15 - 1 Thermochemical Definitions Thermochemistry studies the relationship between chemical...

15 - 1

Thermochemical DefinitionsThermochemical Definitions

Thermochemistry studies the relationshipbetween chemical reactions and the

resultingenergy changes involving heat.

Heat is a form of energy that naturally flowsfrom an object of higher temperature to anobject of lower temperature.

Heat is measured in calories (cal) or Joules (J).

1.00 cal = 4.19 J

15 - 2

Temperature is a measure of the averagekinetic energy per molecule.

Temperature is measured in °C or K and

is an intensive property.

An intensive property is a characteristic

of matter that does not depend on the amount of matter present.

K = °C + 273

15 - 3

The internal (thermal) energy of a substanceis the sum of the potential energy (PEe) andkinetic energy (KE) possessed by themolecules.

Thermal energy is measured in joules (J) and is an extensive property.

PEe is the energy due to composition or position.

PEe results from the attractive and repulsive forces found in atoms and molecules which is stored as energy in bonds.

15 - 4

An extensive property is a characteristic of matter that depends on the amount of matter present.

Internal energy is a state function which means that it is path independent.

Heat will always pass from an substance at ahigher temperature to that of a lowertemperature.

Heat will not always pass from a substance ofhigh thermal energy to that of a lower thermalenergy.

15 - 5

We are unable to determine the absoluteinternal energy of a system so weconcentrate on the change in internal

energy.

The change in internal energy is given by,

ΔE = q + w

where q is the heat and w is the work.

Work is a force acting over a distance and is

equal to the change in energy.

15 - 6

The sign convention can be somewhatconfusing but keep in mind that everything

ismeasured or compared to the system.

When energy is input with respect to thesystem, ΔE = +, otherwise ΔE = -.

When work is done on the system by thesurroundings, w = +, otherwise w = -.

Both heat and work are path functions which

mean they are dependent on pathway.

15 - 7

To clarify the difference between a statefunction and a path function, let’s review

theirdefinitions.

A state function refers to a property of a system that only depends on the

present state of the system and not how it got to that state.

Familiar quantities that are state functions are pressure, volume, and soon to become familiar is enthalpy.

15 - 8

A path function refers to a property of a system that is dependent on how the system attained that state (the path followed).

Examples of a path function is heat and work.

To illustrate the difference between a statefunction and a path function consider goingaround a circle one time.

If you arrive at the point from which youstarted and stop, your distance traveled anddisplacement are not the same.

15 - 9

The distance you travel is equal to onecircumference, C = 2πr.

Your displacement equals zero because youend up where you started from.

In this simplified example, distance is a pathfunction because if you go around thecircumference five times, your distancetraveled equals 5C = 10πr.

At the end of five revolutions, if you stop atyour original starting point, then yourdisplacement is zero because it is a statefunction.

15 - 10

Two terms that are frequently used inthermochemistry are system andsurroundings.

The system is the portion of the universe that you choose to focus your attention on.

In the process of dissolving a soluble salt such as KBr, the system can be considered to be the KBr and the water.

The surroundings include everything else in the universe.

15 - 11

EnthalpyEnthalpy

Enthalpy is defined as the sum of the internalenergy and PV work as shown below:

ΔH = ΔE + w = ΔE + PΔV

where E is the internal (thermal) energy andw is the PΔV work, P is the pressure, and V isthe volume.

E, P, V, and ΔH are state functions which donot depend on the pathway between the twostates.

15 - 12

At constant pressure, only PΔV work isallowed and the change in enthalpy, ΔH, isequal to the heat transferred, ΔH = qp.

PV work is given by,

w = -PΔV

with units of L•atm.

When a gas expands, ΔV = +, giving anegative value for the work.

15 - 13

When a gas expands, it uses some of itsinternal energy to expand, and the energycontent of the gas decreases (as indicated

bythe - sign).

When a gas is compressed, ΔV = -, giving apositive value for the work.

