Energy

Energy

• Energy – the ability to do work or produce heat

• Energy exists in two different forms – kinetic energy & potential energy

Potential Energy

• Potential energy – energy due to composition or position of an object

• Potential energy is stored energy that results from the attractions or repulsions of other objects

Kinetic Energy

• Kinetic energy – the energy of motion

• Kinetic energy depends on as objects mass & its velocity

• Atoms has mass & they are in motion; therefore they will have kinetic energy

Energy

• A roller coaster at the top of a hill has a great amount of potential energy.

• As the rollercoaster begins to speed down the hill, the potential energy is turned into kinetic energy

Energy

• The SI unit for energy is the joule (J)

• 1 J = 1 Kgm2 / s2

• Another unit of energy that you may be more familiar with is the calorie

• calorie – amount of energy required to raise 1 g of water 1°C

• 1 cal = 4.18 J

Energy

• The calories that you eat are actually kilocalories or Calories (with a big C)

• 1000 calories = 1 Kilocalorie = 1 Calorie

Energy Conversions

• Convert 15,500 joules into Calories

• 15500 J x 1 cal x 1 Cal =

• 4.18 J 1000 cal

• 3.71 Cal

Formulas – Kinetic Energy

• Kinetic energy

• KE = ½ mv2

• KE = kinetic energy (joules)

• m = mass (must be in Kg)

• V = velocity (must be in m/s)

Formulas – Potential Energy

• Potential Energy

• PE = mgh

• PE = Potential Energy (J)

• m = mass (Kg)

• g = gravitational constant = 9.8 m/s2

• h = height (m)

Formulas - Work

• Work (w) – the energy used to move an object against a force

• Force (f) – a push or pull on an object

• W = mgd = fd = PE

• Work and potential energy can be looked at in the same light

Work

• It is important to understand that if there is no movement, there is no work done

• If I push and push on the demonstration table with all of my might. I may get hot and sweaty and feel like I have done a TON of work, but in reality I have done NO work because the table has not moved

Examples

• A bowler lifts a 5.4 kg bowling ball 1.6m and then drops it to the ground.

• How much work was required to raise the ball?

• W = mgd

• W = (5.4 kg)(9.8 m/s2)(1.6m)

• 85 Kgm2/s2 = 85 J

Examples

• How much potential energy does that ball have at this height?

• 85 J

Examples

• If the mass is dropped and we assume that all of the potential energy is turned into kinetic energy, at what velocity will the bowling ball hit the ground?

• KE = PE = 85J

• m = 5.4 Kg

• V = ?

Examples

• KE = ½ mv2

• 85 J = ½ (5.4 Kg) v2

• v2 = 31.5

• v = 5.6 m/s

More examples

• What is the kinetic energy of 1 atom of Ar moving at 650 m/s?

• KE = ½ mv2

• 1atom Ar x 1mol Ar x 39.95 g Ar x 1 Kg Ar =• 6.02 x 1023 atoms 1 mol Ar 1 x 103 g Ar

• 6.64 x 10-26 kg Ar

More Examples

• KE = ½ mv2

• KE = ½ (6.64 x 10-26 Kg)(650m/s) 2

• KE = 1.4 x 10 -20 J

1st Law of Thermodynamics

• 1st Law of Thermodynamics – energy is conserved

• The law of conservation of energy states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed.

1st Law of Thermodynamics

• Since energy can neither be gained nor lost, the change in E can be calculated using:

E = Ef – Ei

• In a chemical reaction i indicates reactants and f indicated products

E

E has 3 parts:1. A # indicating the magnitude

2. A sign (+/-) indicating the direction

3. A unit

Thermochemistry

• Thermochemistry is the study of heat changes that accompany chemical reactions and phase changes.

• In thermochemistry, the system is the specific part of the universe that contains the reaction or process you wish to study.

Thermochemistry

• Everything in the universe other than the system is considered the surroundings.

