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THERMOCHEMISTRY[MH5; Chapter 8]

• Thermochemistry is the study of the heat flow that accompanieschemical reactions.

PRINCIPLES OF HEAT FLOW [MH5; 8.1]• Heat is a process whereby energy is transferred from a warmer

object to a colder object. • When we speak of heat flow, we must distinguish between the

system and the surroundings.• The System is the part of the universe that is under observation. • The Surroundings is the rest of the universe.• A system may exchange energy with its surroundings.

Sign convention (a “system-centric” viewpoint):• The symbol commonly used to denote heat flow is “ q ”.• For gain of heat energy by a system , q is positive. • This occurs when heat flows into the system from its surroundings.• For loss of heat energy by a system, q is negative.• This occurs when heat flows out of the system into the

surroundings.• Consider a beaker of water sitting on a hot plate........

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• We can look at chemical reactions using the same logic.......• The reaction mixture (reactants and products) is usually thought of

as the system.• If heat flows into the reaction system from the surroundings, the

reaction is endothermic. • In these cases, q > 0. (A positive value)

• If heat flows from the reaction system to the surroundings, thereaction is exothermic.

• In an exothermic reaction, q < 0. (A negative value)

Magnitude of Heat Flow• As well as knowing the direction of heat flow, chemists usually also

want to know how much heat has been transferred.• Heat flow, or q, is usually measured in joules or kilojoules.

1000 joules (J) = 1 kilojoule (kJ)• You may still see calories (especially in US literature); a calorie is

defined as “Amount of heat required to raise the temperature of 1 gof water through 1EC” ....................

• 1 calorie = 4.184 J• The Calorie used for food energy is actually a kilocalorie. (1000

calories = 1 kilocalorie)• The simplest calculation of the magnitude of heat flow involves the

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amount of heat required to change the temperature of a system.• This can be calculated using the equation:

q = C ∆T

where C is the heat capacity of the system.• The heat capacity is the amount of heat required to raise the

temperature of the system by 1 oC or 1 K. (Units are J/ oC or J/K )

• If the system is a pure substance with a mass of “ m ” grams......q = m c ∆T

where c is the specific heat of the substance. • The specific heat is the amount of heat required to raise one gram

of the substance by 1 o C or 1 K. (Units are J/ g C oC or J/ g C K)• Specific heat is an intensive property; it can be used to identify a

substance.

EXAMPLES: (More in MH5; Table 8.1)

• Sometimes you will see heat capacities or specific heats given withthe temperature in units of K.

• Note that the size of the Celsius degree is exactly the same as theKelvin degree !!!!

• So ∆T, the temperature change, will be the same for both scales !!

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EXAMPLES:

Calculations Involving Heat Flow

EXAMPLE 1:How much heat, in kJ, is required to raise the temperature of 120 g Alfrom 25EC to 100EC ? (Specific Heat of Al is 0.902 J g — 1 K — 1)

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EXAMPLE 2:The specific heat of water is 4.184 J g—1 K—1. A piece of iron (Fe) weighing 40 g is heated to 80EC and dropped into100 g of water at 25EC. (Specific heat of Fe is 0.446 J g—1 K—1)What is the temperature when thermal equilibrium has been reached ?

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• Heat energy may be used not only to change the temperature ofsomething, but also for changes of state at constant temperature.

• This is called latent heat.

EXAMPLE 1:1 g ice at 0EC ! 1 g water at 0EC [H2O(s) ! H2O (R)]

Latent heat of fusion (melting): q = + 333 J g—1

EXAMPLE 2:1 g H2O(R) at 100EC ! 1 g H2O(g) at 100EC [H2O(R) ! H2O(g)] Latent heat of vaporization (evaporation):

q = +2,260 J g—1

• How do we use latent heat values ?

EXAMPLE: Consider a 750 g block of ice at a temperature of - 15oC.How much heat energy, in kJ, would be required to melt the ice, andheat the water formed to 22oC ?Data Needed: Specific Heat of Ice: 2.01 J g—1 K—1

Heat of Fusion of Ice: 6.01 kJ mol —1

Specific Heat of Water: 4.184 J g —1 K —1

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MEASUREMENT OF HEAT FLOW; CALORIMETRY [MH5; 8.2]• Measurement of heat flow in a reaction is carried out in an

apparatus known as a calorimeter.• The calorimeter is insulated and therefore will absorb heat. • The amount of heat that the calorimeter can absorb is the heat

capacity of the calorimeter.• The calorimeter usually contains water, or another solution of known

specific heat.• The insulation ensures that no heat gets out of the calorimeter....

