = ½ (mv22)) Energy can be convert from one form to anotherEnergy can be convert from one form to another For example, consider the two balls in the following For example, consider the two balls in the following

figurefigure

44

B

A

P.E. of ball A > P.E. of ball B (because its higher position).When (A) moves down the hill (P.E. of (A) has decreased and change to K.E.) and strike (B), part of this K.E. is then transfer to (B) causing it to raised to higher final position.Thus P.E. of (B) increased. The final position of (B) has been increased but lower than the original of position (A).Both balls in their final position are at rest, so the missing energy can not be due to their motion.

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What happened to the remaining energy?What happened to the remaining energy?As ball (A) rolls down the hill some of its K.E. is transferred As ball (A) rolls down the hill some of its K.E. is transferred

to the surface of hill as to the surface of hill as heatheat which is called frictional which is called frictional heating. The heating. The temperaturetemperature of the hill increase very of the hill increase very slightly.slightly.

What is the difference between heat and temperature?What is the difference between heat and temperature?Temperature: Temperature: is a property that reflect the random motion is a property that reflect the random motion

of the particles in a particular substance.of the particles in a particular substance.Heat:Heat: transfer of energy between two objects due to a transfer of energy between two objects due to a

temperature difference.temperature difference.From the previous figure: in going from the initial to final From the previous figure: in going from the initial to final

arrangement, ball (B) gain P.E. because arrangement, ball (B) gain P.E. because workwork was done was done by ball (A) on ball (B).by ball (A) on ball (B).

WorkWork is define as: force acting over a distance. is define as: force acting over a distance.

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Work is required to raise (B) from its original position to Work is required to raise (B) from its original position to its final position.its final position.Part of stored energy as P.E. in (A) has been transferred Part of stored energy as P.E. in (A) has been transferred through work to (B), increasing the P.E. of (B)through work to (B), increasing the P.E. of (B)Thus there are two ways to transfer energy:Thus there are two ways to transfer energy:

- Through work- Through work - through heat- through heatBall (A) will always lost the same amount of P.E.Ball (A) will always lost the same amount of P.E.The way that this energy transfer as work or heat depends The way that this energy transfer as work or heat depends on the specific conditions – the pathway.on the specific conditions – the pathway.e.g.: e.g.: surface of the hill so rough, more frictional heating, less surface of the hill so rough, more frictional heating, less work done on ball (B) and can not move to next level, no work work done on ball (B) and can not move to next level, no work donedoneThe amount of heat and work will differ. The total energy The amount of heat and work will differ. The total energy transferred will be constant. Energy change is independent transferred will be constant. Energy change is independent of the pathway, work and heat depend on the pathway. of the pathway, work and heat depend on the pathway.

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This brings us to a very important concept: the This brings us to a very important concept: the state state functionfunction or or state propertystate property..State function: a property of the system that: State function: a property of the system that: depends only on its present state.depends only on its present state.does not depend in any way of the system’s past or futuredoes not depend in any way of the system’s past or future its value does not depend on how the system arrived at its value does not depend on how the system arrived at the present statethe present state a change in this function in going from one state to a change in this function in going from one state to another state is independent of the particular pathway another state is independent of the particular pathway taken between two states.taken between two states.Energy is a state function, work and heat is not a state Energy is a state function, work and heat is not a state function…. it’s a path function function…. it’s a path function

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Chemical EnergyChemical EnergyThe ideas we have just illustrated also apply to chemical The ideas we have just illustrated also apply to chemical systems.systems.To discuss the chemical reactions, the universe is divided into To discuss the chemical reactions, the universe is divided into two halves.two halves.The system: The system: is the part of the universe on which we wish to is the part of the universe on which we wish to focus our attention.focus our attention. The surrounding: The surrounding: include everything else in the universe.include everything else in the universe.For the reaction: For the reaction:

CHCH4(g)4(g) + 2 O + 2 O2(g)2(g) CO CO2(g)2(g) + 2 H + 2 H22OO(g)(g) + + energyenergyThe system: is the reactants and products.The system: is the reactants and products.The surrounding: everything else … container, room..ect.The surrounding: everything else … container, room..ect.

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When the reaction evolve heat, it is said to be When the reaction evolve heat, it is said to be ExothermicExothermic reactions (exo- mean “out of ’’). Energy flow reactions (exo- mean “out of ’’). Energy flow out of the out of the systemsystem. e.g.: combustion of CH. e.g.: combustion of CH44

Endothermic reactions : Endothermic reactions : absorb energy from the absorb energy from the surroundings. Energy flow surroundings. Energy flow into a systeminto a system..

Where does the energy come from in an exothermic reactions? Where does the energy come from in an exothermic reactions? It come from the difference in P.E. between the products and the It come from the difference in P.E. between the products and the

reactants.reactants.In an exothermic reactions, some of the P.E. stored in the chemical In an exothermic reactions, some of the P.E. stored in the chemical

bonds is being converted to thermal energy via heat.bonds is being converted to thermal energy via heat.In an endothermic reactions, the situation is reversed. Energy that In an endothermic reactions, the situation is reversed. Energy that

flows into the system as heat is used to increase the P.E. of the flows into the system as heat is used to increase the P.E. of the products.products.