• Therefore, the universe is defined as the system plus the surroundings.

universe = system + surroundings

Relating E to heat & work

• The system can exchange energy with its surroundings in 2 ways: as heat or work

E = q + wE = change in energy

• q = heat

• w = work

q & w

• Don’t forget q & w must have signs

• In order to get the sign you must look at the system as a box and the surroundings as everything else

SystemSurroundings

q & w

• Anything going INTO the box will be +

• Anything going OUT of the box will be –

+-

q & w

• If heat is transferred from the surroundings to the system and work is done on the system what are the signs for q & w?

q = +

w = +

q & w

• If heat is lost to the surroundings and work is done on the system what are the signs for q & w?

q = -

w = +

Summary for q & w

• q + = heat into system

• q - = heat into surroundings

• w + = work done on the system

• w - = work done on the surroundings

Examples

• A system loses 1150 J of heat to the surroundings and does 480 J of work on the surroundings. Calculate E.

E = q + wE = (-1150J) + (-480J)E = -1630 J

Examples

• A system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings. Calculate E.

E = q + wE = (+ 140J) + (-85J)E = + 55 J

Endothermic & Exothermic

• Endothermic– system absorbs heat– Heat flows into the system– Temperature goes down

• Exothermic – Heat flows out of the system and into the

surroundings– Temperature goes up

• Only look at heat (q) to determine if the system is endo or exo

for example,

1. Ice melting at room temperature

2. Dissolving sugar in hot coffee

3. Na(s) + H2O(l) Na+(aq) + OH-

(aq) + H2(g)

Entropy

1st Law of Thermodynamics

• Energy is neither created nor destroyed– The energy of the universe is constant– Energy just changes from one form to another

• This law lets us know the energy of the system, but does not give us any information about the direction of the energy flow

Spontaneity

It tells us NOTHING about the speed of the reaction

Spontaneous process – a process that occurs

without intervention

Spontaneous processes can be fast or slow

Spontaneity tells us the direction of the energy

flow

For example…

• A ball spontaneously rolls down a hill– It does not spontaneously roll up

• If iron is exposed to air, it spontaneously rusts– The rust does not spontaneously turn back

into air & iron

Entropy

• Entropy (S) – the measure of molecular randomness or disorder– Think of entropy as the amount of chaos

• The driving force for a spontaneous process is an increase in entropy

Entropy

• The natural progression of things is from order to disorder

• Or from lower entropy to higher entropy– Think of a deck of cards…when you drop one it goes

from order to disorder– Think of your room… it goes from order to disorder

• Think if entropy as a probability…not a certainty– Nature spontaneously proceeds toward that states

that have the highest probability of existing

2nd Law of Thermodynamics

• 2nd Law of Thermodynamics – In any spontaneous process there is always an increase in entropy of the universe

• Energy is conserved…entropy is NOT conserved!• The entropy of the universe is always increasing

S univ = + = spontaneous S univ = - = not spontaneous (would be spontaneous in the

opposite direction) S univ = 0 = no tendency to occur (system is at equilibrium)

S univ = S sys + S surr

Temperature & Spontaneity

Ssurr = _ H T

T must be in Kelvin H is usually given in KJ/mol S will be in KJ/K, but is usually changed to

J/K

Example

Sb4O6 (s) + 6C (s) 4Sb (s) + 6CO (g) H = 778 kJ

Calculate Ssurr for this reaction at 25C and 1 atm

S = - (H/T)

S = - (778KJ/298K)

S = -2.61 KJ/K

S = -2610 J/K

3rd Law of Thermodynamics

• The entropy of a perfect crystal at 0K is zero

The Universe

The System

The Surroundings

Lecture - Entropy

Spontaneous Processes

spontaneous change - no action from outside the system is

necessary

the system

1. 2 Fe(l) + Al2O3(s) 2 Al(s) + Fe2O3(s)

2. Boiling water at 1 atm, 50oC

3. H2O(l) H2(g) + ½ O2(g) at 25oC

Non-spontaneous reactions...

What about the reverse reactions?