• Heat evolved in reaction = Heat absorbed by water + heat absorbed by calorimeter

or....

qreaction = ( –qwater) + ( –qcalorimeter )

• A calorimeter can be as simple as a styrofoam coffee cup, or ascomplex as a “bomb calorimeter”.

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EXAMPLE 1:When a sample of sucrose, C12H22O11 is burnt in a bomb calorimeter, thetemperature rises from 25.00EC to 30.80EC. The mass of water is 1.50 x 103 g, and the heat capacity of the bomb is837 J K —1. a) Calculate the heat evolved in this reaction.

b) If the mass of the sucrose was 2.50 grams, calculate the heatevolved per mole of sucrose.

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ENTHALPY [MH5; 8.3]• Enthalpy is a type of chemical energy; sometimes it is referred to as

“heat content”.• It is defined so that for a constant pressure process..... ∆ H = q• ∆ H is the difference in enthalpy between reactants and products.

• So...... ∆ H = Hproducts - Hreactants

• We can apply this to our two types of chemical reactions:

• If heat is lost from the system (so q would be < 0), ∆H is negative;the reaction is EXOTHERMIC.

• If heat is gained by the system, (so q would be > 0), ∆H is positive,and the reaction is ENDOTHERMIC.

EXAMPLE: For the combustion of octane:

C8H18(R) + 25/2 O2(g) ! 8 CO2(g) + 9 H2O(R)

∆H = –5476 kJ ; highly exothermic !!!

Notes:1) ∆H is a State Property; it depends only on the initial and final

states, and not how the system got from one to the other. (In thermodynamics, state includes T, P, V)

2) A ∆H value written beside an equation is based on the numbers ofmoles of all substances present as the equation is written. This means we do not write “ mol —1 ".

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EXAMPLE:

4 K(s) + 4 Cr(s) + 7 O2(g) ÿ 2 K2Cr2O7(s) ∆H = –4123 kJ

• In this case, ∆H is the enthalpy change for the reaction as written.• If the ∆H value is associated with one species only it would be

written as “per mole” of that species.

EXAMPLE: The heat of combustion of octane, C8H18, is –5476 kJ mol —1

C8H18(R) + 25/2 O2(g) ÿ 8 CO2(g) + 9 H2O(R) ∆H = –5476 kJ

• In this case, ∆H for the reaction is numerically the same as the∆Hcombustion for 1 mole of octane; we have associated the ∆H valuewith one particular species in the reaction.

• From Note (1) It follows that, if a reaction is carried out in a seriesof steps, the overall ∆H is the sum of the ∆H values of thecomponent steps.

• This relationship is called Hess’s Law.

EXAMPLE:∆H1

C(s) + O2 (g) ! CO2 (g)

9 ∆H2 9 ∆H3 8 ∆H5

9 9 ∆H4 8

C (g,atomic) 2 O(g,atomic) ! CO (g) + O(g,atomic)

∆H1 = ∆H2 + ∆H3 + ∆H4 + ∆H5

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• It also follows that the ∆H value for the reverse reaction has theopposite sign.

EXAMPLE:C(s) + O2(g) ! CO2(g) ∆H = –394 kJ exothermic

CO2(g) ! C(s) + O2(g) ∆H = +394 kJ endothermic

An alternative way of expressing Hess’s Law• Thermochemical equations may be added and subtracted like

algebraic equations. • A thermochemical equation is a chemical equation together with the

appropriate ∆H.

EXAMPLE:Given: (1) C(s) + ½ O2(g) ! CO(g) ∆H = -111 kJ

(2) C(s) + O2(g) ! CO2(g) ∆H = -394 kJ

What is ∆H for: (3) CO(g) + ½ O2(g) ! CO2(g) ?

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Standard States

Diatomic gases: H2, N2, O2, F2, CR2Noble gases: He, Ne, Ar, Kr, Xe, RnLiquids: Br2, HgSolids: Everything else !

ENTHALPIES OF FORMATION [MH5; 8.5]• ∆H of reaction when 1 mol of compound (in its Standard State) is

formed from its component elements in their Standard States. • Symbol is ∆HfE• Standard State is the physical state in which an element is found

at 25EC and 1 atm pressure.

• In the case of an element known in more than one form (allotropy)the standard state is that of lowest energy.

EXAMPLES: C (graphite), not diamond; O2 (g) not O3 (g) (ozone).• There is a sole exception to the element being the most stable

allotrope: white Phosphorus (P4(s)) is chosen as the standard stateof phosphorus; the more stable forms are difficult to get in pureform but white phosphorus is easily made pure.