This is shown in the following figures for exo- and endothermic This is shown in the following figures for exo- and endothermic reactions reactions

1010

SystemSurroundings

Energy released to surrounding as heat

E <0Pote

ntia

l en

ergy

∆(P.E.)

2 mol O2

1 mol CH4

Reactants

2 mol H2O1 mol CO2

(products)

CH + 2O CO + 2H O + Heat4 2 2 2

1111

SystemSurroundings

Energy absorbed from the surroundings

E >0

N + O 2NO2 2 + heat Po

tent

ial e

nerg

y

∆(P.E.)

2 mol NO(Products)

1 mol N21 mol O2(Reactants)

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The study of energy and its interconversions is called The study of energy and its interconversions is called thermodynamics.thermodynamics.The law of conservation of energy is called the The law of conservation of energy is called the first law first law of thermodynamics of thermodynamics and is stated as follows: and is stated as follows: the energy the energy of the universe is constant.of the universe is constant.The The internal energy (E) internal energy (E) of a system can be define as: of a system can be define as: the the sum of the K.E. and P.E. of all “particles” in the system.sum of the K.E. and P.E. of all “particles” in the system.The internal energy can be changed by a flow of The internal energy can be changed by a flow of work, work, heat, or bothheat, or both. That is:. That is: ∆∆E = q + wE = q + w ∆E: change in the system’s internal energy

q: the heat. w: work

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Every thermodynamic quantity has three parts.Every thermodynamic quantity has three parts.1.1. A unit ( Joules of calories).A unit ( Joules of calories).2.2. A number: giving the magnitude of the change.A number: giving the magnitude of the change.3.3. A sign: indicate the direction of the flow.A sign: indicate the direction of the flow.For exothermic process:For exothermic process:

q: q: negative, -ve sign indicates that the system’s negative, -ve sign indicates that the system’s energy is decreasing.energy is decreasing. Heat flow out the systemHeat flow out the systemw: negative, system does work on the surrounding.w: negative, system does work on the surrounding.energy flow out of the system.energy flow out of the system.

For endothermic process:For endothermic process:q: positive, +ve sign indicates that the system’s q: positive, +ve sign indicates that the system’s energy is increasing. Heat flow into the system.energy is increasing. Heat flow into the system.w: positive, work done on the system by the w: positive, work done on the system by the surrounding. Energy flow into of the system.surrounding. Energy flow into of the system.

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SummarySummary Heat given off is negative.Heat given off is negative. Heat absorbed is positive.Heat absorbed is positive. Work done by system on surroundings is Work done by system on surroundings is

positive.positive. Work done on system by surroundings is Work done on system by surroundings is

negative.negative. q , w : positive: energy enter system.q , w : positive: energy enter system. q , w : negative: energy leaves system.q , w : negative: energy leaves system.

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Example: Example: A system released 125 kJ of heat while 104 kJ A system released 125 kJ of heat while 104 kJ of work is done on it, calculate ∆Eof work is done on it, calculate ∆Esys.sys. , ∆E , ∆Esurr.surr.

q = - 125 kJ w = + 104 kJq = - 125 kJ w = + 104 kJ

∆∆EEsys.sys. = q + w = - 125 kJ + 104 kJ = - 21 kJ= q + w = - 125 kJ + 104 kJ = - 21 kJ

∆∆EEsurr. surr. = + 21 kJ= + 21 kJ

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A common type of work associated with chemical process A common type of work associated with chemical process is work done by a gas (through expansion) or work done is work done by a gas (through expansion) or work done on a gas (through compression).on a gas (through compression).Suppose we have a gas in Suppose we have a gas in a cylindrical containera cylindrical containerwith movable piston aswith movable piston asshown in this figure whereshown in this figure whereF: force acting on pistonF: force acting on piston of area (A)of area (A)Pressure is define as forcePressure is define as force per unit areaper unit area P = F/AP = F/A

∆h ∆h ∆V

Initial state Final state

Area = A

P = F/AP = F/A

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Work is defined as a force acting over a distance.Work is defined as a force acting over a distance. w= force x distance = F x ∆hw= force x distance = F x ∆h P = F/ A or F = P x AP = F/ A or F = P x A w= (P x A x ∆ hw= (P x A x ∆ h Volume of the cylinder (V) = A x hVolume of the cylinder (V) = A x h w = P x ∆Vw = P x ∆V

where ∆V = final volume – initial volumewhere ∆V = final volume – initial volume P: the external pressureP: the external pressure

units of work is liter - atm (L-atm)units of work is liter - atm (L-atm) to change this unit to joule (J) (1.0 L. atm. = 101.3 J)to change this unit to joule (J) (1.0 L. atm. = 101.3 J) If the volume of a gas increases, the system has done work on If the volume of a gas increases, the system has done work on

the surroundings. work is negative.the surroundings. work is negative. If the volume of a gas decreases, the surroundings If the volume of a gas decreases, the surroundings

has done work on the system. work is positivehas done work on the system. work is positive

Volume of the cylinder

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Example: A) Example: A) What amount of work is done when 15 L What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure?of gas is expanded to 25 L at 2.4 atm pressure?

