Chapter 5
1

• Thermochemistry :
Relationships between chemical reactions and energy changes
• Thermodynamics :
• Therm : heat Dynamics : power
2

Energy: t he capacity to do work
• Radiant energy
• Thermal energy
• Chemical energy
• Nuclear energy
• Potential energyrelative to position- stored in bonds electrostatic potential energy (C) = kQ
1
Q
2 d
• Kinetic energy Ek = ½ mv 2
• Units of Energy: J = 1kg-m 2 /s 2 1 cal= 4.184 J
3

Energy Changes in Chemical Reactions
Which one has more thermal energy?
90 0 C
Coffee cup?
90 0 C
Bathtub?
4

Energy Changes in Chemical Reactions
Which one has more thermal energy?
90 0 C
What additional information is needed?
How does it relate to heat?
40 0 C
5

Heat: the transfer of _______ between two bodies that are at ________temperatures.
Temperature is a measure of the ________ .
Measure of the ______________
Temperature = Thermal Energy
6

The system is the specific part of the universe that is of interest in the study.
Exchange: open mass & energy closed energy isolated nothing
7

8

Exothermic process : is any process that gives off heat – transfers thermal energy from the system to the surroundings.
2H
2
( g ) + O
2
( g ) 2H
2
O ( l ) + energy
H
2
O ( g ) H
2
O ( l ) + energy
Endothermic process is any process in which heat is supplied to the system from the surroundings. energy + 2HgO ( s ) energy + H
2
O ( s )
2Hg ( l ) + O
2
( g )
H
2
O ( l )
9

• Enthalpy : a thermodynamic quantity used to describe heat changes(absorbed/released) at constant pressure (most reactions occur at constant pressure).
H = E + PV
Equation:
ΔH reaction
= ΔH products
ΔH reactants
10

D
H = H
(products)
– H
(reactants)
D
H = heat given off or absorbed during a reaction at constant pressure
H products
< H
D
H < 0 reactants
H products
> H
D
H > 0 reactants
11

Enthalpy: Enthalpy Diagrams
ΔH rxn
= ΔH products
– ΔH reactants
12

Indicate the sign of enthalpy change in the following processes carried under atmospheric pressure, and indicate whether the process is exothermic or endothermic:
1.
An ice cube melts
2.
1 g of butane (C
4
H
10
) is combusted in sufficient oxygen to give complete combustion to CO
2 and
H
2
O
3.
A bowling ball is dropped from a height of 8 ft into a bucket of sand.
13

Is
D
H negative or positive?
System absorbs heat
Endothermic
D
H > 0
6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm.
H
2
O ( s )
→ H
2
O ( l )
D
H = 6.01 kJ
14
6.3

Is
D
H negative or positive?
System gives off heat
Exothermic
D
H < 0
890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm.
CH
4
( g ) + 2O
2
( g)
→ CO
2
( g) + 2H
2
O ( l )
D
H = -890.4 kJ
15
6.3

Stoichiometric coefficients : equal the number of moles of a substance
H
2
O ( s ) H
2
O ( l )
D
H = 6.01 kJ
If you reverse a reaction, the sign of
D
H changes
H
2
O ( l ) H
2
O ( s )
D
H = 6.01
kJ
If you multiply both sides of the equation by a factor n , then
D
H must change by the same factor n .
2H
2
O ( s ) 2H
2
O ( l )
D
H = 12.0 kJ
2 x 6.01kJ/mol
=12.0 KJ/2 mol
16

Thermochemical Equations: Rules
When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes.
17

The physical states of all reactants and products must be specified in thermochemical equations.
H
2
O ( s ) → H
2
O ( l ) ΔH = 6.00 kJ
H
2
O (l) → H
2
O (g) ΔH = 44.0 kJ
• What will be the ΔH when 1 mole of ice at
0ºC is changed into one mole of steam at the boiling point?
18