• It follows from the definition of standard heat of formation and ofthe standard state of an element that this definition that:

∆HfEof an element in its standard state is zero

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EXAMPLES of Formation Reactions:Write the reaction that corresponds to the Standard Heat ofFormation, ∆HfE, for the following compounds.a) C2H5OH (R)

b) Mg(NO3)2 (s)

c) KBrO4 (s)

• It is important to realize that the Standard Heat of Formation of acompound cannot be calculated from the Formation Equation.......

• It may be determined experimentally, and found on a table ofStandard Heat of Formation values. OR......

• It may be calculated using a reaction in which the compound inquestion takes part (but NOT the Formation Reaction!)

• Note that (with the sole exception of phosphorus, when there are acouple of allotropes more stable than the standard state) ∆HfE forother forms of the element must be positive....

EXAMPLES:

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H2(g) ! 2 H (g, atomic); ∆H = +436 kJ,

• So for H(g, atomic); ∆HfE = 436 /2 = 218 kJ mol—1

C(graphite) ! C(diamond); ∆HfE = + 1.895 kJ mol—1

C(graphite) ! C(atomic,gas); ∆HfE = + 717 kJ mol—1

Enthalpy of Formation and Enthalpy of Reaction

• Imagine going from reactants to products via the elements......

∆H reaction Reactants ! Products

–Total ∆HfE + Total ∆HfE of reactants of products

Elements in Standard States

By Hess’s Law..................

∆H(Reaction) = 3∆HfE(Products) – 3∆HfE(Reactants)

This is an alternative way of stating Hess’s Law!!

∆Ho Reaction = 3∆HfE (Products) – 3 ∆hfE(Reactants)

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• This is a useful equation if one knows all of the ∆HfE values for allthe species taking part in the reaction.................

• Otherwise, it is necessary to set up all the intermediate equationsas in the example on p. 223.

EXAMPLE 1: Calculate the heat of combustion of methane, CH4; Using ∆HfE values (in kJ mol—1) [MH5; p. 605-606]CH4(g) = –75; CO2(g) = –394; H2O(R) = –286

The combustion reaction is:

∆Ho Reaction = 3∆HfE (Products) – 3 ∆HfE(Reactants)

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• We can also solve this problem by setting up the actual equationsfor the formations of the species involved in the combustionreaction..............

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EXAMPLE 2: Benzene, C6H6(R), has enthalpy of combustion; ∆Hcomb = –3271 kJ mol —1.

Given the following ∆HfE’s (in kJ mol –1 ) : CO2(g) = –394; H2O(R) = –286,what is ∆HfEfor benzene?

• Note the difference in the ∆HfE values:

• Methane, CH4(g):

C(s) + 2 H2(g) ! CH4(g); ∆HfE = –75 kJ mol —1 exothermic

Benzene, C6H6(R):

6 C(s) + 3 H2(g) ! C6H6(R); ∆HfE = +49 kJ mol —1 endothermic

• Benzene is less stable than its component elements. It is anendothermic compound.

• Most compounds have a negative ∆HfE value and are more stablethan their elements (exothermic compounds).

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BOND ENERGY [MH5; 8.6]• Bond Energy (more correctly termed Bond Enthalpy) is defined as

the “enthalpy change (in kJ/mol) when a particular type of bond isbroken in the gas phase.”

• Bond Energy is always expressed as a positive value; the process ofbreaking a bond is always an endothermic process.

• A more positive bond energy indicates a stronger bond.

EXAMPLES:H2 (g) ! 2 H (g, atomic) ∆H = + 436 kJmol —1

F2 (g) ! 2 F (g, atomic) ∆H = + 153 kJmol —1

• When the bond in a diatomic molecule is broken, two atoms form.• ∆Ho

f for an atomic species (H(g,at) or F(g,at)) is ½ the bond energyfor breaking the bond of the diatomic species; H2(g) or F2(g).

• So it follows that the bond energy of a diatomic gaseous elementmust be 2 x the ∆Ho

f of the gaseous atoms formed by breaking thebond.

EXAMPLE:∆Ho

f for CR (g,at) = 122 kJmol -1

What is the energy of the CR — CR bond ? The equation for the formation of CR (g,at) is:

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A selection of Bond Dissociation Energies, kJ mol G1

H C N O S F CR

H 436 414 389 464 339 565 431

C 347 293 351 259 485 331

N 159 222 272 201

O 138 184 205

S 226 285 255

F 153 255

CR 244

For Multiple Bonds [ MH5; Table 8.4]

C/C 820 C/N 890 N/N 941C=C 612 C=O 715 C=N 615C=S 477 N=N 418 O=O 498S=O 498 N=O 607

• Bond strength between like atoms (ie, 2 C atoms) increases: triple > double > single.