1. Water freezing at room temperature?

(ice melts)

2. Sugar precipitating out of hot coffee?

(sugar dissolves)

3. Na+(aq) + OH-

(aq) + H2(g) Na(s) + H2O(l) ?

(reverse reaction occurs)

Entropy

• Predict the sign of the entropy change for the following…

1. Sugar is added to water to form a solution

+

2. Iodine vapor condenses on a cold surface to produce a liquid

-

so why does this reaction work?

NH4Cl(s) + H2O(l) NH4+

(aq) + Cl-(aq)

Ho = + 15.1 kJ/mol

or this one?

H2O(l) H2O(g)

H = + 44 kJ/mol

C3H8(l) C3H8(g) H = +16.7 kJ/mol

l

gcold!

Heat vs. Energy

spontaneous change = heat released?

NOT ALWAYS!

spontaneous change = energy released?

ALWAYS!

Entropy, S

- a measure of the disorder of the system

Entropy tends to increase

breaking a glass

using the library

not cleaning up your house

1st law of thermodynamics:

The energy of the universe is constant

(The best you can do is break even)

2nd law of thermodynamics:

The entropy of the universe is increasing

(You can’t break even)

Low entropy is less probable

according to Boltzmann,S = k ln W

k = Boltzmann constant

= 1.38 x 10-23 J K-1

W = number of ways that state can be achieved

royal flush: 4 hands possible

S = 1.38 x 10-23 J K-1 ln(4)

= 1.9 x 10-23 J K-1

pair: 1,098,240 hands possible

S = 1.38 x 10-23 J K-1 ln(1098240)

1.9 x 10-22 J K-1

Entropy is a State Function

S = Sfinal - Sinitial

path taken is irrelevant

rate of change is irrelevant

S > 0 for:

- melting

- vaporizing

- making a solution

- a reaction that produces

an increased number of moles

- heating a substance

H2O(s) H2O(l)

ordered, low S

less ordered, high S

S > 0

S > 0 for:

- melting

- vaporizing

- making a solution

- a reaction that produces

an increased number of moles

- heating a substance

H2O(l) H2O(g)

high entropy

low entropy

S > 0 for:

- melting

- vaporizing

- making a solution

- a reaction that produces

an increased number of moles

- heating a substance

low entropy

high entropy

Very unlikely!

More likely!

Benzene Toluene

recall this one?

NH4Cl(s) + H2O(l) NH4+

(aq) + Cl-(aq)

H>0 (unfavorable)

S>0 (favorable!!)

S > 0 for:

- melting

- vaporizing

- making a solution

- a reaction that produces

an increased number of moles

- heating a substance

Ba(OH)28H2O(s) + 2 NH4NO3(s) Ba(NO3)2(aq) + 2 NH3(aq) + 10 H2O(l)

H = +80.3 kJ (unfavorable)

3 moles 13 moles

S > 0 (favorable)

S > 0 for:- melting

- vaporizing

- making a solution

- a reaction that produces

an increased number of moles

- heating a substance

Temperature

Ent

ropy

S

L

G

Sfusion

Svaporization

Calculating S for a reaction

Srxn =(i Soi)

absolute entropy of species i

stoichiometriccoefficient of species i

for example,

C8H18(g) + 12.5 O2(g)8 CO2(g) + 9 H2O(g)

13.5 moles 17 moles

(expect S > 0)

Srxn=(i Soi)

Entropy

• Predict which has the highest entropy

1. CO2 (s) or CO2 (g)CO2 (g)

2. 1 mol of N2 at 1 atm or 1 mol of N2 at 0.001 atm

1 mol of N2 at 0.001 atm

for example,

= 8 Soi(CO2(g)) + 9 So

i(H2O(g)) - So

i(C8H18(g)) - 12.5 Soi(O2(g))

= 8(213.6) + 9(188.6) - 463.2 - 12.5(204.8)

= +383.0 J K-1 mol-1

The Food Calorie

1 kcal = 1000 cal

1 Cal = 1 kcal

1 Cal = 4180 J

2,000 Cal/day is a meager diet.

what if you drank gasoline?