B)B) If 2.36 J of heat are absorbed by the gas above. what If 2.36 J of heat are absorbed by the gas above. what is the change in energy?is the change in energy?

C)C) How much heat would it take to change the gas How much heat would it take to change the gas without changing the internal energy of the gas? without changing the internal energy of the gas? A) A) w = P ∆V = 2.4 atm. x (25 – 15) = 24 L. atm.w = P ∆V = 2.4 atm. x (25 – 15) = 24 L. atm.

w = - 24 L. atm. x 101 (J/L. atm) = - 2431.2 Jw = - 24 L. atm. x 101 (J/L. atm) = - 2431.2 JB) ∆E = w + q = - 2431.2 + 2.36 J = - 2428.84 JB) ∆E = w + q = - 2431.2 + 2.36 J = - 2428.84 JC) ∆E = 0C) ∆E = 0

q = w so q = + 2431.2 Jq = w so q = + 2431.2 J

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Enthalpy and CalorimetryEnthalpy and Calorimetry Enthalpy Enthalpy A less familiar property of a system is its enthalpy, A less familiar property of a system is its enthalpy,

abbreviated H, which is define asabbreviated H, which is define asH = E + PV H = E + PV

E: internal energyE: internal energy P: pressure of the systemP: pressure of the systemV: volume of the systemV: volume of the system

Since (E, P, V) are all state functions, Since (E, P, V) are all state functions, enthalpy is also a enthalpy is also a state function.state function.

What exactly is enthalpy? What exactly is enthalpy? To answer this question, consider a process carried out To answer this question, consider a process carried out

at constant pressure, the only work allowed is pressure-at constant pressure, the only work allowed is pressure-volume work (w = -P∆V)volume work (w = -P∆V)

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Under these conditions:

∆E = qp + w = qp - P∆V

qp = ∆E + P ∆V (qp: heat at constant pressure)

Since : H = E + PV then change in H = (change in E) + (change in PV)

∆H = ∆E + ∆(PV) ∆H = ∆E + P∆V (since P is constant) identical So : ∆H = qp At constant pressure, the change in the enthalpy ∆H of the

system is equal to the energy flow as heat.

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This means that for a reaction studied at constant This means that for a reaction studied at constant pressure, the flow of heat is measured as ∆H pressure, the flow of heat is measured as ∆H

∆∆H = HH = Hproductsproducts – H – Hreactantsreactants

HHproducts products >> H Hreactantsreactants , ∆H will be positive, , ∆H will be positive,

heat absorbed by the system,heat absorbed by the system,the reaction is endothermic the reaction is endothermic

HHproducts products < H< Hreactantsreactants , ∆H will be negative, , ∆H will be negative, heat evolved by the heat evolved by the

system,system,the reaction is the reaction is

exothermic exothermic

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Example:Example: consider the following reaction: consider the following reaction: CHCH4(g)4(g) + 2 O + 2 O2(g)2(g) CO CO2(g)2(g) + 2 H + 2 H22OO(ℓ)(ℓ) ∆H= -981kJ ∆H= -981kJ

Calculate the enthalpy change for each of the following Calculate the enthalpy change for each of the following cases: A) 1.0 g cases: A) 1.0 g CHCH44 of is burned in enough of is burned in enough OO22

B) 1x10B) 1x1033 L L CHCH44 of gas at 740 torr (0.97 atm.)at of gas at 740 torr (0.97 atm.)at 25ºC is burned in excess 25ºC is burned in excess OO22

A) mole of CA) mole of CHH4 4 = 1.0 g/ (16g/mol) = 0.0625 mol= 1.0 g/ (16g/mol) = 0.0625 mol

from the equation: 1 mol Cfrom the equation: 1 mol CHH4 4 evolve 981 kJevolve 981 kJ qqpp = ∆H = (- 981kJ/mol) x 0.0625 mol = -55.69 kJ = ∆H = (- 981kJ/mol) x 0.0625 mol = -55.69 kJ

B) n =PV/RT = 0.97 atm.x1x10B) n =PV/RT = 0.97 atm.x1x1033 L/(0.0821 L.atm/mol K)298K L/(0.0821 L.atm/mol K)298K n = 39.65 moln = 39.65 mol ∆ ∆H = (- 891 kJ/mol) x 39.65 mol = - 35325.6 kJH = (- 891 kJ/mol) x 39.65 mol = - 35325.6 kJ

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CalorimetryCalorimetryCalorimetry: is the science of measuring heat. It based on Calorimetry: is the science of measuring heat. It based on

observing the temperature change when body absorbed or observing the temperature change when body absorbed or released energy as heatreleased energy as heat

The device used experimentally to determine the heat The device used experimentally to determine the heat associated with chemical reaction is called a associated with chemical reaction is called a calorimeter.calorimeter.