• Similarly with diatomic molecules;

F—F O=O N / N

153 498 941 kJ molG1

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• The Total Bond Energy (TBE) of a molecule is the energy requiredto break every bond in a mole of the molecules in the gas phase.

• Think about what happens if you break every bond in a molecule; youget a bunch of atoms...in this case they will be gaseous atoms.

• Depending on the information given, there are two different waysthat you could express what is happening.

1) Using Heats of Formation:

2) Using the Structure and Individual Bond Energies:

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Total Bond Energy and Reaction Enthalpy Change for a Gas-PhaseReaction

• Consider:

• This gives us the relationship:

∆H = G (Energy of Bonds Broken) - G (Energy of Bonds Formed)

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EXAMPLE 1:Calculate ∆HE for the reaction:

C2H4Cl2(g) + F2(g) ÿ C2H4F2(g) + Cl2(g)

using Bond Energies. (p. 231)

∆Ho of the reaction = G (Energy of Bonds Broken) - G (Energy of Bonds Formed)

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EXAMPLE 2:

For the hydrogenation of acetylene, C2H2:

HC / CH (g) + 2 H2 (g) ! C2H6 (g) ∆H = - 311 kJ

Using the bond energies:C - C: 347 kJ, H - H: 436 kJ, C - H: 414 kJ, estimate the bond energy of C / C, a triple bond.

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• Total Bond Energy of a molecule may also be calculated using Heatsof Formation.....

EXAMPLE:SF6 contains six identical S–F bonds. Given the following ∆HfE’s, in kJ mol—1 : SF6(g), –1209; S(atomic,g), 275; F(atomic,g), 79;Calculate the average S–F bond energy of SF6, in kJ mol—1 .

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Bond Energy and Heat of Formation• We have now seen that there are two ways to calculate the Total

Bond Energy in a molecule. Recall that.............• The molecule must be in the gas phase.• The “reaction” involved is the one in which all the bonds in the

molecule are broken and gaseous atoms are formed.• When we calculate ∆H for this “reaction”, we are calculating Total

Bond Energy.

1) Calculate ∆H using:

∆Ho Reaction = G∆HfE (Products) – G ∆HfE (Reactants)

To use this method, we must know all of the Heats of Formation ofthe reactant molecule and the gaseous atoms formed.

2) Calculate ∆H using:

∆Ho Reaction = G (Energy of Bonds Broken) - G (Energy of Bonds Formed)

To use this method, we must know the structure of the molecule andthe energies of all the types of bonds present; these energies addedtogether will give us the “Energy of Bonds Broken”.

There are no bonds formed in this “reaction”, so the “Energy ofBonds Formed” will be 0.

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• Having these two methods to calculate Total Bond Energy is usefulbecause they will both give us the same value (within someexperimental error) for the Total Bond Energy of the molecule inquestion.

• So...........we can substitute what we calculated using 1) into 2) andvice versa!!

• Recall Hess’s Law; ∆H for an overall reaction does not depend onthe path taken.

EXAMPLE: Propane, H3C–CH2 –CH3, has ∆HfE !104 kJ mol G1

∆HfE C (g) = 717 kJ mol —1 ; ∆HfE H (g) = 218 kJ mol —1

a) What is the TBE in propane ?

b) If the C–C single bond energy is 347 kJ molG1 what is the averageC–H bond energy in propane ?

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Estimating ∆HfE values from Bond EnergyEXAMPLE 1: Calculate the ∆HfE for hexane, C6H14 (g).DATA: E(C-H) = 414 kJmol —1 E(C-C) = 347 kJ mol G1

∆HfE of H (g, at) = 218 kJmol — 1 ∆HfE of C (g, at) = 717 kJmol —1

• We do need the structure of hexane:

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EXAMPLE 2: Calculate the energy of C = O in gaseous acetic acid, CH3COOH.DATA: ∆HfE for CH3COOH (g) = - 315 kJmol —1,∆HfE of C(g, at) = 717 kJmol —1, ∆HfE of O(g, at) = 249 kJmol —1

∆HfE of H(g, at) = 218 kJmol —1

Other bond energy data may be found on p. 231.• Recall the structure of acetic acid: