5.4 enthalpy of reactions 5.5 calorimetry 5.6 hess’s … whether heat is absorbed endothermic or...

Chapter 5

THERMOTHERMOTHERMOTHERMO chemistry

5.4 Enthalpy of Reactions

5.5 Calorimetry

5.6 Hess’s Law

5.7 Enthalpies of Formation

Chemical Equations

1st WRITE the Chemical Equation

2nd BALANCE the Chemical Equation

3rd INTERPRET the Chemical Equation

Now when Interpreting Chemical

Equations must also include ENERGY

Involved in Chemical Processes

Physical “Process”

H2O(gas) →→→→ H2O(liquid) + HEAT

or Chemical “Process

CH4(g) + 2O2(g) ���� CO2(g) + 2H2O + HEAT

Chemical reactions can RELEASE

_______ when they occur

CH4(g) + 2 O2(g) � CO2(g) + 2 H2O + HEAT

or

Chemical reactions can ABSORB

________when they occur

2 C (s) + 2 H2 (g) + HEAT+ HEAT+ HEAT+ HEAT → C2H4(g)

predict whether heat is absorbed endothermic or

released exothermic by the system for :

(1) An ice cube melts _____________

(2) butane is burned ______________

Exo thermic and Endo thermic

Processes

_______thermic: transfers heat TO THE surroundings

An exothermic process feels HOT

_____thermic: absorbs heat FROM THE surroundings

An endothermic process feels cold

Two (2) Ways to express heat

1. Heat can be expressed as q

- or -

2. Expressed as H called Enthalpy

- or -

∆H Change in ________

Table 5.1 Sign Convention

EXO thermic

Heat is transferred FROM SYSTEM to the

surroundings: q > 0 ; ∆ H < 0

ENDO thermic

Heat is transferred FROM SURROUNDINGS

to the system: q < 0 ; ∆ H > 0

System & Surroundings

I. SYSTEMSYSTEMSYSTEMSYSTEM the portion of the universe that

is singled out for study

II. SURROUNDINGSSURROUNDINGSSURROUNDINGSSURROUNDINGS everything outside the

system

III. CONSERVATION OF ENERGY

________________________________

________________________________

Open , Closed, & Isolated

Systems

HEAT LOSTHEAT LOSTHEAT LOST = HEAT GAINHEAT GAINHEAT GAINHEAT GAIN

Something is gaining Heat

While Something else looses Heat.

IF YOU KNOW ONE OF THESE IF YOU KNOW ONE OF THESE IF YOU KNOW ONE OF THESE IF YOU KNOW ONE OF THESE

{Heat Lost - or - Heat Gained}

THEN YOU KNOW ____________THEN YOU KNOW ____________THEN YOU KNOW ____________THEN YOU KNOW ____________

THERMOTHERMOTHERMO chemistry

Two parts

I. EXPERIMENTAL

II. MATHEMATICAL

Energy Changes Involved With

Part 1. PHYSICAL Changes

a) Phase Changes

b) WITHIN a Phase

Part 2. CHEMICAL Changes

Part 1a

Energy Changes Involved Physical Changes

PHASE Changes

Gas ↔ Liquid ↔ Solid

Energy Expressed In Terms of qfor a Change Of State

H2O (solid) � H

2O (liq) - Energy

H2O(solid) � H

2O(liq) - 6.01 kJ/mole

The ___________Sign Means Heat is Needed

Energy Can Also Be Expressed As A

CHANGE In ENTHALPY (∆∆∆∆ H)

H2O (solid) � H

2O (liq) - Energy

H2O (solid) � H

2O (liq) ∆H = + 6.01 kJ/mole

The _________Sign Means Heat is Needed

!!!! NOTE THE SIGN CHANGE !!!!

Example 1: How much energy is needed to melt

[ H2O (solid) � H

2O (liq) +/- Heat ]

18 grams of H2O at 0 oC ?

[The Heat of fusion of H2O = 6.008 KJ / mole]

18 grams of water = 1.0 mole of water

therefore need _______KJ of heat

Example 2: What quantity of Heat is required to

vaporize 18 grams of H2O at 100 oC ?

H2O (liq) � H

2O (gas) +/- Heat

[The Heat of vaporization = 40.67 KJ / mole ]

18 grams of water = 1.0 mole of water

therefore need _______ KJ of heat

Part 1b

[NO CHANGE IN PHASE]

Energy Change WITHIN A State

Gas ↔ Gas

Liquid ↔ Liquid

Solid↔ Solid

Specific Heat

Specific Heat- The heat required to raise the

temperature of 1 gm of a substance by 1oC

EVERY SUBSTANCE HAS ITS OWN

UNIQUE SPECIFIC HEAT (SH)

T) ( (Grams)

JOULES 4.18 S.H. : Water

∆=For

UNIQUE SPECIFIC HEAT

change) re(temperatu x substance) of (grams

ansferredheat tr of quantity heat Specific =

How much heat energy is required to heat one

pound of water from 25 oC (room temp) to its

boiling point (100 oC) ?

LET THE UNITS SOLVE THE PROBLEM.LET THE UNITS SOLVE THE PROBLEM.LET THE UNITS SOLVE THE PROBLEM.LET THE UNITS SOLVE THE PROBLEM.

Energy = (Specific Heat)x(grams)x(change in Temp)

T) ( (Grams)

JOULES 4.18 S.H. : Water

∆=For

Energy required to heat one pound of

water from 25 to 100 oC ?

Heat Energy = 142,329 Joules

How many Significant Figures ?

Therefore answer is ?????? Joules

)( )(( ) )75(45418.4 Cg

Tg

JoulesJoules

o

∆=

Large beds of rocks are used in some

solar-heated homes to store heat

Calculate the quantity of heat absorbed

by 50.0 kg of granite if the temperature

increases by 12.0°C

[The specific heat of granite is 0.79 J/g-K]

How much heat is absorbed by 50.0 kg of

cement if the temperature increases 12.0°C

[The specific heat of cement is 0.88 J/g-K]

Let UNITS solve the problem

Joules = (Specific Heat)x(grams)x(change in Temp)

Joules = (0.88 J / g-K)(50.0x103 g)(12.0°C)

Joules = ________

How much heat is absorbed by 50.0 kg of

• Rocks ………………. 4.7 × 105 J

• Cement ……………... 5.3 × 105 J

• Water ……………… 25.1 × 105 J

• What statement can be made about specific heat in terms of substances absorbing or releasing heat?

Heat Flow

Heat spontaneously flows from a hot

object to a cold one until the temperature

of the two objects are the same

Every substance has its own unique specific heat

Use that information to

identify an unknown

1. Put into a Styrofoam cup

¾ full of water a piece of

hot metal

2. Measure temperature

change of water

A 2.61 gram block of

metal was heated to

100.0 oC and put into an

insulated cup containing

28.00 g of water. The

water temperature rose

from 25.00 to 26.48 oC

When the Hot Metal is dropped into the cup of

water HEAT Flows from the metal to the water

Heat Lost by Metal = Heat Gain by Water

Heat spontaneously flows from a hot object to a

cold one until the temperature of the two

objects are the same

Metal: 2.61 grams ; initial temp = 100oC

Water: 28.00 grams; initial temp = 25.00oC

final temp = 26.480C

1. What was the final temperature of the Metal ?

2. How much energy did the water gain ?

3. How much energy did the Metal lose ?

4. What is the specific heat of Metal ?

Heat GAINGAINGAINGAIN by Water

= 173 Joules {Heat Gained by Water

equals Heat LOSTLOSTLOSTLOST by Metal

Metal LOST 173 Joules

FOR THE METAL:

173 Joules = S.H x (2.61 g) x (100 – 26.48)

)( )(( ) )00.2548.26(0.2818.4 Cg

Tg

JoulesJoules

o−

∆=

WHAT WAS THE UNKNOWN METAL ?

)( )(????

48.2610061.2

173=

−=

g

JoulesatSpecificHe

Review of Part 1 Energy Changes Involved With

PHYSICAL Changes

a) Phase Changes

b) WITHIN a Phase

How much Energy required to heat 1.0

gram of ice at –10oC to steam at 110oC

Ice � Ice � Liq � Liq � Gas � Gas

-10oC 0oC 0oC 100oC 100oC 110oC

What information do you need to work this problem

Data Required for problem

• Heat of fusion = 6.008 kJ / mole

• Heat of vaporization = 40.67 kJ / mole

• Specific heat: Ice 2.092 J / g - K

• Liq 4.184 J / g – K

• Steam 1.841 J / g - K

•Energy required to heat 1.0 gram of ice at –10oC

to ice at 0oC ?

• Heat required to melt 1.0 gram of ice?