Substances respond differently to being heated, this property is measured by the heat capacity, C, which defined as: C = (heat absorbed/ increase in temp.) = (J/ºC)

Heat capacity is the energy required to raise the temperature of a substance by one degree Celsius.

Energy required depend on the amount of substance present.

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In defining the heat capacity of a substance, the amount of substance must be specified.

If the heat capacity is given per gram of substance, it is called specific heat capacity, and its unit is (J/ºC. g)

If the heat capacity is given per mol of substance, it is called molar heat capacity, and its unit is (J/ºC. mol)

The specific heat capacities of some common substances are given in the

following table:substance Specific heat capacity

(J/ºC.g)H2O(ℓ) 4.18H2O(s) 2.03Al(s) 0.89Fe(s) 0.45Hg(ℓ) 0.14C(s) 0.71

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Two kinds of calorimeters are usedTwo kinds of calorimeters are used1. Constant pressure calorimeter (called a coffee-cup 1. Constant pressure calorimeter (called a coffee-cup

calorimeter)calorimeter) - - Since theSince the pressure remain constantpressure remain constant (atmospheric pressure) during the (atmospheric pressure) during the process, it is used to determineprocess, it is used to determine the enthalpy, the enthalpy, ∆H∆H(heat) of the (heat) of the

reactionreaction - C= heat absorbed/ ∆T = ∆H/ ∆T- C= heat absorbed/ ∆T = ∆H/ ∆T - specific heat capacity = C/mass - specific heat capacity = C/mass - heat =∆H∆H=specific heat x mass x ∆T

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Example:Example: The specific heat of graphite is 0.71 J/ The specific heat of graphite is 0.71 J/ºC.ºC.g. g. Calculate the energy needed to raise the temperature of Calculate the energy needed to raise the temperature of 75 kg of graphite from 21º C to 75ºC.75 kg of graphite from 21º C to 75ºC.

heat = specific heat x mass x ∆Theat = 0.71 J/ºC.g x 75x10J/ºC.g x 75x1033 g x (75-21)ºC g x (75-21)ºCheat = 2875500 J = 2875.5 kJheat = 2875500 J = 2875.5 kJ

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Example:Example: If 50.0 ml of 1.0 M HCl solution at 25ºC mixed with If 50.0 ml of 1.0 M HCl solution at 25ºC mixed with 1.0 M NaOH solution also at 25ºC in calorimeter, after the 1.0 M NaOH solution also at 25ºC in calorimeter, after the reactants are mixed by stirring, the temperature is observed to reactants are mixed by stirring, the temperature is observed to increased to 31.9 ºC. Calculate the heat of neutralization of the increased to 31.9 ºC. Calculate the heat of neutralization of the reaction. HCl + NaOHreaction. HCl + NaOH H H22O + NaCl O + NaCl

Assuming that: Assuming that: specific heat capacity of solution=4.18specific heat capacity of solution=4.18 J/ºC.gJ/ºC.gdensity of the solution = 1.0g/ml density of the solution = 1.0g/ml

mass of solution = 100 ml x 1 g/mlmass of solution = 100 ml x 1 g/ml energy released by the reaction = energy absorbed by solutionenergy released by the reaction = energy absorbed by solutionHeat released by the reaction = specific heat x (mass)solution x ∆T

heat = 4.18 J/ºC.g x 100 g x (31.9 - 25)ºCJ/ºC.g x 100 g x (31.9 - 25)ºCheat = 2.9 x 10heat = 2.9 x 1033 J J (heat released when 0.05 mol H(heat released when 0.05 mol H22O produced)O produced)

heat (kJ/mol Hheat (kJ/mol H22O) = 2.9 x 10O) = 2.9 x 103 3 J/0.05 mol = 58 kJ/molJ/0.05 mol = 58 kJ/mol

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Example:Example: A 46.2 g sample of copper is heated to 95.4ºC A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of and the copper is 21.8ºC. What is the specific heat of copper?copper?heat absorbed by water = heat lost by the metalheat absorbed by water = heat lost by the metal

(specific heat x mass x ∆T)water = (specific heat x mass x∆T)metal

4.18 J/ºC.g x 75.0g x (21.8 – 19.6) ºC = specific heat x 46.2 g x (95.4

– 21.8) ºC (specific heat)metal = 0.203 J/ºC.g

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22. Constant volume calorimeter is called a bomb . Constant volume calorimeter is called a bomb calorimeter.calorimeter.

Material is put in a container with pure oxygen. Wires Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put are used to start the combustion. The container is put into a container of water.into a container of water.

The heat capacity of the calorimeter is known and The heat capacity of the calorimeter is known and tested.tested.

