chapter 5 thermochemistry. energy of objects objects can possess 2 kinds of energy. ke= energy of...

Chapter 5

Thermochemistry

Energy of objects

• Objects can possess 2 kinds of energy.

• KE= energy of motion• Ek= ½ mv2

• PE= stored energy (energy of position)

Units of heat

• Joule= 1 kg-m2/s2

• 1 cal= 4.184 Joule

The universe

• is divided into two halves.• the system and the surroundings.• The system is the part you are concerned

with.• The surroundings are the rest.• Exothermic reactions release energy to

the surroundings.• Endo thermic reactions absorb energy

from the surroundings.

Energy is...• The ability to do work or transfer

heat.• Energy can be transferred by doing

work• Force- any kind of push or pull

exerted on an object• Work= any energy used to move an

object against a force• W=f x d

• Energy can be transferred as heat.• Always in the direction from hotter

object to a colder one. When you apply an ice pack your body part feels cooler because your body is transferring heat to the ice pack.

Energy is conserved!

• First law of thermodynamics- any energy that is lost by a system must be gained by its surroundings, and vice versa.

• The change in energy that occurs during a reaction can be calculated

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

CH + 2O 4 2

CO + 2 H O 2 2

Pote

nti

al en

erg

y

Heat

N + O2 2

Pote

nti

al en

erg

y

Heat

2NO

N + O 2NO2 2 + heat

Energy changes

• Every energy measurement has three parts.

1. A unit 2. A number (indicating magnitude)3. and a sign to tell direction.• negative – exothermic (system loses

energy)• positive- endothermic (system gains

energy)

System

Surroundings

Energy

E <0

System

Surroundings

Energy

E >0

• LOOK AT FIGURE 5.5 on page 150.

Calculating energy changes

• System can change energy by work or heat

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

surroundings is positive.

• Work done on system by surroundings is negative.

• Thermodynamics- The study of energy and the changes it undergoes.

First Law of Thermodynamics

• Law of conservation of energy.• q = heat• w = workE = q + w• Take the systems point of view to

decide signs.• Look at Table 5.1

• The internal energy of a system is a state function, it is independent of it past history.

• ΔE= state function (q and w are not though)

Practice

• Hydrogen and oxygen gases are in a cylinder. As a reaction occurs between them, the system loses 1150 joules of heat to the surroundings. The reaction also causes a piston to rise as the hot gas expands. The expanding gas does 480 J of work on the surroundings. What is the change in internal energy of the system?

Enthalpy

• Most reactions occur at constant pressure and so energy is transferred mainly in the form of heat.

• Enthalpy is the heat absorbed or released by a system under constant pressure

• abbreviated H

More about enthalpy

• State function• Cannot be measured, but the change

can.

• ΔH equals the heat (qp) gained or lost

by the system when the process occurs

under constant pressure.

• ΔH = Hfinal – Hinitial = qp

• Since H is a state function it

depends only on the starting and

ending states of the system, not

how they got there.

Enthalpies of reactions

• ΔHrxn = H(products)- H(reactants)

• Thermochemical equations

• 2H2 + O2 = 2H2O ΔH= -483.6 kj

• Enthalpy is an extensive property (dependent on amount)

Calorimetry

• Measuring heat.• Use a calorimeter.• Two kinds• Constant pressure calorimeter (called

a coffee cup calorimeter)• heat capacity for a material, C is

calculated • C= heat absorbed/ T = H/ T• specific heat capacity = C/mass

Calorimetry

• molar heat capacity = C/moles• heat = specific heat x m x T• heat = molar heat x moles x T• Make the units work and you’ve done

the problem right.• A coffee cup calorimeter measures H.• An insulated cup, full of water. • The specific heat of water is 1 cal/gºC• Heat of reaction= H = sh x mass x T

Examples

• How much heat is released when 4.5 g of methane is burned in a constant pressure system.

CH4 + 202 CO2 + H2O ΔH=-890 kJ

Calorimetry

• Constant volume calorimeter is called a bomb calorimeter.

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

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

• Since V = 0, PV = 0, E = q

Bomb Calorimeter

• thermometer

• stirrer

• full of water

• ignition wire

• Steel bomb

• sample

Properties

• intensive properties not related to the amount of substance.

• density, specific heat, temperature.• Extensive property - does depend

on the amount of stuff.• Heat capacity, mass, heat from a

reaction.

Hess’s Law

• Enthalpy is a state function.• It is independent of the path.• We can add equations to to come up with

the desired final product, and add the H• Two rules• If the reaction is reversed the sign of H

is changed• If the reaction is multiplied, so is H

N2 2O2

O2 NO2

68 kJ

NO2180 kJ

-112 kJ

H (

kJ)

Standard Enthalpy

• The enthalpy change for a reaction at standard conditions (25ºC, 1 atm , 1 M solutions)

• Symbol Hº• When using Hess’s Law, work by

adding the equations up to make it look like the answer.

• The other parts will cancel out.

• Use the following data to calculate the enthalpy of combustion of C CO

( C + ½ O2 CO )

1. C + O2 CO2 ΔH= -393.5 kJ/mol

2. CO + ½ O2 CO2 ΔH=-283.0 kJ/mol

Standard Enthalpies of Formation

• Made a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states.

• Standard states are 1 atm, 1M and 25ºC

• ΔHrxn= ΣnΔHf (products) – ΣmΔHf (reactants)

• Use appendix C to calculate the standard enthalpy change for the combustion of 1 mol of benzene (C6H6).

Since we can manipulate the equations

• We can use heats of formation to figure out the heat of reaction.

• Lets do it with this equation.

• C2H5OH +3O2(g) 2CO2 + 3H2O

• which leads us to this rule.

Since we can manipulate the equations

• We can use heats of formation to figure out the heat of reaction.

• Lets do it with this equation.

• C2H5OH +3O2(g) 2CO2 + 3H2O

• which leads us to this rule.

( H products) - ( H reactants) = Hfo

fo o

• You are responsible for all of chapter this includes the part we did not talk about on foods and fuels. Make sure you study it. It will be on the test!