When work is done on a gas, the energycontent of the gas increases (as indicated

bythe + sign).

15 - 14

Another Look at EnthalpyAnother Look at Enthalpy

Enthalpy is a measure of the differencebetween the potential energy of the

productsand the potential energy of the reactants.

ΔH° = ∑nΔH°prod - ∑nΔH°reac

The ° indicates the standard state of a substance.

The standard state of a substance is its state at a P = 1.00 atm and T = 25°C = 298 K.

15 - 15

The heat of reaction, ΔH, is the heat energyabsorbed (endothermic) or released(exothermic) in a chemical reaction.

The following two slides show the energyprofile of an endothermic and exothermicreaction.

15 - 16

In an exothermic reaction, the chemicalbonds in the products are stronger thanthose of the reactants.

When ΔH << 0, reactions tend to be spontaneous in the forward direction.

In an endothermic reaction, the chemicalbonds in the reactants are stronger those ofthe products.

When ΔH >> 0, reactions are not spontaneous in the forward direction

but are in the reverse direction.

15 - 17

Energy Profile for Endothermic RxnEnergy Profile for Endothermic Rxn

Energy vs Reaction Pathway

Reaction Pathway

En

erg

y

Energy Content of Products

Energy Content of Reactants

ΔH = +

15 - 18

Energy Profile for Exothermic RxnEnergy Profile for Exothermic Rxn

Energy vs Reaction Pathway

Reaction Pathway

En

erg

y

Energy Content of Reactants

Energy Content of Products

ΔH = -

15 - 19

A chemical reaction is most likely to goforward if it is very exothermic, ΔH° < 0.

When two elements react exothermically toform a compound, the compound will berelatively stable.

Such a reaction is the formation of carbondioxide,

C(s) + O2(g) CO2(g) ΔH = -393.5 kJ

For an endothermic reaction, ΔH° > 0, it isunlikely to proceed without the application

ofexternal energy.

15 - 20

Exothermic reactions result in the new bonds

(products) being more stable.

Endothermic reactions result in the newbonds (products) being less stable.

15 - 21

Enthalpy ProblemEnthalpy Problem

The combustion of 1.00 mol of benzene,C6H6(l), gives off 3.268 × 103 kJ.

(a)What is the thermochemical equation for the reaction?

(b) Calculate the change in enthalpy when 25.0 g of C6H6 is burned?

(c) How many grams of C6H6 need to be burned to liberate 2.0 kJ of heat?

15 - 22

(a)C6H6(l) + 7½O2(g) 6CO2(g) + 3H2O(l) ΔH = -3.268 x

103 kJ

Note how a thermochemical equation isdifferent from an ordinary balanced equation.

Fractional coefficients are permissible in thermochemical equations because the question asked for one mole of C6H6.

All thermochemical equations include the enthalpy term, - for an exothermic and a + for an endothermic reaction.

15 - 23

(b) ΔH = 25.0 g C6H6 × ×

3.268 x 103 kJ1 mol C6H6

1 mol C6H6

78.12 g C6H6

= 1.05 × 103 kJ

(c)m = 2.0 kJ × 1 mol C6H6

3.268 x 103 kJ×

1 mol C6H6

78.12 g C6H6

m = 0.048 g C6H6

15 - 24

Hess’s LawHess’s Law

Hess’s law says that when a chemicalreaction can be expressed as the algebraicsum of two or more other reactions, then

theheat of reaction is the algebraic sum of theheats of those reactions.

Because enthalpy is a state function, thechange in enthalpy is the same whether thereaction takes place in one step or a series

ofsteps.

15 - 25

Sometimes it is easier to start with the finalreaction and work backwards to decide howto manipulate the other reactions.

Calculate ΔHf° for the formation of dinitrogen

pentoxide given the followingthermochemical equations.