∆∆E = q + w = q + p ∆VE = q + w = q + p ∆V Since ∆V = 0, P∆V = 0, ∆E = qSince ∆V = 0, P∆V = 0, ∆E = qvv

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Bomb CalorimeterBomb Calorimeter thermometerthermometer stirrerstirrer full of waterfull of water ignition wireignition wire Steel bombSteel bomb samplesample

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Example:Example: A 1.65 g of Mg (molar mass = 24.3 g/mol) was A 1.65 g of Mg (molar mass = 24.3 g/mol) was oxidized to MgO in calorimeter with heat capacity = oxidized to MgO in calorimeter with heat capacity = 6.27 kJ/ºC, if the temperature increase from 21.3ºC to 6.27 kJ/ºC, if the temperature increase from 21.3ºC to 28.56ºC. Calculate ∆E in kJ for the reaction:28.56ºC. Calculate ∆E in kJ for the reaction:2 Mg + O2 Mg + O22 MgO MgO

Heat (kJ) = heat capacity x ∆THeat (kJ) = heat capacity x ∆THeat (kJ) = 6.27Heat (kJ) = 6.27 kJ/ºC x (28.56 – 21.3)kJ/ºC x (28.56 – 21.3) ºC ºC = 45.52 kJ = 45.52 kJ (heat released when 1.65g Mg oxidized or(heat released when 1.65g Mg oxidized or 0.0679 mol Mg) 0.0679 mol Mg)

Moles of Mg = 1.65 g/ (24.3g/mol) = 0.0679 molMoles of Mg = 1.65 g/ (24.3g/mol) = 0.0679 mol heat for the above reaction (2 mol Mg) = heat for the above reaction (2 mol Mg) =

45.52 kJ x (2 mol Mg/ 0.0679 mol Mg)45.52 kJ x (2 mol Mg/ 0.0679 mol Mg)= 1.34 x 10= 1.34 x 1033 kJ kJ

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Example:Example: A) The combustion of 0.1584 g benzoic acid increase A) The combustion of 0.1584 g benzoic acid increase the temperature of bomb calorimeter by 2.54ºC, calculate the the temperature of bomb calorimeter by 2.54ºC, calculate the heat capacity of calorimeter (energy released by combustion of heat capacity of calorimeter (energy released by combustion of benzoic acid = 26.42 kJ/g)benzoic acid = 26.42 kJ/g)

B) A 0.2130 g of vanillin (molar mass = 152 g/mol) is burned in B) A 0.2130 g of vanillin (molar mass = 152 g/mol) is burned in the same calorimeter and the temperature increased by 3.25ºC. the same calorimeter and the temperature increased by 3.25ºC. Calculate the heat of combustion per gram of vanillin…… per Calculate the heat of combustion per gram of vanillin…… per molmol

A) A) (Heat capacity)(Heat capacity)calorimetercalorimeter = (26.42kJ/g) x 0.1584 g / 2.54ºC = (26.42kJ/g) x 0.1584 g / 2.54ºC

= 1.65 kJ/ºC = 1.65 kJ/ºC B) B) ∆E = q∆E = qvv = heat capacity x ∆T = = heat capacity x ∆T = - -1.65 kJ/ºC x 3.25ºC1.65 kJ/ºC x 3.25ºC

= - 5.36 kJ= - 5.36 kJ= - 5.36 kJ/0.2130 g = - 25.18 kJ/g= - 5.36 kJ/0.2130 g = - 25.18 kJ/g = - 5.36 kJ/ 0.0014 mol = - 3828.57 kJ/mol= - 5.36 kJ/ 0.0014 mol = - 3828.57 kJ/mol

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Hess’s LawHess’s Law Enthalpy is a state function.Enthalpy is a state function. It is independent of the path.It is independent of the path. This is mean: This is mean: in going from a particular set of in going from a particular set of

reactants to a particular set of products, the change in reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place enthalpy is the same whether the reaction takes place in one step or in a series of steps. in one step or in a series of steps. This principle is This principle is known as known as Hess’s lawHess’s law

We can add equations to come up with the desired We can add equations to come up with the desired final product, and add the ∆H.final product, and add the ∆H.

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Hess’s law can be illustrated in the following example:Hess’s law can be illustrated in the following example:NN2(g)2(g) + 2 O + 2 O2(g)2(g) 2 NO 2 NO2(g)2(g) ∆H∆H11 = 68 kJ = 68 kJ

The reaction also can also be carried out in two distinct steps:The reaction also can also be carried out in two distinct steps: NN2(g)2(g) + O + O2(g)2(g) 2 NO 2 NO(g)(g) ∆H∆H22 = 180 kJ = 180 kJ

2 NO2 NO(g)(g) + O + O2(g)2(g) 2 NO 2 NO2(g)2(g) ∆H∆H33 = -112 kJ = -112 kJNet reaction: Net reaction: NN2(g)2(g) + 2 O + 2 O2(g)2(g) 2 NO 2 NO2(g)2(g) ∆H∆H11=68 kJ = ∆H=68 kJ = ∆H2 2 +∆H+∆H33 Two characteristics of Two characteristics of ∆H for a reaction:∆H for a reaction: If the reaction is reversed the sign of ∆H is also reversedIf the reaction is reversed the sign of ∆H is also reversed If the coefficient in a balanced reaction is multiplied by an If the coefficient in a balanced reaction is multiplied by an

integer, the value of ∆H is multiply by the same integerinteger, the value of ∆H is multiply by the same integer

