Chapter 5 THERMO THERMO THERMO THERMO chemistry 5.4 Enthalpy of Reactions 5.5 Calorimetry 5.6 Hess’s Law 5.7 Enthalpies of Formation
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