How much heat is evolved when 266 g of white phosphorus (P
4
) burn in air?
P
4
( s ) + 5O
2
( g ) P
4
O
10
( s )
D
H = -3013 kJ
266 g P
4 x
1 mol P
4
123.9 g P
4 x
-3013 kJ
1 mol P
4
[ -6470 kJ]
19

4. The complete combustion of liquid octane, C
8
H
18
, to produce carbon dioxide and liquid water at 25 ºC and at constant pressure gives off 47.9 kJ per gram of octane. Write a chemical equation to represent this information.
[2C
8
H
18
(l) + 25 O
2
(g) → 16 CO
2
(g) + 9H
2
O(l)
ΔH = -1.09 x 10 4 kJ]
20

5.
Given the thermochemical equation:
SO
2
(g) + ½ O
2
(g) →SO
3
(g) ΔH = -99.1 kCal
Calculate the heat evolved when 74.6 g of SO
2
(MM = 64.07 g/mol) is converted to SO
3
.
[ -115 kJ]
6.
Hydrogen peroxide can decompose to water and oxygen by the following reaction:
2H
2
O
2
(l) → 2H
2
O(l) + O
2
(g) ΔH = -196 kJ
Calculate the value of q (heat evolved or absorbed) when
5.00 g of H
2
O(l) decomposes at constant pressure.
[ -14.4 kJ]
21

7.
Given the equation
H
2
(g) + I
2
(s) → 2HI(g) ΔH = +52.96 KJ
Calculate ΔH for the reaction
HI(g) → ½ H
2
(g) + ½ I
2
(s)
[ -26.48 kJ ]
8.
Given the equation:
3O
2
(g) → 2O
3
(g) ΔH = +285.4 kJ
Calculate ΔH for the following reaction:
3/2 O
2
(g) → O
3
(g)
[ 142.7 kJ]
22

• Read Section 6.4 (Pages 213-2200)
• Examples: pp. 216 – 220: 6.1 – 6.3
• Exercises: pp. 240: 6.19, 6.23, 6.26, 6.28
23

Calorimetry is performed in devices called calorimeters
Calorimetry is based on the
Law of
Conservation of
Energy
24

9.
The specific heat of aluminum is 0.895 J/g
º
C.
Calculate the heat necessary to raise the temperature of 40.0 g of aluminum from -20.0
º
C to 32.3
º
C. Specific heat of aluminum is
0.215-cal/g
º
C.
[1.87 x 10 3 Joules]
25

• Calorimetry :
Experimental measurement of heat produced in chemical and physical processes
• Heat Capacity, C, of a system: the quantity of heat needed to raise the temperature of matter 1 K.
•
•
Units: J/ºC C
 q
D
T
Molar Heat Capacity :
The heat required to raise 1 mole of a substance, 1 K.
26

• Specific heat of a substance
(also represented by c p
): the amount of heat ( q ) required to raise the temperature of one gram of the substance by one degree Celsius ( or 1 K).
Specific heat
 heat capacity mass

C m but C
 q
D
T
Specific h eat : c p

C m
 q
D
T
(
1 m
)
 q m
D
T q
 mc p
D
T
27

Heat ( q ) absorbed or released: q = mc p
D t q = C
D t
D t = t final
t initial
When q>0 endothermic reaction
When q<0 exothermic reaction
28

10.
How much heat, in joules and kilojoules, does it take to raise the temperature of 225 g of water from
25.0 ºC to 100.0 ºC
[ 7.05 x 10 4 J; 70.5 kJ]
11. Calculate the heat capacity, specific heat and the molar heat capacity of an aluminum block that has mass of 5.7 g that must absorb 629 J of heat from its surroundings for its temperature to rise from 22
ºC to 145 ºC.
[C = 5.11 J/ºC; c p
= 0.90 J g -1 ºC -1 , 24 J mol -1 ºC -1 ]
29