• Heat required to heat 1.0 gram of water at 0oC

to water at 100oC ?

• Heat required to vaporize 1.0 grams of H2O ?

• Heat required to heat 1.0 gram of steam at 100oC to

steam at 110oC ?

How much Energy required to heat 1.0

gram of ice at –10oC to steam at 110oC

Add all the numbers

21 J + 0.33 kJ + 418 J + 0.59 kJ + 18 J

How many significant figures in answer ?

Part 2

CHEMICAL REACTIONS

Reactants ���� Products +/-ENERGY

Chemical reactions can release or

absorb heat

Energy is “stored” in Chemical BONDS

It TAKES Energy To BREAK Bonds--------------------------------------------------------------------------------------------------------------------------------------------------------------------

Energy Is RELEASED When A Bond Is

Formed

• For a reaction:

Enthalpies of ReactionEnthalpies of Reaction

reactantsproducts

initialfinal

HH

HHH

−=

−=∆

HEAT OF REACTIONHEAT OF REACTIONHEAT OF REACTIONHEAT OF REACTIONHEAT OF REACTIONHEAT OF REACTIONHEAT OF REACTION

CH4 + 2 O2 � CO2 + 2 H2O + HEAT

The PLUS Sign Means Heat Is Given Off

OR

CH4 + 2 O2 � CO2 + 2 H2O ∆∆∆∆H = - #

The MINUS Sign Means Heat Is Given Off

DETERMINATION

of HEATS of REACTIONS

1. THE DIRECT METHOD

EXPERIMENTAL

Use A Calorimeter

2. THE INDIRECT METHOD

MATHEMATICAL

Use HESS’S Law

1. The Direct Method

EXPERIMENTAL

Go to Lab

and Use A Calorimeter

HEAT LOST = HEAT GAINEDHEAT LOST = HEAT GAINEDHEAT LOST = HEAT GAINEDHEAT LOST = HEAT GAINED

Two (2) types of

CALORIMETERS

1. OPEN { to the atmosphere

and

2. CLOSED {to the atmosphere

OPEN CALORIMETER

Also called a

CONSTANT PRESSURE

CALORIMETER

Styrofoam cup

100 mL of 0.5 M HCl added to

100 mL of 0.5 M NaOH

Data Collected from experiment

Volume of 0.500 M NaOH(aq) = 1.00 x 102 mL

Volume of 0.500 M HCl(aq) = 1.00 x 102 mL

Initial Temperature = 22.50 oC

Final Temperature = 25.92 oC

Detemine the heat of the reaction PER MOLE

(the Heat of neutralization)

What GIVES OFF THE HEAT ?

The reaction !

1 HCl(aq) + 1 NaOH (aq) � H2O + NaCl (aq) + HEAT

What ABSORBS THE HEAT ?

The “solution”

How do you determine the Heat of the reaction?

From the HEAT GAINED by the “SOLUTION”

Joules = (Specific Heat) x (grams) x (Temp Change)

of the solution

from density

from experiment

Number of GRAMS = ?

100 mL HCl (aq) + 100 mL NaOH (aq)

1. 100 mL of each = _____ mL total

2. Relationship between weight & volume ?

3. DENSITY of solution

4. If Density = 1.00 g / mL

5. Have _____ grams of solution

Let UNITS solve the problem

Joules = (Specific Heat) x (grams) x (change in T)

change in Temp = ?

grams = 200 grams of solution

Specific Heat =

Joules = (Specific Heat) x (grams) x (change in T)

Joules = (4.184)(200)(3.42) =

change x tempgram

Joules 18.4

Joules = (4.184)(200)(3.42) = 2.866 kJ

1 HCl(aq) + 1 NaOH (aq) � H2O + NaCl (aq) + Heat

2.8661 kJ of heat given off when

1.00 x 102 mL of 0.500 M HCl (aq) is mixed with

1.00 x 102 mL of 0.500 M NaOH (aq)

How much heat given off per mole of water formed ?

heat of neutralization per mole ?

How many moles of water formed ?