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N2 , 2O2

O2 , NO2

68 kJ

NO2180 kJ

-112 kJ

H (k

J)

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Example: Example: Given the following data:Given the following data:a)a) 2 NH2 NH3(g)3(g) N N2(g)2(g) + 3 H + 3 H2(g)2(g) ∆H = 92 kJ ∆H = 92 kJ

b)b) 2 H2 H2(g)2(g) + O + O2(g)2(g) 2 H2 H22OO(g)(g) ∆H = - 484 kJ∆H = - 484 kJ

Calculate Calculate ∆∆H for the reaction:H for the reaction: 2 N2 N2(g)2(g) + 6 H + 6 H22OO(g)(g) 3 O 3 O2(g)2(g) + 4 NH + 4 NH3(g)3(g)

Solution:Solution:Reverse equation (a) and multiply it by the coefficient (2)Reverse equation (a) and multiply it by the coefficient (2)Reverse equation (b) and multiply it by the coefficient Reverse equation (b) and multiply it by the coefficient

(3)(3) 2 2 NN2(g)2(g) + 6H + 6H2(g)2(g) 4NH4NH3(g)3(g) ∆H = -184 kJ∆H = -184 kJ

6 H6 H22OO(g) (g) 6 H6 H2(g)2(g) + 3 O + 3 O2(g)2(g) ∆H = + 1452 kJ ∆H = + 1452 kJ

2 N2 N2(g)2(g) + 6 H + 6 H22OO(g)(g) 4NH 4NH3(g) 3(g) ++ 3 O3 O2(g) 2(g)

∆∆H = + 1268 kJH = + 1268 kJ

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Standard EnthalpyStandard Enthalpy The enthalpy change for a reaction at standard The enthalpy change for a reaction at standard

conditions (25ºC, 1 atm , 1 M solutions)conditions (25ºC, 1 atm , 1 M solutions) Symbol ∆HºSymbol ∆Hº When using Hess’s Law, work by adding the When using Hess’s Law, work by adding the

equations up to make it look like the answer. equations up to make it look like the answer. The other parts will cancel out.The other parts will cancel out.

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Example:Example: Given the following thermochemical equation: Given the following thermochemical equation:a) Ca) C22HH2(g)2(g) + 5/2 O + 5/2 O2(g)2(g) → 2 CO → 2 CO2(g)2(g) + H + H22OO(ℓ) (ℓ) ∆Hº = - 1300 kJ ∆Hº = - 1300 kJ

b)b) CC(s)(s) + O + O2(g)2(g) → CO → CO2(g)2(g) ∆Hº = - 394 kJ ∆Hº = - 394 kJ

c)c) HH2(g)2(g) + ½ O + ½ O2(g)2(g) → H → H22OO(ℓ)(ℓ) ∆Hº = - 286 kJ ∆Hº = - 286 kJ

Calculate ∆Hº for the reaction:Calculate ∆Hº for the reaction:2 C2 C(s)(s) + H + H2(g)2(g) → C → C22HH4(g)4(g)

Solution:Solution: Reverse equation (a). Multiply equation (b) by the Reverse equation (a). Multiply equation (b) by the coefficient 2. equation (c) remain as it is.coefficient 2. equation (c) remain as it is.2 CO2 CO2(g)2(g) + H + H22OO(ℓ) (ℓ) → C→ C22HH2(g)2(g) + 5/2 O + 5/2 O2(g)2(g) ∆Hº = +1300 kJ∆Hº = +1300 kJ

2 C2 C(s)(s) + 2 O + 2 O2(g)2(g) → 2 CO → 2 CO2(g)2(g) ∆Hº = - 788 kJ ∆Hº = - 788 kJ

HH2(g)2(g) + ½ O + ½ O2(g)2(g) → H → H22OO(ℓ)(ℓ) ∆Hº = - 286 kJ ∆Hº = - 286 kJ

2 C2 C(s)(s) + H + H2(g)2(g) → C → C22HH4(g)4(g) ∆Hº = +229 kJ∆Hº = +229 kJ

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Example:Example: Given the following thermochemical equation: Given the following thermochemical equation:a) Ha) H2(g)2(g) + 1/2 O + 1/2 O2(g)2(g) → H → H22OO(ℓ) (ℓ) ∆Hº = - 285.8 kJ ∆Hº = - 285.8 kJ

b)b) NN22OO5(g)5(g) + + HH22OO(ℓ)(ℓ) → 2 HNO → 2 HNO3(3(ℓℓ)) ∆Hº = - 76.6 kJ ∆Hº = - 76.6 kJ

c)c) 1/2 N1/2 N2(g)2(g) + 3/2 O + 3/2 O2(g)2(g) + 1/2 + 1/2 H H2(g) 2(g) → → HNOHNO3(ℓ) 3(ℓ) ∆Hº = - 174.1 kJ ∆Hº = - 174.1 kJ