How much would you need?1. Assume gasoline is octane (C8H18(l))

2. Find Hcombustion

MW = 114.2 g/mol = 0.70 g/mL

How much would you need?

1. Assume gasoline is octane (C8H18(l))

2. Find Hcombustion

C8H18(l) + 12.5 O2(g) 8 CO2(g) + 9 H2O(g)

Hcombustion = 8(-393)+9(-285.6) - (-208.2)

= -5506.2 kJ/mol

How much would you need?

2000 Cal

dayx

1000 cal

Calx

4.18 J

calx

x1 mol gas

5506 x 103 Jx

1 mol gas

114.2 g gasx

0.7 g gas

1.0 mL gas

= 248 mL gas / day

0102030405060708090

100

1850 1900 1940 1980 1990

Wood

Coal

Petroleum/NaturalGas

Hydro andNuclear

U.S. Energy Usage

buried plant material

Coal

heat,pressure,time

empirical formula = CH2O MW 500,000

coal

Coal gets better with age

% C

Lignite 71

Subbituminous 77

Bituminous 80

Anthracite 92

increasing H

combustion

Burning coal

Coal + Air CO2(g) + H2O(g) + heat

+ CO(g) + SO2(g) + ash + NOx(g)

Other Energy Sources

1. Coal conversion

- convert to gaseous fuels

- large molecules small molecules

- must break C-C bonds

- C-H and C-O bonds are made

... or chemical2 CaBr2 + 4 H2O(l) 2 Ca(OH)2 + 4 HBr

H>0

2 Hg + 4 HBr 2 HgBr2 + 2 H2(g)

H<0

2 HgBr2 + 2 Ca(OH)2 2 CaBr2 + 2 HgO

+ 2 H2O(l) H>0

2 HgO 2 Hg + O2(g)

H>0

TiHx(s)Ti(s) + x/2 H2(g)

favored at highertemperature

favored at lowertemperature

Free Energy

• Free energy (G) – a thermodynamic function equal to the enthalpy minus the product of the entropy and the Kelvin temperature

G = H – TS

• A process is only spontaneous in the direction where G is negative

Example

• At what temperatures is the following process spontaneous at 1 atm?– Br2(l) Br2(g) H = 31.0 KJ/mol 31000 J/mol S = 93.0 J/ K mol

G = H – TS• 0 = 31000 – T(93.0)• T = 333K• Above 333K the reaction is spontaneous

Example

Sb2S3 (s) + 3Fe (s) 2Sb (s) + 3FeS (s) H = -125 kJ

Calculate Ssurr for this reaction at 25C and 1 atm

S = - (H/T)

S = - (-125KJ/298K)

S = 0.419 KJ/K

S = 419 J/K

Dependence of H & S on Spontaneity

H S Result

- +

+ +

- -

+ -

G = H-TS

Spontaneous at all temperatures

Spontaneous at high temperatures

Spontaneous at low temperatures

Not spontaneous at any temperature

Hydrocarbons and Heat

• Most hydrocarbons are used as fuels.

• For example, the amount of energy a food releases when burned, can tell us about it’s caloric content (fats release lots of energy).

• Knowing how much energy a fuel provides, can tell us if it is useful for a certain application.

• A “bomb calorimeter” is shown. It includes water in a heavily insulated container, a stirrer, valve, bomb chamber, ignition wires, & a thermometer

• Heat energy released during combustion can be measured with a calorimeter.

Exothermic and Endothermic changes

• An alternative to the bomb calorimeter is a “coffee cup” calorimeter, where two nested polystyrene cups take the place of the container

• In either case, the change in heat of the water tells us about the reaction of the chemicals.

• We will see that heat is measured in Joules (J) or kiloJoules (kJ). Before we do any heat calculations, you should know several terms …

• In an “endothermic” reaction, water temperature decreases as the chemicals absorb energy.