1.00 x 102 mL of both 0.500 M of HCl(aq) & NaOH

Moles = Molarity x Volume = 0.0500

0.05 H+ (aq) + 0.05 OH- (aq) � 0.05 H2O + 2.866 kJ

1.00 H+ (aq) + 1.00 OH- (aq) � 1.00 H2O + _______

Calorimeter[Used For Gas Reactions]

CONSTANT VOLUME CALORIMETER

It is called a “BOMB” CALORIMETER

Because the Chemical Reaction Occurs

in a CLOSED Container

See Text Fig 5.19 Page 186

Chemical Reaction takes place in “Bomb”

• HEAT IS GIVEN off by Reaction (SYSTEM)

• HEAT IS ABSORBED by SURROUNDINGS

Surroundings are

– 1. WATER In Calorimeter

– 2. Everything else {thermometer, stirrer,

metal bomb itself, etc

Example 1 2.431 grams Of Mg Was BurnedIn a Constant Volume Calorimeter

Write & Balance the COMBUSTION

Reaction

__Mg + __ O2 � __ MgO + HEAT

Heat Lost by = Heat Gain by

Chemical 1. Water +

Reaction 2. Calorimeter

Data Collected from experiment

The Calorimeter had a Heat Capacity = 1769 J/ 0C

Calorimeter Contained 3.00x102 mL of Water

Initial Temperature = 22.5 oC

Final Temperature = 42.4 oC

1. Heat Gain by Water = S. H. x grams x Temp

qwater = (4.184 x 300 x 19.9 )

2. Heat Gain by Calorimeter = Heat Cap x Temp Change

qCalorimeter = (1769 x 19.9 )

Total Heat Gained = Water + Calorimeter

= (1769 x 19.9 ) + (4.184 x 300 x 19.9 )

= (35,203.1) + (24,978.48) Joules

= ______ Joules for 2.431 grams of Mg

HEAT LOST = HEAT GAIN

How do you convert from grams to moles?

Record Data

Record results in Tabular Form

Heats of Combustion (- ∆ H)per mole in kJ at 25oC

• HC2H3O2 (aq) { Acetic Acid (aq)

• CH4 (g) { Methane

• C2H6 (g)

• C6H6 (liq)

• C6H12 (liq)

• Sucrose (s)

874.5

890.4

1541.4

3277.7

3928.8

5690.2

Enthalpy is an EXTENSIVE property

(∆H is directly proportional to amount):

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(liq) ∆ H = - 890 kJ

2 CH4(g) + 4 O2(g) → 2 CO2(g) + 4 H2O(liq) ∆ H = -1780 kJ

Phase (solid, liquid, gas) is Important

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(gas) ∆ H = - 802 kJ

4 CH4(g) + 8 O2(g) → 4 CO2(g) + 8 H2O(g) ∆ H = -3208 kJ

Note the state of water

Next, DETERMINATION OFHEATS OF REACTIONS

Using

2. THE INDIRECT METHOD

MATHEMATICAL

Use Of HESS’S Law

I. COMBUSTION of hydrocarbons

CH4 (g) + 2 O2 (g) � CO2 (g) + 2 H2O (g)

II. FORMATION Reactions

e.g., CH4 (Methane)

C(s) + 2 H2(g) � CH4 (g)

Standard State

• The Standard State of a substance is the state the pure substance is in at atmospheric pressure ( 1 atm) and 25 oC

• The standard state of carbon is graphiteand not diamond

• The standard state of hydrogen is H2 not H

STANDARD ENTHALPIES

OF FORMATION

For example H2O (liq) is formed from its

elements as they exist in nature

H2 (gas) + ½ O2 (gas)���� H2O (liq) ∆ H =-285.8 kJ

STANDARD ENTHALPIES

OF FORMATION

THE STANDARD ENTHALPY OF

FORMATION (∆Hf ) OF ANY ELEMENT

IN ITS MOST STABLE FORM IS ZERO

(BY DEFINATION)

HESS’S LAW

ENTHALPY CHANGES ARE ADDITIVE

Example 1 Calculate [using Hess’ Law] the heat

of reaction for CO(g) + ½ O2(g) → CO

2(g)

What DATA Do You Need From Table ?