Calculate ∆Hº for the reaction:Calculate ∆Hº for the reaction:2 2 NN2(g)2(g) + 5 O + 5 O2(g)2(g) → N → N22OO5(g)5(g)

Solution:Solution: Reverse equation (a), and multiply it by 2. Reverse equation (a), and multiply it by 2. Reverse Reverse equation equation (b), (b), multiply it by multiply it by 2. multiply equation (c) by 42. multiply equation (c) by 422 /H /H22OO(ℓ) (ℓ) → H→ H2(g)2(g) + 1/2 O + 1/2 O2(g)2(g) ∆Hº =+∆Hº =+ 285.8 285.8 kJ kJ

2/2/ 2 HNO 2 HNO3(ℓ)3(ℓ) → → NN22OO5(g)5(g) + H + H22OO(ℓ)(ℓ) ∆Hº = + 76.6 kJ ∆Hº = + 76.6 kJ

4/ ½ N4/ ½ N2(g)2(g) + 3/2 O + 3/2 O2(g)2(g) + 1/2 H + 1/2 H2(g) 2(g) → HNO→ HNO3(ℓ)3(ℓ) ∆Hº = - 174.1 kJ∆Hº = - 174.1 kJ

2 N2 N2(g)2(g) + 5 O + 5 O2(g)2(g) → N → N22OO5(g) 5(g) ∆Hº = +28.8 kJ∆Hº = +28.8 kJ

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Standard Enthalpies of FormationStandard Enthalpies of FormationStandard enthalpy of formationStandard enthalpy of formation (∆H (∆Hƒƒ

ºº): ): is the heat is the heat change change that results when that results when one moleone mole of a compound is formed from its of a compound is formed from its elementselements in their standard states. in their standard states.

Standard states are 1 atm, 1M and 25ºCStandard states are 1 atm, 1M and 25ºC For an elements: For an elements: ∆H∆Hƒƒ

º º = = 00 ∆ ∆HHƒƒ

ºº (O (O22) = 0) = 0The formation of NOThe formation of NO22 from its elements in their standard states: from its elements in their standard states:

½ N½ N2(g)2(g) + O + O2(g)2(g) → NO → NO2(g)2(g) ∆H∆Hƒƒº = 34 kJ/molº = 34 kJ/mol The The ∆H∆Hƒƒº values for common substances are º values for common substances are

present in certain tables (present in certain tables (∆H∆Hƒƒºº for some substances for some substances

are shown in the following table)are shown in the following table)..

∆Hƒº (C, graphite) = 0

4141

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Write the equation for the formation of methanol Write the equation for the formation of methanol CHCH33OH.OH.

CC(s)(s) + 2 H + 2 H2(g)2(g) + ½ O + ½ O2(g)2(g) → CH → CH33OHOH(ℓ) (ℓ) ∆Hº ∆Hºƒƒ = -239kJ/mol. = -239kJ/mol.

Several important characteristics of the definition of ∆HºSeveral important characteristics of the definition of ∆Hºƒƒ : : The reaction is written so that elements are in their The reaction is written so that elements are in their

standard states.standard states. 1.0 mol of product is formed so ∆Hº1.0 mol of product is formed so ∆Hºƒƒ are always given are always given

in kJ/mol.in kJ/mol. For most compounds ∆HºFor most compounds ∆Hºƒƒ is negative, because you are is negative, because you are

making bonds; making bonds is exothermic process.making bonds; making bonds is exothermic process.

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Enthalpy is a state function, so we can invoke Hess’s law and Enthalpy is a state function, so we can invoke Hess’s law and choose any convenient pathway from reactants to products and choose any convenient pathway from reactants to products and then sum the enthalpy changes along the chosen pathway.then sum the enthalpy changes along the chosen pathway.A convenient pathway involve the use of ∆HºA convenient pathway involve the use of ∆Hºƒƒ . .

For example For example we will calculate ∆Hº for the combustion of CHwe will calculate ∆Hº for the combustion of CH44: :

CHCH4(g)4(g) + 2 O + 2 O2(g)2(g) → CO → CO2(g)2(g) + 2 H + 2 H22OO(ℓ)(ℓ)

By using ∆HºBy using ∆Hºƒƒ for each compound in the reaction for each compound in the reaction

1)1) CC(s)(s) + 2 H + 2 H2(g)2(g) → CH → CH4(g)4(g) ∆Hº ∆Hºƒƒ = - 75 kJ/mol = - 75 kJ/mol

2)2) CC(s)(s) + O + O2(g)2(g) → CO → CO2(g)2(g) ∆Hº ∆Hºƒƒ = - 394 kJ/mol = - 394 kJ/mol

3)3) HH2(g)2(g) + ½ O + ½ O2(g)2(g) → H → H22OO(g)(g) ∆Hº ∆Hºƒƒ = - 286 kJ/mol = - 286 kJ/mol

