Energy, EnthalpyCalorimetry & Thermochemistry

Energy is...• The ability to do work.• Conserved.• made of heat and work.• a state function.• independent of the path, or how you get from point A

to B.• Work is a force acting over a distance.• Heat is energy transferred between objects because of

temperature difference.

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.• Endothermic reactions absorb energy from the

surroundings.

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

CH + 2O 4 2

CO + 2 H O 2 2Pote

nti

al en

erg

y

Heat

N + O2 2

Pote

nti

al en

erg

y 2NO

N + O 2NO2 2 + heat

Heat

Units of Heat

• Calorie (cal)– The quantity of heat required to change the

temperature of one gram of water by one degree Celsius.

• Joule (J)– SI unit for heat

1 cal = 4.184 J

Direction

• Every energy measurement has three parts.1. A unit ( Joules or calories).2. A number how many.3. and a sign to tell direction.• negative - exothermic• positive- endothermic

System

Surroundings

Energy

E <0

System

Surroundings

Energy

E >0

First Law of Thermodynamics

• The energy of the universe is constant.• Law of conservation of energy.• q = heat• Take the systems point of view to decide signs.• q is negative when the system loses heat energy• q is positive when the system absorbs heat energy

Enthalpy

• Heat content of a substance• Depends on many things• Symbol is “H”• Can’t be directly measured• Changes in enthalpy can be calculated, H =

Hfinal - Hinitial

• Changes in heat (q) can be used to find H

Same rules for Enthalpy

• Heat energy given off by the system is negative, H < 0

• Heat energy absorbed by the system is positive, H > 0

q and H

q is the “heat flow for a system” It can be for any amount of mass or moles of substance

H is “change in heat”It is for a specific chemical or physical change and it refers

to that reaction or process.

Heat Capacity• The quantity of heat required to change the temperature

of a system by one degree.

• Three “types” with different “systems:– Specific Heat Capacity– Molar Heat Capacity– Heat Capacity

Heat Capacity Types

– Specific heat capacity• System is one gram of substance• Units are J/goC• q = heat change• m = mass• C = Specific Heat Capacity• final temperature – initial temperature

q = (m)(C)(T)

Heat Capacity Types

– Molar heat capacity.• System is one mole of substance.• Units are J/mol oC• q = heat change• n = moles• C = molar heat capacity• final temperature – initial temperature

q= nCT

Heat Capacity Types

– Heat capacity of an object• Specific for a given amount or unit

so it is not mass dependent• Units are J/oC• q = heat• C = heat capacity• final temperature – initial temperature

q = CT

Heat Capacity Types

• So how do you know which one to use when?• Look at the known information• Select the value with appropriate units to

cancel

Example

• The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.

Calorimetry

• Measuring temperature changes to calculate heat changes.

• Use a calorimeter.• Two kinds

– Constant pressure calorimeter (called a coffee cup calorimeter)

– Constant volume calorimeter (called a bomb calorimeter)

Calorimetry

• A coffee cup calorimeter measures temperature and calculates q.

• An insulated cup, full of water (constant pressure, variable volume)

• Water is the surroundings in which the system changes• Calculate the heat change of water. • The specific heat of water is 4.184 J /gºC• Heat of water qH2O = CH2O x mH2O x T

qsystem + qsurroundings = 0

• qH2O + qrxn = 0

• qH2O = - qrxn

In interactions between a system and its surroundings the total energy remains constant— energy is neither created nor destroyed.

Law of Conservation of Energy

Coffee Cup Calorimeter

• A simple calorimeter.– Well insulated and therefore isolated.– Measure temperature change.

qrxn = -qcal

Determination of Specific Heat

ExampleDetermining Specific Heat from Experimental Data.

