Thermochemistry © 2009, Prentice-Hall, Inc. Unit 5 Thermochemistry.

Thermochemistry

© 2009, Prentice-Hall, Inc.

Unit 5 Thermochemistry

Thermochemistry


Energy

• Energy is the ability to do work or transfer heat.– Energy used to cause an object that has

mass to move is called work.– Energy used to cause the temperature of

an object to rise is called heat.

Thermochemistry


The Nature of Energy

• Two forms of energy exist, potential and kinetic.

• Potential energy is due to composition or position.

• Chemical potential energy is energy stored in a substance because of its composition.

• Kinetic energy is energy of motion.

Thermochemistry


Kinetic-Molecular TheoryEnergy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.

Thermochemistry


Kinetic-Molecular Theory

The average kinetic energy of the molecules is proportional to the absolute temperature.

Thermochemistry


Units of Energy

• The SI unit of energy is the joule (J).

• An older, non-SI unit is still in widespread use: the calorie (cal).

1 cal = 4.184 J

1 J = 1 kg m2

s2

Thermochemistry


Heat

• Energy can also be transferred as heat.

• Heat flows from warmer objects to cooler objects.

Thermochemistry


Definitions:System and Surroundings

• The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).

• The surroundings are everything else (here, the cylinder and piston).

Thermochemistry


First Law of Thermodynamics• Energy is neither created nor destroyed.• In other words, the total energy of the universe is

a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Thermochemistry


Internal EnergyThe internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

Thermochemistry


Internal EnergyBy definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:

E = Efinal − Einitial

Thermochemistry


Changes in Internal Energy

• If E > 0, Efinal > Einitial

– Therefore, the system absorbed energy from the surroundings.

– This energy change is called endothermic.

Thermochemistry


Changes in Internal Energy

• If E < 0, Efinal < Einitial

– Therefore, the system released energy to the surroundings.

– This energy change is called exothermic.

Thermochemistry


Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the surroundings, the process is endothermic.

Thermochemistry


Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the surroundings, the process is endothermic.

• When heat is released by the system into the surroundings, the process is exothermic.

Thermochemistry


Calorimetry

Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

Thermochemistry


Constant Pressure Calorimetry

By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

Thermochemistry


Specific Heat

• The specific heat of any substance is the amount of heat required to raise one gram of that substance one degree Celsius.

Heat Capacity - the amount of heat needed to increase the temperature of an object exactly 1oC Depends on both the object’s mass and its chemical composition

Thermochemistry


Specific Heat (cont.)

• Calculating heat absorbed and released

– q = c × m × ΔT

– q = heat absorbed or released (in Joules)

– c = specific heat of substance

– m = mass of substance in grams

– ΔT = change in temperature in Celsius

Thermochemistry


• Examples:• How much heat does a 20.0 g ice cube absorb

as its temperature increases from (-27.0oC) to 0.0oC? Give your answer in both joules and calories.

• q = c × m × ΔT

• Specific Heat of Ice = 2.03 J/goC

• 1 calorie = 4.184 J


Thermochemistry


• Example Cont. q = ? c = 2.03 J/goC m = 20.0 grams ΔT = FinalTemp(0.0oC) – InitialTemp (-27.0oC) = Change (27.0oC)

q = c × m × ΔT

q = (2.03 J/goC)(20.0g)(27.0oC)= 1.10 x 103 J

263 cal


Thermochemistry


• Example 2: • A 5.00 gram sample of a metal is initially at 55.0

ºC. When the metal is allowed to cool for a certain time, 98.8 Joules of energy are lost and the temperature decreases to 11.0 ºC. What is the specific heat of the metal? What metal is it?

• q = c × m × ΔT

0.449 J/g ºCIron

23Thermochemistry

The solid temperature is rising from -20 to 0 oC (use q = mass x ΔT x C)

The solid is melting at 0o C; no temperature change (use q = moles x ΔHfus.)

The liquid temperature is rising from 0 to 100 oC (use q = mass x ΔT x C)

The liquid is boiling at 100o C; no temperature change (use q = moles x ΔHvap.)

The gas temperature is rising from 100 to 120 oC (use q = mass x ΔT x C)The Heat Curve for Water, going from -20 to 120 oC,

120

Thermochemistry


State Functions

Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

Thermochemistry


State Functions• However, we do know that the internal energy

of a system is independent of the path by which the system achieved that state.– In the system below, the water could have reached

room temperature from either direction.

Thermochemistry


State Functions• Therefore, internal energy is a state function.• It depends only on the present state of the

system, not on the path by which the system arrived at that state.

• And so, E depends only on Einitial and Efinal.

Thermochemistry


Enthalpy

• If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system.

