Chapter 6

Thermochemistry

Energy is...– the capacity to do work or produce heat– conserved, amount of energy in universe never

changes– Temperature is measure of how hot or cold an

object is–Heat ALWAYS move from hot to cold– State Function or Property - something that is

independent of the path, or how you get from point A to B–Work* (w) = force acting over a distance–Heat* (q) = energy transferred between objects

* Not a State Function

State Function or Property

• All gases are at 1 atm• All liquids are pure• All solids are pure• All solutions are at 1 M concentrations• The energy of formation of an element in its

normal state is defined as zero• The temperature used for standard state values

is almost invariably room temperature: 25oC (298 K)

Kinetic Energy• Energy an object possesses by virtue of

its motion.21

2KE mv

Potential EnergyEnergy an object possesses by virtue of its position or chemical composition.

The Universe for Chemists

• Divided into two halves: 1. System2. 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

System and SurroundingsThe System includes the molecules we want to study (here gases in system)The Surroundings are everything else (here, the cylinder and piston).

C H + 2O C O + 2H O + Heat4 2 2 2

C H + 2O 4 2

CO + 2 H O 2 2

Inte

rnal Energ

y (

E)

Heat

Exothermic

Chemical Energy

Inte

rnal En

erg

y (

E)

Heat

2NO

N + O 2N O2 2 + heat

Endothermic

Chemical Energy

Energy Measurement• Every energy measurement has three parts:

1. A unit ( Joules or calories).2. A number, it’s magnitude.3. A sign to tell direction.

• Exothermic measurements are negative • Endothermic measurements are positive• Sign reflects system point of view

Units of Energy

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

• An older, non-SI unit is still in widespread use: calorie (cal) 1 cal = 4.184 J

2

2

1kg

Jsm

First Law of Thermodynamics

• Energy of the universe is constant.• Energy can neither be created nor destroyed*

• Law of conservation of energy.

• ΔE = q + w• w = work• q = heat• ΔE = KE + PE = Change in Energy

* You can’t win – all you can do is break even

Internal Energy• The internal energy of a system is the sum of all

kinetic and potential energies of all components of the system; we call it E.

• Internal energy can be changed by the flow of work, heat or both.

• ∆ means change in the systems internal energy

E q wΔ

Sign Conventions for q, w, and ∆E

+ (plus) - (minus)q sys. gains heat sys. loses heat

w work done on sys. work done by sys.

∆E net gain of energy by sys.

net loss of energy by sys.

Exothermic Reaction

Endothermic Reaction

increases

Some rules for heat and work

• Heat given off is negative. (Exothermic)• Heat absorbed is positive. (Endothermic)• Work done by system on surroundings, heat flows

out of system, heat is product.• Work done on system by surroundings, heat flows

into system, heat is reactant.• Thermodynamics- The study of energy and the

changes it undergoes.

System View Heat q Work w

Endothermic(Heat Reactant) + + Exothermic

(Heat Product) - -

Take The Systems Point Of View

Calculate ∆E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.

Time to plug and chug

Remember to watch signs

E q wΔ

Piston System & Surroundings

Work Done By A Piston1. Since pressure is defined as force per unit area

pressure of gas is

2. Work is defined as force applied over a distance and piston moves ∆h

3. Work = force x distance = F x ∆h 4. Use equations in 1 & 3, then w = P x A x ∆h 5. Since Volume = A x ∆h = ∆V can be substituted

into work in step 4 and get

or F=PxAF

PA

w P V Δ

Work Done By A Piston

• If gas is expanding it is work done on the surroundings and work has to be a negative number and ΔV is positive (Vfinal – Vinitial)

• w and PΔV must have opposite signs• Hence the negative in formula

Calculate the work associated with expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm.

w P V Δ

PracticeA balloon is being inflated to its full extent by heating the air inside it. In the final states of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ∆E for the process.

Work involved in the expansion or compression of gases is called pressure-volume work.

Equality you should know: 1 101.3L atm J

E q w

w P V

Δ Δ

Enthalpy (H)A measure of total energy in system

Enthalpy is defined as H = E + PVWhen a change occurs at constant pressure the

equation becomes:ΔH = ΔE + PV) = ΔE + P∆V

the heat at constant pressure qp can be calculated

ΔE = qp + w and w= – PΔV

Solve for qp

qp = ΔE + P ΔV = ΔH

End result of calculation: qp =ΔH

Enthalpy (H)

• For a chemical reaction, enthalpy change is∆H = Hproducts – Hreactants

• If Hproducts > Hreactants +∆H Endothermic

• If Hreactants > Hproducts -∆H Exothermic

Equality you should know: 1 mole X = (value) ∆Hx kJ

Determine the Sign of ∆H• Indicate the sign of ∆H in each of the following

process carried out under atmospheric pressure and indicate if the process is endothermic or exothermic:

1.An ice cube melts2.1 g of butane (What is the formula?) is combusted in

sufficient O2 to give complete

Calorimetry• Device used to measure heat associated with a

chemical reaction is a calorimeter• Calorimetry is the Science Of Measuring Heat• Two kinds of devices:– Constant Pressure Calorimeter – Coffee cup calorimeter– Bomb Calorimeter – constant volume