Use the data presented on the last slide to calculate the specific heat of lead.

qlead = -qwater

qwater = mcT = (50.0 g)(4.184 J/g °C)(28.8 - 22.0)°C

qwater = 1.4x103 J

qlead = -1.4x103 J = mcT = (150.0 g)(c)(28.8 - 100.0)°C

clead = 0.13 Jg-1°C-1

Example

• When a piece of copper (5.0 g) is heated for 2.0 seconds, and 100 J of heat energy is transferred to the copper, the temperature increases from 20.0 ˚C to 71.9 ˚C.

• What is the specific heat of the copper?

Examples

If 10.0 g of Cu is heated for 2.0 seconds from 20.0 ˚C and 200 J of heat are absorbed, what is the final temperature of the block?

a. 71.9 ˚C d. 123.9 ˚Cb. 100 ˚C e. 719 ˚Cc. 46 ˚C

Example

Equal masses of liquid A, initially at 100 ˚C, and liquid B, initially at 50 ˚C, are combined in an insulated container. The final temperature of the mixture is 80 ˚C. Which has the larger specific heat capacity, A or B?•A•B•A and B have the same specific heat capacity.

Example• When 86.7 grams of water at a temperature of 73.0 ˚C is mixed with

an unknown mass of water at a temperature of 22.3 ˚C the final temperature of the resulting mixture is 61.7 ˚C. What was the mass of the second sample of water?

a. 24.9 g c. 48.2 g b. 302 g d.419 g

States of Matter

General Heating Curve

http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/031_ChangesState.MOV

Changes of State of Water

H2O (l) → H2O(g)

Hv = 40.65 kJ/mol @ 373.15 K

Molar enthalpy of vaporization:

Molar enthalpy of fusion (melting):

H2O (s) → H2O(l)

Hf = 6.01 kJ/mol @273.15 K

Hv = (40.65 kJ/mol)(1 mol/18.02 g) = 2256 J/g

Hf = (6.01 kJ/mol)(1 mol/18.02 g) = 333 J/g

qphase change = DHphase change (mass or moles)

Heating Curve for Water

Total heat absorbed by the water as it is warmed can be determined by finding the heat involved in each “step” of the process.

q1+ q2 + q3 …. = qtotal

Note:

Cice= Csteam = 2.09 J/goCq = (m)(C)(T)

qphase change = Hphase change (mass or moles)

Example

Calculate q for the process in which 50.0 g of water is converted from liquid at 10.0°C to vapor at 125.0°C.

Example

• What is the heat of fusion of lead in J/g if 6.30 kilojoules of heat are required to convert 255 grams of solid lead at its melting point into a liquid?

a. 0.0250 J/g c. 1.61 J/g b.24.7 J/g d. 40.5 J/g

Example• What quantity of heat is required to heat 1.00 g of lead from 25 ˚C to

the melting point (327 ˚C) and melt all of it? (The specific heat capacity of lead is 0.159 J/g • K and it requires 24.7 J/g to convert lead from the solid to the liquid state.)

a. 2.47 J c. 39.4 J b. 48.0 J d. 72.7 J

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.

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.

Bomb Calorimeter

• thermometer

• stirrer

• full of water

• ignition wire

• Steel bomb

• sample

Bomb Calorimeterqrxn = -qcal

qcal = qbomb + qwater + qwires +…

Define the heat capacity of the calorimeter:

qcal = miciT = CcalT

heat

q and H revisited

• If you know “q” for a process for a known amount of sample

• And you have a balanced chemical equation

• “q” can be converted to H

q to H

If you know the heat per amount of substance, find it per gram then use the molar mass to find the heat per mole.

Thermochemical Equations

• Balanced chemical equation

• Includes a term which indicates the change in enthalpy, H

• Enthalpy term can be embedded in the reaction or written after the reaction with H notation.

The combustion of 1.010 g sucrose, in a bomb calorimeter, causes the temperature to rise from 24.92 to 28.33°C. The heat capacity of the calorimeter assembly is 4.90 kJ/°C.