• An increase in water temperature indicates that the chemicals released energy when they reacted. This is called an “exothermic” reaction.

Specific heat capacity

The heat needed to the temperature of 1 g of a substance by 1 C. Symbol: c, units: J/(gC).+

The heat needed to the temperature of an object by 1 C. Symbol: C (=c x m), units: J/C

Heat capacity

The heat released during a chemical reaction per gram of reactant. Symbol: h, units: J/g.

Specific heat (of reaction)

The heat released during a chemical reaction. Symbol: none, units: J.

Heat of reaction

Molar heat of reaction

The heat released during a chemical reaction per mole of reactant. Symbol: H, units: J/mol.

Heat Calculations• To determine the amount of heat a substance

produces or absorbs we often use q = cmT• q: heat in J, c: specific heat capacity in J/(gC),

m: mass in g, T: temperature change in C,• This equation makes sense if you consider units

J = JgC

x g x C

For a list of c values see page 568 (table 3)Sample problem: (must know water = 4.18 J/gC)When 12 g of a food was burned in a calorimeter,

the 100 mL of water in the calorimeter changed from 20C to 33C. Calculate the heat released.

= 4.18 J/(gC) x 100 g x 13C = 5.4 kJq=cmT

More practice1. 5 g of copper was heated from 20C to 80C.

How much energy was used to heat the Cu?

2. If a 3.1 g ring is heated using 10.0 J, it’s temp. rises by 17.9C. Calculate the specific heat capacity of the ring. Is the ring pure gold?

q=cmT = 0.38 J/(gC) x 5 g x 60 C = 114 J

q=cmT10.0 J

3.1 g x 17.9C = 0.18 J/(g°C)c = q/mT=

The ring is not pure. Gold is 0.13 J/(gC) -pg. 568

5. q=cmT = 4.18 J/(gC) x 2570 g x 92 C= 988319 J = 988 kJ = 0.988 MJ

q=cmT1 750 000 J

4.18 J/(gC) x 12500 g= 33.5°CT = q/cm =

6.

Since the temperature started at 5.0°C, the final temperature is 38.5°C.

7. q=cmT = 0.86 J/(gC) x 2500 g x 335C= 720 kJ or 0.72 MJ

8. Total heat = water heat + pot heat = 4.18 J/(gC) x 1200 g x 53.0C = 265 848 J= 0.510 J/(gC) x 450 g x 53.0C = 12163.5 J

= 278 kJ

How much heat is required to change 50.0 g of ice at 0 degrees to steam at 100 degrees

Ice at 0 C

water at 0 C

water at 100 C

steam at 100 C

Hfmi + mwcwT + Hvmw

Hf for water = 334 j/g = 80 cal

Hv = 2257 j = 540 cal

You are looking for q

(334j)(50.0 g) g

+ (50.0g)(4.18 j)(100 C) g C

+ (2257 j) ( 50.0 g) g

16700 j + 20900 j + 112850 j

Heat Capacity Calculations• Recall that heat capacity (J/C) is different from specific heat capacity

(J/gC).• Heat capacity is sometimes a more useful value• For example, because a calorimeter includes wires, the stirrer,

thermometer, etc. some heat will be transferred to these other materials.• Rather than having to calculate q for each material a J/C value is used.

Sample problem:A calorimeter has a heat capacity of 2.05 kJ/C.

How much heat is released if the temperature change in the calorimeter is 11.6C?

x 11.6 C = 23.8 kJq = cm T q= 2.05 kJ/C

q=cmT

Thermochemical EquationsThermochemical equations are chemical

equations with an added heat term.• KBrO3(s) + 42 kJ KBrO3(aq)

This is endothermic (heat is absorbed/used)• 2 Mg(s) + O2(g) 2 MgO(s) + 1200 kJ

This is exothermic (heat is produced/released)Read 12.3 (pages 582 – 585). Do 2-6 (pg. 585).Sample problem:

3.00 g of octane was burned in a calorimeter with excess oxygen, the 1000 mL of water in the calorimeter rose from 23.0C to 57.6C. Write the thermochemical equation for octane, representing the molar heat of combustion.