• Enthalpy is the internal energy plus the product of pressure and volume:

H = E + PV

Thermochemistry


Enthalpy

• When the system changes at constant pressure, the change in enthalpy, H, is

H = (E + PV)

• This can be written

H = E + PV

Thermochemistry


Enthalpy

• Since E = q + w and w = -PV, we can substitute these into the enthalpy expression:

H = E + PV

H = (q+w) − w

H = q

• So, at constant pressure, the change in enthalpy is the heat gained or lost.

Thermochemistry


Endothermic and Exothermic

• A process is endothermic when H is positive.

Thermochemistry


Endothermicity and Exothermicity

• A process is endothermic when H is positive.

• A process is exothermic when H is negative.

Thermochemistry


Enthalpy of Reaction

The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants

Thermochemistry


Enthalpy of Reaction

This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

Thermochemistry


The Truth about Enthalpy

1. Enthalpy is an extensive property (depends on how much).

2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.

3. H for a reaction depends on the state of the products and the state of the reactants.

Thermochemistry


Hess’s Law

H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested.

• However, we can estimate H using published H values and the properties of enthalpy.

Thermochemistry


Hess’s Law

Hess’s law states that “[i]f a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

Thermochemistry


Hess’s Law

Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

Thermochemistry


Enthalpies of Formation

An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

Thermochemistry


Standard Enthalpies of Formation

Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure).

Thermochemistry


Calculate the standard enthalpy of formation of CS2 (l) given that:C(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJrxn

S(rhombic) + O2 (g) SO2 (g) H0 = -296.1 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1. Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)

2. Add the given rxns so that the result is the desired rxn.

rxnC(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJ

2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -296.1x2 kJrxn

CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)

H0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJrxn

Thermochemistry


Calculation of H

• Imagine this as occurringin three steps:

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

C3H8 (g) 3 C (graphite) + 4 H2 (g)

3 C (graphite) + 3 O2 (g) 3 CO2 (g)

4 H2 (g) + 2 O2 (g) 4 H2O (l)

Thermochemistry


Calculation of H


C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)



4 H2 (g) + 2 O2 (g) 4 H2O (l)

Thermochemistry


Calculation of H


C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)



4 H2 (g) + 2 O2 (g) 4 H2O (l)

Thermochemistry


C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)



4 H2 (g) + 2 O2 (g) 4 H2O (l)

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

Calculation of H

• The sum of these equations is:

Thermochemistry


Calculation of H

We can use Hess’s law in this way:

H = nHf°products – mHf° reactants

where n and m are the stoichiometric coefficients.

Thermochemistry


H = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)]

= [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)]= (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

Calculation of H

Thermochemistry


Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn nH0 (products)f= mH0 (reactants)f-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ

-5946 kJ2 mol

= - 2973 kJ/mol C6H6

6.6

Thermochemistry


The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy.

H2 (g) H (g) + H (g) H0 = 436.4 kJ

Cl2 (g) Cl (g)+ Cl (g) H0 = 242.7 kJ

HCl (g) H (g) + Cl (g) H0 = 431.9 kJ

O2 (g) O (g) + O (g) H0 = 498.7 kJ O O

N2 (g) N (g) + N (g) H0 = 941.4 kJ N N

Bond Energy

Bond Energies

Single bond < Double bond < Triple bond

9.10

Thermochemistry


Average bond energy in polyatomic molecules

H2O (g) H (g) + OH (g) H0 = 502 kJ

OH (g) H (g) + O (g) H0 = 427 kJ

Average OH bond energy = 502 + 427

2= 464 kJ

9.10

Thermochemistry


Bond Energies (BE) and Enthalpy changes in reactions

H0 = total energy input – total energy released= BE(reactants) – BE(products)

Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products.

9.10

Thermochemistry


Use bond energies to calculate the enthalpy change for:H2 (g) + F2 (g) 2HF (g)

H0 = BE(reactants) – BE(products)

Type of bonds broken

Number of bonds broken

Bond energy (kJ/mol)

Energy change (kJ)

H H 1 436.4 436.4

F F 1 156.9 156.9

Type of bonds formed

Number of bonds formed

Bond energy (kJ/mol)

Energy change (kJ)

H F 2 568.2 1136.4

H0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ

9.10

Thermochemistry

© 2009, Prentice-Hall, Inc.9.3

Lattice energy (E) increases as Q increases and/or

as r decreases.

cmpd lattice energyMgF2

MgO

LiF

LiCl

2957

3938

1036

853

Q= +2,-1

Q= +2,-2

r F < r Cl

Electrostatic (Lattice) Energy

E = kQ+Q-r

Q+ is the charge on the cation

Q- is the charge on the anionr is the distance between the ions

Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.

Thermochemistry

© 2009, Prentice-Hall, Inc.9.3

Born-Haber Cycle for Determining Lattice Energy

Hoverall = H1 + H2 + H3 + H4 + H5o ooooo

Thermochemistry © 2009, Prentice-Hall, Inc. Unit 5 Thermochemistry.

Documents

energy change

heat change

energy of motion

initial energy

final energy

total energy

exchange of heat

surroundingswhen heat