12. What will be the final temperature if a 5.00-g silver ring at 37.0 ºC gives off 25.0 J of heat to the surroundings? Specific heat of silver = 0.235
J g -1 ºC -1
[15.7
ºC]
13. A 15.5 g sample of a metallic alloy is heated to
98.9 ºC and then dropped into 25.0 g of water in a calorimeter. The temperature of the water rises from 22.5 ºC to 25.7 ºC. Calculate the specific heat of the alloy.
[0.29 J g -1 ºC -1 ]
30

14. Suppose a piece of gold with a mass of 21.5 g at temperature of 95.00
º
C is dropped into an insulated calorimeter containing 125.0 g of water at 22.00 º C.
What will be the final temperature of the water?
[22.4
º
C]
31

32

Isolated system:
No exchange of matter and energy
Uses Cp of water
33

Constant-Volume Calorimetry: Isolated system q sys
= q water
+ q bomb
+ q rxn
= 0
There is no heat exchange with the surroundings q rxn
= - ( q water
+ q bomb
)
C bomb
= m bomb x c bomb q bomb
= C bomb
D t q water
= m w c p
Δ t q rxn =
-(q water
+ C bomb
Δ t) = -(m w c p
Δt + C bomb
Δt) q rxn
= - (m w c p
+ C bomb
) Δt
Measured heat is not enthalpy: constant volume 34

Constant Volume Calorimetry: Problems
15. In a preliminary experiment, the heat capacity of a bomb calorimeter assembly is found to be 5.15 kJ/ °C. In a second experiment. A 0.480 g sample of graphite
(carbon) is placed in the bomb with an excess of oxygen. The water, bomb, and other contents of the calorimeter are in thermal equilibrium at 25.00 °C. The graphite is ignited and burned, and the water temperature rises to 28.05 °C. Calculate ΔH for the reaction.
C(graphite) + O
2
(g) → CO
2
(g)
[ ΔH = -393 kJ]
35

Measuring Enthalpy Changes for
Chemical Reactions
Coffee-Cup Calorimeter:
Constant pressure
36

Constant-Pressure Calorimetry q sys
= q water
+ q cal
+ q rxn q sys
= 0 q rxn
= - ( q water
+ q calor
) q water
= m c p
D t q cal
= C cal
D t
Reaction at Constant P
No heat enters or leaves!
Insulated system!
Pressure is constant.
Δ H = q rxn
= - ( q water
+ q calor
)
Measured heat is the enthalpy
37

38

Constant Pressure Calorimetry: Problems
16. (a) A 50.0 mL sample of 0.250 M HCl at 19.50 °C is added to 50.0 mL of 0.250 M NaOH, also at 19.50 °C, in a calorimeter. After mixing, the solution temperature rises to 21.21 °C. Calculate the heat of this reaction.
Assume te volumes of the solutions are additive. Ignore the heat absorbed by the calorimeter.
[-715 J]
(b) Determine the value of ΔH that should be written in the equation for the neutralization reaction:
HCl(aq) + NaOH(aq) → NaCl(aq) + H
2
O(l) ΔH = ?
[-57.2 kJ]
39

17. A quantity of 1.00 x 10 2 mL of 0.500M HCl is mixed with 1.00 x 10 2 mL of 0.500 M NaOH in a constant pressure calorimeter having a heat capacity of 335 J/ ºC. The initial temperature of the
HCl and NaOH solutions is the same, 22.50 ºC, and the final temperature of the mixed solution is 24.90
ºC. Calculate the heat change for the neutralization reaction and the molar heat of the neutralization reaction. Assume s = 4.18 J/gºC for the solution.
[-2.81 kJ; -56.2 kJ/mol]
40

Chemistry in Action :
Fuel Values of Foods and Other Substances
C
6
H
12
O
6
( s ) + 6O
2
( g ) 6CO
2
( g ) + 6H
2
O ( l )
D
H = -2801 kJ/mol
1 cal = 4.184 J
One food Calorie:
1 Cal = 1000 cal = 4184 J
41