2. Specific refers to mass in grams.3. Specific heat of reaction: J/g or kJ/g, etc.

Molar heat of reaction: J/mol or kJ/mol, etc.4. J/mol = J/g x g/mol (molar mass is used)5. Exothermic.

E.g. CH4 + 2O2 CO2 + 2H2O + x kJ

Other examples include propane, octane, Mg6.

49.90 kJg

= 1299 kJ/mol26.04 gmol

x

C2H2 + 2.5O2 2CO2 + H2O + 1299 kJor 2C2H2 + 5O2 4CO2 + 2H2O + 2598 kJ

Sample: First, calculate the heat released by the combustion reaction via q=cmT …

= 4.18 J/(gC) x 1000 g x 34.6 C = 144 628 J

h =

We can determine the molar heat of reaction 1) via the specific heat of reaction or 2) directly

Jg

144 628 J3.00 g

=

48209 J/g = 48.2 kJ/g= = 5508 kJ/mol

48.2 kJg

114.26 gmol

x

Octane (C8H18) has a molar mass of 114.26 g/mol

H =

CC88HH1818 + 12.5O + 12.5O22 8CO 8CO22 + 9H + 9H22O + 55O + 55008 kJ8 kJ

= 5508 kJ/molJmol

144 628 J0.0263 mol

=H = or

For more lessons, visit www.chalkbored.com

Entropy & Chemical Reactions

2nd Law of Thermodynamics

• A process will be spontaneous is the entropy of the universe increases

• Now we will look at entropy regarding to chemical reactions

What is the sign for S?

• N2(g) + 3H2(g) 2NH3(g)• First look at the states…if you go from a solid or

a liquid to a gas, you will have a + entropy• In this case, all of the states are the same, so

we look at the number of moles• N2(g) + 3H2(g) 2NH3(g)• 1 + 3 vs. 2• The entropy decreases because you go from 4

moles to 2 S is negative

What is the sign for S?

• 4NH3 + 5O2 4NO + 6H2O

• 9 moles vs. 10 molesS increases (+)

Calculating S

• Calculating S is just like calculating H

• Simply use the Appendix…just look at the column for S instead of H

S° of any element or diatomic molecule is NOT zero.

• You must look these up!

Example

• Calculate S for the following reaction:

• 2NiS(s) + 3O2(g) 2SO2(g) + 2NiO (s)

• 2(-53) + 3(-205) + 2(248) + 2(38)

• -149 J/K

Example

• Calculate S for the following reaction:

• Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)

• (-51) + 3(-131) + 2(28) + 3(189)

• 179 J/K

Gibbs Free Energy & Chemical Reactions

• You can calculate G in 3 ways…

1. Like Hess’s Law

2. Like H°

3. With the equation G = H - T S

Example

• Calculate H, S, & G at 25°C using the following data…

• 2SO2 + O2 2SO3

Substance H (KJ/mol) S (J/K mol)

SO2 -297 248

SO3 -396 257

O2 0 205

Example

2SO2 + O2 2SO3

H = 2(+297)+(0) + 2(-396) = -198 KJ

S = 2(-248)+ (-205) + 2(257) = -187J/K

G = H – T S

=(-198) – (298 x -0.187)

=-142 KJ

Calculate G

• Using the following data at 25°C• Cdiamond + O2(g) CO2 (g) G = -397KJ• Cgraphite + O2(g) CO2 (g) G = -394KJ• Calculate G for the reaction:• C diamond C graphite

G = -3KJ

Calculating G

• Methanol is a high octane fuel used in high performance racing engines. Calculate G for the following reaction

• 2 CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g)

• Given the following free energies of formation:

Substance G°(KJ/mol)

CH3OH (g) -163

O2 (g) 0

CO2 (g) -394

H2O (g) -229

Calculating G

• 2 CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g)