2H2(g) + O2(g) 2H2O(l) ΔH = -571.6 kJ

N2O5(g) + H2O(l) 2HNO3(l) ΔH = -73.7 kJ

1/2N2(g) + 3/2O2(g) + 1/2H2(g) HNO3(l)

ΔH = -174.1 kJ

15 - 26

2HNO3(l) N2O5(g) + H2O(l) ΔH = +73.7 kJ

N2(g) + 3O2(g) + H2(g) 2HNO3(l) ΔH = -

348.2 kJH2O(l) H2(g) + 1/2O2(g) ΔH = +285.8 kJN2(g) + 5/2O2(g) N2O5(g) ΔH = +11.3 kJ

There are many things to take note of here:

Remember ΔHf°’s are always computed for one mole.

The reaction containing N2O5 as a reactant must be reversed.

15 - 27

When a reaction is reversed, the sign is also reversed giving ΔH = +73.7 kJ.

To cancel the HNO3(l), you must multiply the equation by 2.

Because enthalpy is an extensive property, you must also multiply that by 2 giving -348.2 kJ.

To cancel the H2O(l), that equation must be reversed as well as multiplied by ½, giving +285.8 kJ.

15 - 28

Standard Enthalpy of FormationStandard Enthalpy of Formation

Standard enthalpy of formation is also knownas heat of formation and is represented byΔH.

Standard enthalpy of formation is the heatabsorbed or liberated when a compound isformed by its constituent elements in theirstandard state.

Remember that the standard state of an element is the state or phase of that element at P = 1.00 atm and T = 25°C = 273K.

15 - 29

Remember that systems want to achieve thelowest possible energy level possible.

The greater the amount of heat released during the formation of a compound, the lower the potential energy and the more stable the product.

The greater the amount of heat absorbed during the formation of a compound, the higher the potential energy and the less stable the product.

15 - 30

Heat of Formation ProblemHeat of Formation Problem

Given the following reaction,

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

(a)Determine the heat of formation of H2O.

(b)Calculate the for the formation of 11.45 g H2O.

(c) How would the problem change if H2O(g) was replaced by H2O(l)?

ΔH°= 483.64 kJ

ΔH°

15 - 31

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

H2(g) + ½O2(g) H2O(g)

Before moving on to (b):

When asked for the heat of formation of

a compound that compound must always

be a product.

The heat of formation is always for one mole of the compound.

ΔH = -483.64 kJ

ΔH = -241.82 kJ

15 - 32

It makes sense that if a chemical reaction is endothermic in the forward direction then it must be exothermic in the reverse direction.

When reverse the direction of a reaction,

you reverse the sign.

Remember that enthalpy is an extensive property, therefore, the ΔH needs to be divided by the coefficient of the formed compound.

15 - 33

(b) ΔH°= 11.45 g H2O ×1 mol H2O

18.02 g H2O

×1 mol H2O-241.82 kJ

ΔH°= -153.7 kJ

(c) To form one mole of H2O(g) (-241.82 kJ) requires a greater amount of heat than to form one mole of H2O(l) (-285.83 kJ).

-285.83 kJ would replace -241.82 kJ in the calculations.

15 - 34

When ammonia reacts with calciumcarbonate, calcium cyanamide, CaCN2, andwater are formed. The reaction absorbs90.1 kJ.

(a)What is the thermochemical equation?

(b) Determine the standard molar enthalpy of formation of calcium cyanamide.

(a) 2NH3(g) + CaCO3(s) CaCN2(s) + 3H2O(l) ΔH = +90.1 kJ

15 - 35

(b) ΔH° = ∑nΔHf°prod - ∑nΔHf°reac

90.1 kJ = 1 mol CaCN2×

ΔHf°CaCN2

1 mol CaCN2

+ 3 mol H2O×-285.8 kJ

1 mol H2O

- 1 mol CaCO3×

-1206.9 kJ

mol CaCO3

- 2 mol NH3×

-46.1 kJ

1 mol NH3

ΔHf°CaCN2 = -351.6 kJ

15 - 36

Calorimetry and Specific HeatCalorimetry and Specific Heat

Specific heat is the amount of heat needed to

raise or lower the temperature of one gram of

a substance 1°C or 1 K.

