Chapter 6 Energy Thermodynamics. 6.1 Nature of Energy The ability to do work or produce heat. Conserved - can be converted from one form to another but.

Chapter 6

Energy

Thermodynamics

6.1 Nature of Energy • The ability to do work or produce heat.• Conserved - can be converted from one form to

another but can neither be created nor destroyed.

• Work is a force acting over a distance.• Potential: due to position or composition - can

be converted to work.• Kinetic: due to motion of the object.

KE = 1/2 mv2

(m = mass, v = velocity)

Heat and Temperature• Temperature reflects random motion of

particles in a substance.• Heat is the measure of energy content.• Heat is energy transferred between

objects because of temperature difference.

• State Function - property of a system that depends only on its present state.

• Independent of the path, or how you get from point A to B.

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 2

Potential energy

Heat

N + O2 2

Potential energy

Heat

2NO

N + O 2NO2 2 + heat →

Direction

• Every energy measurement has three parts.

1. A unit ( Joules or calories).

2. A number - how many.

3. A sign to tell direction.

• Negative - exothermic

• Positive- endothermic

System

Surroundings

Energy

E <0

System

Surroundings

Energy

E >0

Same rules for heat and work

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

is negative.• Work done on system by surroundings

is positive.• Thermodynamics - The study of energy

and the changes it undergoes.

• (YDVD)

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First Law of Thermodynamics

• The energy of the universe is constant.

• Law of conservation of energy.

• q = heat

• w = workE = q + w

• Take the systems point of view to decide signs.

What is work?

• Work is a force acting over a distance.• work = force distance• since pressure = force / area, • work = pressure volume• Work can be calculated by multiplying

pressure by the change in volume at constant pressure.

• units of L•atm

Work needs a sign• If the volume of a gas increases, the system

has done work on the surroundings.• work is negative

• wsystem = PV

• Expanding work is negative.• Contracting, surroundings do work on the

system w is positive.• 1 L•atm = 101.3 J

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Examples• Calculate the 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. EX. 6.1

• What amount of work is done when 46L of gas is expanded to 64 L at 15 atm pressure? EX. 6.2

• If 2.36 J of heat are absorbed by the gas above. What is the change in energy? 1 L•atm = 101.3 J

6.2 Enthalpy• Abbreviated H• H = E + PV (that’s the definition), at constant

pressure. H = E + PV

• the heat at constant pressure qp can be

calculated from: E = qp + w = qp - PV

qp = E + P V = H

• Where qP = H at constant pressure. H = energy flow as heat (at constant

pressure).

Examples• When 1 mole of methane (CH4) is burned

at constant pressure, 890 kJ of energy is released as heat. Calculate the H for a process in which 5.8 g sample of methane is burned at a constant pressure. EX 6.4

• Consider the following reaction:

2H2(g) + O2(g) 2H2O(l) H=-572kJ

How much heat is evolved when 2.56 g of hydrogen is reacted with excess oxygen? #34b

Calorimetry

• Measuring heat. We use a calorimeter.• The heat capacity for a material, C, is

calculated.• C = heat absorbed/T = H/ T

• specific heat capacity

heat capacity per gram = J/°C•g or J/K•g

• molar heat capacityheat capacity per mole = J/°C•mol or J/K•mol

Calorimetry• Constant pressure calorimeter (coffee

cup calorimeter, used for solutions).• 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.• The specific heat of water is 1 cal/gºC

(4.184 J/gºC)• Heat of reaction= H = s x mass x T

Examples

• 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.

• A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?

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.

• Ex. density, specific heat, temperature.

• Extensive property - does depend on the amount of stuff.

• Ex. heat capacity, mass, heat from a reaction.

6.3 Hess’s Law

• Enthalpy is a state function.• The change in enthalpy is the same whether

the reaction takes place in one step or a series of steps.

• We can add equations to to come up with the desired final product, and add the H.

• Two rules:

1. If the reaction is reversed the sign of H is changed.

2. If the reaction is multiplied, so is H.

Rules

1. If a reaction is reversed, H is also reversed.

N2(g) + O2(g) 2NO(g) H = 180 kJ

2NO(g) N2(g) + O2(g) H = 180 kJ

2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer.

6NO(g) 3N2(g) + 3O2(g) H = 540 kJ

Examples

• When using Hess’s Law, work by adding the equations up to make it look like the answer.

• Make the other compounds cancel out.

• N2(g) + 2O2(g) 2NO2(g) H1 =68kJ

• Above reaction is carried out in two steps below:

• N2(g) + O2(g) 2NO(g) H2 =180kJ

• 2NO(g) + O2(g) 2NO2(g) H3 =-112kJ

N2 O2

2NO, O2

68 kJ

2NO2180 kJ

-112 kJ

H (kJ)

Practice

• Given(BDVD)

C H (g) + 5

2O (g) 2CO (g) + H O( ) 2 2 2 2 2→ l

C(s) + O (g) CO (g) 2 2→

H (g) + 1

2O (g) H (l) 2 2 2O→

2C(s) + H (g) C H (g) 2 2 2→

Hº= -1300. kJ

Hº= -394 kJ

Hº= -286 kJCalculate Hº for this reaction

Practice

O (g) + H (g) 2OH(g) 2 2 →O (g) 2O(g)2 →H (g) 2H(g)2 →

O(g) + H(g) OH(g) →

Given

Calculate Hº for this reaction

Hº= +77.9kJHº= +495 kJHº= +435.9kJ

Practice

• P4(s) + 6Cl2(g) 4PCl3(g) H = -1225.6kJ

• P4(s) + 5O2(g) P4O10(s) H = -2967.3kJ

• PCl3(g) + Cl2(g) PCl5(g) H = - 84.2kJ

• PCl3(g) + 1/2O2(g) Cl3PO(g) H = -285.7 kJ

• Calculate the H for the reaction #58

• P4O10(s) + 6PCl5(g) 10Cl3PO(g)

6.4 Standard Enthalpies of Formation

Standard States• Compound

For a gas, pressure is exactly 1 atmosphere. For a solution, concentration is exactly 1 molar. Pure substance (liquid or solid), it is the pure

liquid or solid.

• Element The form [N2(g), K(s)] in which it exists at 1 atm

and 25°C.

Standard Enthalpy of Formation• The enthalpy change that occurs in the formation of

one mole of a compound for a reaction at standard conditions (25ºC, 1 atm, 1 M solutions).

• Symbol Hºf

• There is a table in Appendix 4 (pg A21) It is 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 Enthalpies of Formation

• Need to be able to write the equations.• What is the equation for the formation of

NO2 ?

• ½N2 (g) + O2 (g) NO2 (g)• Have to make one mole to meet the

definition.• Write the equation for the combustion of

methanol CH3OH.• What is the equation for the formation of

solid aluminum oxide?

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. Ex.6.11

• CH3OH(l) + 3O2(g) 2CO2 (g) + 3H2O(l)

€

Horxn = ∑(ΔH f

oproducts) - ∑(ΔH foreactants)

Thermite Reaction• Using enthalpies of formation, calculate the

standard change in enthalpy for the thermite reaction.Ex 6.10

2Al(s) + Fe2O3(s) --> Al2O3(s) + 2Fe(s)(YDVD)

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It’s a gas!

• Methanol (CH3OH) is often used as a fuel in high performance engines in race cars. Using the data in table 6.2, compare the standard enthalpy of combustion per gram of methanol with that per gram of gasoline. Gasoline is actually a mixture of compounds, but assume for this problem that gasoline is pure liquid octane (C8H18)