• heat capacity ( C ) for a substance• The heat capacity of an object is the amount of

heat required to raise its temperature by 1 K

heat absorbed

increase in temperature

Calorimetry• Specific Heat Capacity

• heat capacity is given as per gram of substance• Units are J/oC*g or J/K*g

• Molar Heat Capacity• heat capacity is given as per mole of substance• Units are J/oC*mol or J/K*mol

• Basic Principle behind the Calorimeter isEnergy Released by Reaction (Joules)

q = Energy Absorbed by Solutionq = mass of solution (m) * Specific heat capacity (C) * Change

in temperature (∆T)

q = m * C * ∆T

Practice Constant Pressure Calorimeter

1. How much heat is needed to warm 250g of water from 22oC to 98oC? The specific heat of water is 4.18J/gK.

2. What is the molar heat capacity of water?

Practice Constant Pressure Calorimeter

• A hot water bottle filled with 0.75kg of water is at 80.0OC at the start of the night and cools to 20.0OC by morning. How much heat was given out? CH2O(l) = 4180 J/kgOC

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 calorimeters are known and tested

• ΔE = q + w, and and V=0, then w=0 and ΔE = q in a constant volume process.

qrxn = -Ccalorimeter x ∆T

PΔVw

Bomb Calorimeter

• thermometer

• stirrer

• full of water

• ignition wire

• Steel bomb

• sample

Measuring qrxn Using Bomb Calorimeter

• Methylhydrazine (CH6N2) is commonly used as a liquid rocket fuel. Yea – finally some rocket science! The combustion of methylhydrazine with O2 produces N2(g), CO2(g), and H2O(l):

CH6N2 + O2 → N2(g) + CO2(g) + H2O(l)

When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00OC to 39.50OC. In a separate experiment the heat capacity of the calorimeter is measured to be 7.794kJ/OC. What is the heat of reaction for the combustion of a mole of CH6N2 in this calorimeter?

Properties

• Intensive Properties not related to the amount of substance.– density, melting point, temperature.

• Extensive Properties - does depend on the amount of stuff.– Heat capacity (C), mass, heat of reaction (q),

volume

Hess’s Law

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

desired final product, and find ΔH• Two rules

• If the reaction is reversed the sign of ΔH is changed• If the reaction is multiplied, so is ΔH

• Calculate the ΔH1 for the overall reaction

2 2 2( ) 2 ( ) 2 ( )N g O g NO g

N2 2O2

O2NO

68 kJ

NO2∆H2=180 kJ

∆H3 = -112 kJ

H (

kJ)

∆H1 = ∆H2 + ∆H3 Hess’s Law

2( ) 2( ) ( )2g g gN O NO

( ) 2( ) 2( )2 2g g gNO O NO 2 180H kJΔ

3 112H kJΔ

2( ) 2( ) 2( )2 2g g gN O NO

1 2 3 68H H H kJΔ Δ Δ

Net Reaction achieved by summing the 2 reactions

Enthalpy achieved by summing enthalpy for 2 step reactions

PracticeTwo forms of carbon are graphite, the soft, black, slippery

material used in “lead” pencils and as a lubricant for locks, and diamonds, the brilliant, hard gemstone that is a lady’s best friend. Using the enthalpies of combustion for graphite (-394kJ/mol) and diamond (-396kJ/mol), calculate ∆H for the conversion of graphite to diamond:

Cgraphite(s) Cdiamond (s)

Two combustion reactions for the forms:

2 2( ) ( ) ( ) H = -394kJgraphiteC s O g CO g Δ

2 2( ) ( ) ( ) H = -396kJdiamondC s O g CO g Δ

Reverse the 2nd equation and change sign of ∆H

2 2( ) ( ) ( ) H = -394kJgraphiteC s O g CO g Δ

2 2( ) ( ) ( ) H = -(-396kJ)diamondCO g C s O g Δ

Do #58 & 60 on p. 269

Standard Enthalpy• Standard Enthalpy of Formation (ΔHf

o) is the change in enthalpy that accompanies the formation of one mole of a pure substance from its elements with all substances in their standard state.

• Degree symbol on ΔHfo indicates standard states:

ΔHfo For Compounds: Gaseous substance, pressure is exactly 1 atm For liquid or solid it is the pure liquid or pure solid For substance in solution, it is the concentration of 1M

ΔHfo for an element: it is the form the element exists

under 1 atm and 25oC, Ex. oxygen is O2(g) at 1 atm

Standard Enthalpies of Formation

• Enthalpy is a state function so could use Hess’s Law• Choose any pathway from reactant to product and

sum the enthalpy changes along the chosen pathway. • Standard states are 1 atm, 1M and 25oC• For an element ΔHf

o = 0, there is no formation reaction needed when element is already in its standard state

• See table in Appendix 4 (p. A19-22)

How can you find ΔH w/out Hess’s Law?

• We can use Heats of Formation ΔHfo to figure out

the heat of reaction.• Ammonia is burned in air to form nitrogen dioxide

and water – Always start with a balance equation4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(l)

And use this formula

o of fH ( H products)- ( H reactants)o

rxn p rn nΔ Δ Δ

Do #68 p. 269

Other Enthalpies

• Enthalpies of vaporization - ∆H for converting a liquid to a gas

• Enthalpies of fusion - ∆H for melting solids

• Enthalpies of combustion - ∆H for combusting a substance in oxygen