A change in temperature of 1°C = 1 K.

H2O has one of the highest specific heats,c = 4.18

JgK

, of all common substances.

Specific heat is an intensive property because it is quantity independent.

15 - 37

Specific heat is mathematically expressed as:

When doing calorimetry problems a fewassumptions are made:

ΔE = 0 is the conservation of mass-energy which is applied to the

universe.

The universe consists of the system and

the surroundings.

c =q

mΔT= J

gK

15 - 38

Energy may be wasted or lost from the system but the total energy of the universe must be conserved.

If ΔE = 0, then it follows that ql = qg, the

heat lost by the warmer object equals the heat gained by the cooler one.

Calorimetry experiments are always

done at constant pressure, such that, ΔH = qp.

15 - 39

Another related term is heat capacity.

Heat capacity is the temperature change that

an object undergoes when it absorbs acertain amount of heat.

Heat capacity is given by: C =

ΔqΔT

15 - 40

Calorimetry ProblemCalorimetry Problem

When 1.27 g of sodium bromide dissolves in67.1 g of water in a calorimeter, thetemperature drops from, 27.30°C to

26.58°C.Assume all the heat absorbed in the

solutionprocess comes from the water.

(a)Write a balanced equation for the solution process.

NaBr(s) Na+(aq) + Br-(aq)

15 - 41

(b) Is the solution process endothermic or exothermic? Justify your answer.

The 1st law of thermodynamics states the conservation of energy which means in an exchange of heat energy, ql = qg. Heat may escape from a system to its surroundings but in the universe Δq = 0.

Taking the water and the sodium bromide as the system, if the water drops in temperature, then heat is transferred from the water into the sodium bromide, making the solution process endothermic.

15 - 42

(c) What is q when 1.27 g of NaBr dissolves?

m = 1.27 g NaBr Ti = 27.30°C m = 67.1 g H2O Tf =

c = 4.18 J/g•K

Δq = 0 ql = qg

(

m = 1.27 g NaBr Ti = 27.30°C m = 67.1 g H2O Tf = 26.58°C

c = 4.18 J/g•K

Δq = 0 ql = qg

qg = mwcwΔT

qg = 67.1 g ×4.18J

g•K ×(27.30°C-26.58°C)

qg = 2.0 × 102 J

15 - 43

(d) What is q when 1.00 mol of NaBr dissolves?

qg=2.0 × 102 J1.27 g NaBr

×102.89 g NaBr1 mol NaBr

qg=1.6 × 104 J

15 - 44

ΔΔH and Bond Enthalpies (Energies)H and Bond Enthalpies (Energies)

Bond energy is the enthalpy change (energy)needed to break one mole of bonds in thegaseous state.

For most molecules, the bond enthalpydepends on its environment.

Environment refers to the electronegativity of bonded atoms.

Environment is also influenced by unshared pairs of electrons.

15 - 45

Bond energies are dependent upon thestrength of a bond.

Bond energy and bond strength decrease inthe following order:

triple > double > single

15 - 46

Bond Enthalpy ProblemBond Enthalpy Problem

Estimate ΔH for the following reaction:

CO(g) + Cl2(g) COCl2(g)

C O(g) + Cl Cl(g) Cl C Cl

O

15 - 47

ΔH = ∑nΔH(broken) - ∑nΔHreac(formed)

D(C O) + D(Cl Cl) - D(C O) - 2D(C Cl)

ΔH =(1072 kJ•mol-1 + 242 kJ•mol-1) -

(799 kJ•mol-1 + 2 × 328 kJ•mol-1)

ΔH = -141 kJ

15 - 48

Noteworthy features from this problem:

Before doing the problem, it is imperative to have the correct Lewis diagram for each compound.

Bond breaking is always an endothermic

reaction and bond formation is always exothermic.