∆Hf From Table

WRITE AND BALANCE REACTIONS

Formation of CO (g) is :

1. C(s)

+ ½ O2(g) → CO(g) ∆H = - 110.5 kJ

Formation of CO2(g) is :

2. C(s)

+ O2(g) → CO

2(g) ∆H = - 393.5 kJ

Want CO (g) + ½ O2(g) → CO2(g)

1. C(s)

+ ½ O2(g) → CO(g) ∆H = - 110.5 kJ

REWRITE Eq 1

1b. CO (g) → C(s)

+ ½ O2(g) ∆H = + 110.5 kJ

also

2. C(s)

+ O2(g) → CO

2(g) ∆H = - 393.5 kJ

CO (g) → C(s)

+ ½ O2(g) ∆H = + 110.5 kJ

C(s)

+ O2(g) → CO

2(g) ∆H = - 393.5 kJ

Add Equations To Get Reaction Wanted:

CO (g) + ½ O2(g) → CO2(g)

Add ∆H ‘s To Get :

∆H = + 110.5 kJ - 393.5 kJ = ________

Example 2 Find the Enthalpy of Formation of C3H8(g)

Given

Enthalpy of combustion of C3H8 (g) = -2043 kJ

Enthalpy of formation of CO2 (g) = -393.5 kJ

Enthalpy of formation of H2O(g) = -241.8 kJ

Write and balance the following reactions

Combustion of one mole of C3H8 (g)

Formation of one mole of CO2 (g)

Formation of one mole of H2O(g)

Formation of one mole of C3H8 (g)

3 C (s) +4 H2(g) � C3H8 (g) ∆H = ???

Example 3 Find the Enthalpy of Formation of C3H8(g)

From

Enthalpy of combustion of C3H8 (g) = -2043 kJ

Enthalpy of formation of CO2 (g) = -393.5 kJ

Enthalpy of formation of H2O(g) = -241.8 kJ

answer

3 C (s) +4 H2(g) � C3H8 (g) ∆H = - 104.7 kJ

Example 4 Find the Heat of Vaporization of Water

From the following HEATS of COMBUSTION

C3H8(g) + 5O2(g) → 3CO2 + 4H2O(g) ∆H = - 2043 kJ

C3H8(g) + 5O2(g) → 3CO2 + 4H2O(l) ∆H = - 2219kJ

NOTE:

These reactions have Nothing

to do with the vaporization of water

Want ∆ H for H2O (liquid) → H2O (gas)

C3H8(g) + 5O2(g) → 3CO2 + 4H2O(g) ∆H = - 2043 kJ

C3H8(g) + 5O2(g) → 3CO2 + 4H2O(l) ∆H = - 2219kJ

Rewrite last equation to get

3CO2 + 4H2O(l) → C3H8(g) + 5O2(g) ∆H = + 2219 kJ

WHY ?

Want ∆ H for H2O (liquid) → H2O (gas)

C3H8(g) + 5O2(g) → 3CO2 + 4H2O(g) ∆H = - 2043 kJ

3CO2 + 4H2O(l) → C3H8(g) + 5O2(g) ∆H = + 2219 kJ

Add Equations getting

4 H2O (liquid) → 4 H2O (gas) ∆H = + 176 kJ

H2O (liquid) → H2O (gas) ∆ H = ____________

Given the data

N2(g) + O2(g) � 2 NO(g) ∆H = +180.7 kJ

2 NO(g) + O2(g) � 2 NO2(g) ∆H = - 113.1 kJ

2 N2O(g) � 2N2(g) + O2(g) ∆H = - 163.2 kJ

Use Hess’s law to calculate ∆H

For the reaction

N2O(g) + NO2(g) � 3 NO(g)

Example 4 Find ∆H for N2O(g) + NO2(g) � 3NO(g)

Given N2(g) + O2(g) � 2 NO(g) ∆H = +180.7 kJ 2 NO(g) + O2(g) � 2 NO2(g) ∆H = - 113.1 kJ 2 N2O(g) � 2N2(g) + O2(g) ∆H = - 163.2 kJ

----------------------------------

Add The Following Equations

N2(g) + O2(g) � 2 NO(g) ∆H = +180.7 kJ

NO2(g) � NO(g) + ½ O2(g) ∆H = ½ [+ 113.1 kJ]

N2O(g) � N2(g) + ½ O2(g) ∆H = ½ [- 163.2 kJ]

Energy From

Foods

Energy in our bodies comes from fats and

carbohydrates (mostly)

Carbohydrates converted into glucose, then:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

∆H = - 2816 kJ

Fats: contain more energy; are not water

soluble, so are good for energy storage.

2 C57H110O6 + 163 O2 → 114 CO2 + 110 H2O

∆ H = - 75,520 kJ

Which releases the greatest amount of energy

per gram upon metabolism

(a) carbohydrates (b) proteins (c) fats

Energy From

Fuels

Which releases the greatest amount of energy per

gram upon combustion

(a) Methane (b) gasoline (c) hydrogen