Reverse equation (1), multiply equation (3) by the coefficient 2Reverse equation (1), multiply equation (3) by the coefficient 2The reactions become:The reactions become:

4444

1)1) CH CH4(g) 4(g) →→ CC(s)(s) + 2 H + 2 H2(g)2(g) ∆Hº ∆Hºƒƒ = + 75 kJ/mol = + 75 kJ/mol

2)2) CC(s)(s) + O + O2(g)2(g) → CO → CO2(g)2(g) ∆Hº ∆Hºƒƒ = - 394 kJ/mol = - 394 kJ/mol

3)3) 2 / H2 / H2(g)2(g) + ½ O + ½ O2(g)2(g) → H → H22OO(g)(g) ∆Hº ∆Hºƒƒ = - 286 kJ/mol = - 286 kJ/mol

CHCH4(g)4(g) + 2 O + 2 O2(g)2(g) → CO → CO2(g)2(g) + 2 H + 2 H22OO(ℓ)(ℓ)

∆∆Hº = 75 + (- 394) + (-286 x 2) = - 891 kJ/molHº = 75 + (- 394) + (-286 x 2) = - 891 kJ/mol ∆ ∆HºHºreactionreaction = - ∆Hº = - ∆Hºƒƒ for CH for CH44 + ∆Hº + ∆Hºƒ ƒ for COfor CO22+∆Hº+∆Hºƒƒ for H for H22OO

reactantsreactants products productsSo:So:HH0 0 == nnpp ∆Hº ∆Hºƒƒ (products) – (products) – n nrr ∆Hº ∆Hºƒƒ (reactants)(reactants)n: number of moles of each substance in the balance n: number of moles of each substance in the balance equation equation

4545

Example:Example: Calculate the enthalpy (∆Hº) for the following Calculate the enthalpy (∆Hº) for the following reactions:reactions:a) 4 NHa) 4 NH3(g)3(g) + 5O + 5O2(g)2(g) → 4 NO → 4 NO(g)(g) + 6H + 6H22OO(g)(g)

b) 3 NOb) 3 NO2(g)2(g) + H + H22OO(ℓ)(ℓ) → 2 HNO → 2 HNO3(aq)3(aq) + NO + NO(g)(g)

HH00reactionreaction

== HH00 (products) – (products) – HH00 (reactants) (reactants)a) a) HH00

rxnrxn= = [6 ∆Hº [6 ∆Hºƒ ƒ HH22O + 4 ∆HºO + 4 ∆Hºƒƒ NO] – [5∆Hº NO] – [5∆Hºƒƒ O O22 + 4 ∆Hº + 4 ∆HºƒƒNHNH33] ]

= [6 x (-242) + 4 x 90] – [5 x 0 + 4 x (- 46)= [6 x (-242) + 4 x 90] – [5 x 0 + 4 x (- 46) = - 908 kJ= - 908 kJ

b) b) HH00rxnrxn= = [2 ∆Hº[2 ∆Hºƒ ƒ HNOHNO33 + ∆Hº + ∆Hºƒƒ NO] – [∆Hº NO] – [∆Hºƒƒ H H22O + 3 ∆HºO + 3 ∆Hºƒ ƒ NONO22]]

= [2 x (- 207) + 90 ] – [ - 286 + 3 x 34]= [2 x (- 207) + 90 ] – [ - 286 + 3 x 34] = - 140 kJ= - 140 kJ

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Example:Benzene (C6H6) burns in air to produce carbon dioxide and liquid water and librates 3271 kJ at 25ºC and 1 atm. Calculate ∆Hºƒ of benzene in kJ/mol.2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(l)

H0rxn H0 (products)f= H0 (reactants)f-

H0rxn 6H0 (H2O)f12 H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]- 3271=[ 12 × -393.5 + 6 × -285.8 ] – [2H0

f (C6H6)]

6.6

H0f = + 49 kJ

4747

Keep in mind the following key concepts when doing Keep in mind the following key concepts when doing enthalpy calculations:enthalpy calculations:When a reaction is reversed, the magnitude of ∆H remains When a reaction is reversed, the magnitude of ∆H remains the same, but its sign changes.the same, but its sign changes.When the balanced equation for a reaction is multiplied by When the balanced equation for a reaction is multiplied by an integer, the value of ∆H for that reaction must be an integer, the value of ∆H for that reaction must be multiply by the same integer.multiply by the same integer.The change in enthalpy for a given reaction can be The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants calculated from the enthalpies of formation of the reactants and products.and products.HH0 0 == nnpp ∆Hº ∆Hºƒƒ (products) – (products) – n nrr ∆Hº ∆Hºƒƒ (reactants(reactantsElements in their standard states are not included in the Elements in their standard states are not included in the ∆H∆Hreactionreaction calculations. That is, calculations. That is, ∆Hº∆Hºƒƒ for an element in its for an element in its standard state is zero.standard state is zero.