(a) What is the heat of combustion of sucrose, expressed in kJ/mol C12H22O11

(a) Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only 19 calories.

Examples

Endothermic Thermochemical Equations

Al2O3(s) 2Al(s) + 3/2 O2(g)

Hrxn = 1,676kJ/mol rxn

• THE EXACT SAME AS WRITING:

• Al2O3(s) + 1,676 kJ 2Al(s) + 3/2 O2(g)

Exothermic Thermochemical Equations

H2(g) + ½ O2(g) H2O(l) + 285.8kJ

Same as:

H2(g) + ½ O2(g) H2O(l)

Hrxn = -285.8 kJ/mol rxn

Stoichiometry and Thermochemical Equations

• Just as the mole ratios can be used to determine reacting and produced amounts of chemicals

• The ratios can be used to relate amounts of chemicals with energy

Example

Al2O3(s) 2Al(s) + 3/2 O2(g)

Hrxn = 1,676 kJ/mol rxn

• How many grams of aluminum would be produced if the aluminum absorbed 3500 kJ of energy?

Hrxn

Hrxn is for ANY general chemical or physical reaction

There are specific types of reactions which have a special designation for their H.

Some Important Types of Enthalpy Changes

Heat of combustion (Hcomb)

C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g)

Heat of formation (Hf)

K(s) + 1/2Br2(l) KBr(s)

Heat of fusion (Hfus)

NaCl(s) NaCl(l)

Heat of vaporization (Hvap)

C6H6(l) C6H6(g)

Standard States and Standard Enthalpy Changes

• Define a particular state as a standard state.

• Standard enthalpy of reaction, H°

– The enthalpy change of a reaction in which all reactants and products are in their standard states.

• Standard State– The pure element or compound at a pressure of 1 bar

and at the temperature of interest.

Indirect Determination of H:Hess’s Law

• H is an extensive property.Enthalpy change is directly proportional to the amount of substance in a system.

N2(g) + O2(g) → 2 NO(g) H = +180.50 kJ

½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ

H changes sign when a process is reversed

NO(g) → ½N2(g) + ½O2(g) H = -90.25 kJ

Hess’s LawHess’s law of constant heat summation

If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps.

½N2(g) + O2(g) → NO2(g) H = +33.18 kJ

½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ

NO(g) + ½O2(g) → NO2(g) H = -57.07 kJ

Example

Find ΔH for 2H2(g) + O2(g) 2H2O(l)

Given:

CH3COOH(l) + 2O2(g) 2CO2(g) + 2H2O(l) ΔH = -1,959.8 kJ

C(graphite) + O2(g) CO2(g) ΔH = -886.5 kJ

CH3COOH(l) 2C(graphite) + 2H2(g) + O2(g) ΔH = 1,100.2 kJ

Example 11

• Find H for 2NH3(g) N2H4 (l) + H2(g)

• Given: • N2H4(l) + CH4O(l) CH2O(g) + N2(g) + 3H2(g) ΔH = -92.5KJ

N2(g) + 3H2(g) 2NH3(g) ΔH = -115KJ

CH2O(g) + H2(g) CH4O(l) ΔH = 162.5KJ

Standard Enthalpies of Formation

• The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states.

• The standard enthalpy of formation of a pure element in its reference state is 0.

fH°

Standard Enthalpies of Formation

Standard Enthalpies of Formation

Writing Formation Equations

• Remember that fHo refers to forming ONE mole of a pure substance FROM ITS ELEMENTS in their standard state

• Write thermochemical formation equations for the formation of:– Al2O3(s)

– NaHCO3(s)

Standard Enthalpies of Reaction

H°overall = -2fH°NaHCO3

+ Hf°Na2CO3

+ Hf

°CO2

+ Hf°H2O

H°rxn

Enthalpy of Reaction

H°rxn = Hf°products- Hf

°reactants

Enthalpies of Formation of Ions in Aqueous Solutions