Standard Enthalpy of Formation
Absolute value of the enthalpy of a substance cannot be measured. We need a frame of reference.
Standard states for elements and compounds has to be established.
Standard state for a solid or liquid substance is the pure element or compound at 1 atm at a given temperature.
Standard state for gas : pure gas behaving as an ideal gas at 1 atm and temperature of interest.
Standard enthalpy of formation : the enthalpy change for a reaction in which the reactants in their standard state yield products in their standard states. Standard state denoted with a superscript ( º)
42

Standard enthalpy of formation ΔHº f formation
(also called heat of enthalpy change that occurs in the formation of 1 mole of the substance from its elements when both products and reactants are in their standard states (f stands for formation).
Reference forms: standard enthalpy of formation of a pure element in its reference form is 0.
standard enthalpy of formation ( ΔH
0
) as a reference point for all enthalpy expressions is the formation of H
= 0 kJ/mol
2
(g)
43

Standard enthalpy of formation (
D
H 0 f
) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm and 25 ºC .
The standard enthalpy of formation of any element in its most stable form is zero .
D
H 0 f
(O
2
) = 0
D
H 0 f
(O
3
) = 142 kJ/mol
H 0 f
(C, graphite) = 0
D
H 0 (C, diamond) = 1.90 kJ/mol f
D
44

standard
45

Determination of Heat of Formation
1. Direct method : From elements when it can be easily measured.
C(graphite) + O
2
(g) →CO
2
(g) ΔH rxn
= -393.5 kJ
ΔHrxn= ΔH o
CO2
ΔH 0
CO2
– ΔH
= -393.5 kJ
0O
2
– ΔH graphite
= -393.5 kJ
2. Indirect method: Using the heat of reaction and heats of formations
Given CH
4
(g) +2O
2
(g) → CO
2
(g) + 2H
2
O(g) ΔHrxn=-890.3 kJ
Calculate the heat of formation of CH
4
[-74.8kJ/mol]
46

The standard enthalpy of reaction (
D
H 0 rxn
) is the enthalpy of a reaction carried out at 1 atm and 25 ºC.
a A + b B c C + d D
D
H 0 rxn
= [ c
D
H 0 f
(C) + d
D
H 0 f
(D) ] [ a
D
H 0 f
(A) + b
D
H 0 f
(B) ]
D
H 0 rxn
=
S n
D
H 0 f
(products) -
S m
D
H f
0 (reactants)
47

Benzene (C
6
H
6
) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.
2C
6
H
6
( l ) + 15O
2
( g ) 12CO
2
( g ) + 6H
2
O ( l )
D
H 0 rxn
=
S n
D
H 0 f
(products) -
S m
D
H f
0 (reactants)
D
H 0 rxn
= [ 12
D
H f
0 (CO
2
) + 6
D
H 0 f
(H
2
O) ] [ 2
D
H f
0 (C
6
H
6
) ]
D
H 0 rxn
= [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ
-5946 kJ
2 mol
= - 2973 kJ/mol C
6
H
6
48

Standard Enthalpy of Reaction
18. Synthesis gas is a mixture of carbon monoxide and hydrogen that is used to synthesize a variety of organic compounds, such as methanol. One reaction for producing synthesis gas is shown here:
3CH
4
(g) + 2H
2
O(l) + CO
ΔHº = ?
2
(g) → 4CO(g) + 8H
2
(g)
[747.5 kJ]
19. The combustion of isopropyl alcohol, common rubbing alcohol, is represented by this equation:
2(CH
3
)
2
CHOH(l) + 9O
2
(g) → 6CO
2
(g) + 8H
2
O(l)
ΔHº = -4011 kJ use this equation and data from Thermodynamic tables to establish the standard enthalpy of formation for isopropyl alcohol.
[ -318 kJ/mol] 49