• 2(163) + 3(0) + 2(-394) + 4(-299)

• -1378 KJ

Hess’s law• Hess’s Law states that the heat of a whole

reaction is equivalent to the sum of it’s steps.• For example: C + O2 CO2 The book tells us that this can occur as 2 steps

C + ½O2 CO H = – 110.5 kJCO + ½O2 CO2 H = – 283.0 kJ

C + CO + O2 CO + CO2 H = – 393.5 kJ I.e. C + O2 CO2 H = – 393.5 kJ

• Hess’s law allows us to add equations.• We add all reactants, products, & H values.• We can also show how these steps add

together via an “enthalpy diagram” …

Steps in drawing enthalpy diagrams1. Balance the equation(s).2. Sketch a rough draft based on H values.3. Draw the overall chemical reaction as an

enthalpy diagram (with the reactants on one line, and the products on the other line).

4. Draw a reaction representing the intermediate step by placing the relevant reactants on a line.

5. Check arrows:Start: two leading awayFinish: two pointing to finishIntermediate: one to, one away

6. Look at equations to help complete balancing (all levels must have the same # of all atoms).

7. Add axes and H values.

C + O2 CO2 H = – 393.5 kJ

Reactants

Intermediate

Products

C + O2

CO2

CO

Ent

halp

y

Note: states such as (s) and (g) have been ignored to reduce clutter on these slides. You should include these in your work.

H = – 110.5 kJ

H = – 283.0 kJ

H = – 393.5 kJ

+ ½ O2

C + ½ O2 CO H = – 110.5 kJCO + ½ O2 CO2 H = – 283.0 kJ

Practice Exercise 6 with DiagramUsing example 5.6 as a model, try PE 6.

Draw the related enthalpy diagram.

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H= –1411.1 kJ

2CO2(g) + 3H2O(l) C2H5OH(l) + 3O2(g) H= +1367.1 kJ

C2H4(g) + H2O(l) C2H5OH(l)

Reactants Products

Intermediate

C2H4(g) + H2O(l)C2H5OH(l)

2CO2(g) + 3H2O(l)

Ent

halp

yH

=

– 14

11.1

kJ

H = +1367.1 kJ

H= – 44.0 kJ

+ 3O2(g)+ 3O2(g)

H= – 44.0 kJ

GeO(s) Ge(s) + ½ O2(g) H= + 255 kJGe(s) + O2(g) GeO2(s) H= – 534.7 kJGeO(s) + ½ O2(g) GeO2(s)

Reactants

Products

Intermediate

GeO(s) + ½ O2(g)

GeO2(s)

Ge(s) + O2(g)E

ntha

lpy

H = + 255 kJH

=

– 53

4.7

kJ

H= – 280 kJ

H= – 279.7 kJ

NO(g) ½ N2(g) + ½ O2(g) H= – 90.37 kJ

½ N2(g) + O2(g) NO2(g) H= + 33.8 kJNO(g) + ½ O2(g) NO2(g)

Reactants

Products

Intermediate

NO + ½ O2(g)

NO2(g)

½ N2(g) + O2(g)

Ent

halp

yH

=

– 90

.37

kJ

H = +33.8 kJ

H= – 56.6 kJ

H= – 56.57 kJ

Hess’s law: ExampleWe may need to manipulate equations further:

2Fe + 1.5O2 Fe2O3 H=?, given

Fe2O3 + 3CO 2Fe + 3CO2 H= – 26.74 kJ CO + ½ O2 CO2 H= – 282.96 kJ

1: Align equations based on reactants/products.2: Multiply based on final reaction.3: Add equations.

2Fe + 1.5O2 Fe2O3

3CO + 1.5 O2 3CO2 H= – 848.88 kJ

2Fe + 3CO2 Fe2O3 + 3CO H= + 26.74 kJ CO + ½ O2 CO2 H= – 282.96 kJ

Don’t forget to add states.

H= – 822.14 kJFor more lessons, visit www.chalkbored.com