Heat of Formation: Examples
20. Pentaborane-9, B
5
H
9
, is a colorless liquid. It is highly reactive substance that will burst into flame or even explode when exposed to oxygen:
2B
5
H
9
(l) + 12O
2
→ 5B
2
O
3
(s) + 9H
2
O(l)
It was considered a potential rocket fuel as it release large amount of heat per gram. Calculate the kJ of heat released per gram of the compound reacted with oxygen.
The standard enthalpy of formation of B
5
H
9 is 73.2 kJ/mol
[ -71.58 kJ/gB
5
H
9]
50

Heat of Reactions (Formations):
Hess’s Law of Constant heat of Summations
Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)
51

Hess’ Law: Example
Given the following reactions:
CO(g) + ½ O
2
(g) → CO
2
(g) ΔH = -283.0 kJ
C
(graphite)
+ O
2
(g) → CO
2
(g) ΔH = -393.5 kJ
Calculate the enthalpy for the formation of CO from its elements
C(graphite) + ½ O2(g) → CO(g) ΔH = ?
52

53

Calculate the standard enthalpy of formation of CS
2 given that:
C (graphite) + O
2
( g ) CO
2
( g )
D
H 0 rxn
= -393.5 kJ
S (rhombic) + O
2
( g ) SO
2
( g )
D
H 0 rxn
= -296.1 kJ
CS
2
( l ) + 3O
2
( g ) CO
2
( g ) + 2SO
2
( g )
( l )
D
H 0 rxn
= -1072 kJ
1. Write the enthalpy of formation reaction for CS
2
C (graphite) + 2S (rhombic) CS
2
( l )
2. Add the given rxns so that the result is the desired rxn.
+
6.6
C (graphite) + O
2
( g )
2S (rhombic) + 2O
2
CO
2
( g ) + 2SO
2
( g )
( g )
CO
2
( g )
2SO
2
( g )
D
H 0 rxn
= -393.5 kJ
D
H 0 rxn
= -296.1
x2 kJ
CS
2
( l ) + 3O
2
( g )
D
H 0 rxn
= +1072 kJ
C (graphite) + 2S (rhombic) CS
2
( l )
D
H 0 rxn
= -393.5 + 2(-296.1) + 1072 = 86.3 kJ 54

21. Given
C(graphite) + O
2
(g) → CO
2
(g) ΔHº rxn
= -393.5 kJ
H
2
(g) + ½ O
2
(g) → H
2
O(l) ΔHº rxn
2C
2
H
2
(g) + 5O
2
(g) → CO
2
(g) + 2H
2
O(l) ΔHº rxn
= -285.8 kJ
= -2598.8 kJ
Calculate the heat of formation of C
2
H
2
(g)
2C(graphite) + H
2
(g) → C
2
H
2
(g)
[52.3 kJ]
55

Hess’ Law: Conceptual Example
Without performing a calculation, determine which of the two substances should yield the greater quantity of heat upon
Complete combustion, on a per mole basis: ethane, C
2
H
6
(g), or ethanol, CH
3
CH
2
OH(l)
56

The Solution Process for NaCl
D
H soln
= Step 1 + Step 2 = 788kJ – 784kJ = 4 kJ/mol 57

The enthalpy of solution (
D
H soln
) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.
D
H soln
= H final soln
H components
Which substance(s) could be used for melting ice?
Which substance(s) could be used for a cold pack?
Which substances could be used for hot pack?
58

Heat of Solution and Dilution
The process: NaCl(s) → Na + (aq) + Cl (aq)
The steps:
1. Separate ions in the crystal into ions in gaseous state
2. Hydration of the gaseous ions
Energy involved:
Lattice Energy, U
Hydration energy, ΔH hydration
ΔH solution= U + ΔH hydration
59

Lattice energy : energy required to completely separate one mole of solid ionic compound into gaseous ions (endothermic process)
U + NaCl(s) →Na + (g) + Cl (g)
Hydration of the gaseous ions
Heat of hydration: energy released when ions interact with water molecules.
Na + (g) + Cl (g) → Na + (aq) + Cl (aq) + Energy
60

The heat change associated with the dilution of a solution.
Can be exothermic or endothermic
61

62