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Chapter 11 - ThermochemistryHeat and Chemical Change
Milbank High School
2
Section 11.1The Flow of Energy - Heat
OBJECTIVES:
• Explain the relationship between energy and heat.
3
Section 11.1The Flow of Energy - Heat
OBJECTIVES:
• Distinguish between heat capacity and specific heat.
4
Energy and Heat Thermochemistry - concerned with
heat changes that occur during chemical reactions
Energy - capacity for doing work or supplying heat
5
Energy and Heat Heat - represented by “q”, is energy
that transfers from one object to another, because of a temperature difference between them.
Exothermic and Endothermic Processes
In studying heat changes, think of defining these two parts:
• the system
• the surroundings
Exothermic and Endothermic Processes
The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed.
• All the energy is accounted for as work, stored energy, or heat.
Exothermic and Endothermic Processes
Fig. 11.3a, p.294 - heat flowing into a system from it’s surroundings:
• defined as positive
• q has a positive value
• called endothermic
–system gains heat as the surroundings cool down
Exothermic and Endothermic Processes
Fig. 11.3b, p.294 - heat flowing out of a system into it’s surroundings:
• defined as negative
• q has a negative value
• called exothermic
–system loses heat as the surroundings heat up
10
Exothemic and Endothermic
Every reaction has an energy change associated with it
Exothermic reactions release energy, usually in the form of heat.
Endothermic reactions absorb energy
Energy is stored in bonds between atoms
11
Heat Capacity and Specific Heat
A calorie is defined as the quantity of heat needed to raise the temperature of 1 g of pure water 1 oC.
• Used except when referring to food
• a Calorie, written with a capital C, always refers to the energy in food
• 1 Calorie = 1 kilocalorie = 1000 cal.
12
Heat Capacity and Specific Heat Joule-- the SI unit of heat and energy
• 4.184 J = 1 cal Specific Heat Capacity - the
amount of heat it takes to raise the temperature of 1 gram of the substance by 1 oC (abbreviated “C”)
13
Heat Capacity and Specific Heat
For water, C = 4.18 J/(g oC), and also C = 1.00 cal/(g oC)
Thus, for water:
• it takes a long time to heat up, and
• it takes a long time to cool off! Water is used as a coolant!
• Note Figure 11.7, page 297
14
Heat Capacity and Specific Heat
To calculate, use the formula: q = mass (g) x T x C heat abbreviated as “q” T = change in temperature C = Specific Heat Units are either J/(g oC) or cal/(g oC) Sample problem 11-1, page 299
15
Section 11.2Measuring and Expressing Heat
Changes
OBJECTIVES:
• Construct equations that show the heat changes for chemical and physical processes.
16
Section 11.2Measuring and Expressing Heat
Changes
OBJECTIVES:
• Calculate heat changes in chemical and physical processes.
Calorimetry Calorimetry - the accurate and
precise measurement of heat change for chemical and physical processes.
Calorimetry For systems at constant pressure,
the heat content is the same as a property called Enthalpy (H) of the system
Calorimetry Changes in enthalpy = H q = H These terms will be used
interchangeably in this textbook Thus, q = H = m x C x T H is negative for an exothermic
reaction H is positive for an endothermic
reaction (Note Table 11.3, p.301)
20
C + O2 CO2E
nerg
y
Reactants Products
C + O2
C O2
395kJ
+ 395 kJ
21
In terms of bonds
COO C
O
O
Breaking this bond will require energy.
CO
OOO C
Making these bonds gives you energy.In this case making the bonds gives you more energy than breaking them.
22
Exothermic The products are lower in energy
than the reactants Releases energy
23
CaCO3 CaO + CO2E
nerg
y
Reactants Products
CaCO3
CaO + CO2
176 kJ
CaCO3 + 176 kJ CaO + CO2
24
Chemistry Happens in
MOLES An equation that includes energy is
called a thermochemical equation CH4 + 2O2 CO2 + 2H2O + 802.2 kJ
1 mole of CH4 releases 802.2 kJ of energy.
When you make 802.2 kJ you also make 2 moles of water
25
Thermochemical Equations A heat of reaction is the heat
change for the equation, exactly as written• The physical state of reactants
and products must also be given.
• Standard conditions for the reaction is 101.3 kPa (1 atm.) and 25 oC
26
CH4 + 2 O2 CO2 + 2 H2O + 802.2 kJ
If 10. 3 grams of CH4 are burned completely, how much heat will be produced?
10. 3 g CH4
16.05 g CH4
1 mol CH4
1 mol CH4
802.2 kJ
= 514 kJ
27
CH4 + 2 O2 CO2 + 2 H2O + 802.2 kJ
How many liters of O2 at STP would be required to produce 23 kJ of heat?
How many grams of water would be produced with 506 kJ of heat?
Summary, so far...
29
Enthalpy The heat content a substance has at a
given temperature and pressure Can’t be measured directly because
there is no set starting point The reactants start with a heat content The products end up with a heat content So we can measure how much enthalpy
changes
30
Enthalpy Symbol is H Change in enthalpy is H (delta H) If heat is released, the heat content of
the products is lower
H is negative (exothermic) If heat is absorbed, the heat content
of the products is higher
H is positive (endothermic)
31
Ene
rgy
Reactants Products
Change is down
H is <0
32
Ene
rgy
Reactants Products
Change is upH is > 0
33
Heat of Reaction The heat that is released or absorbed in a
chemical reaction Equivalent to H C + O2(g) CO2(g) + 393.5 kJ
C + O2(g) CO2(g) H = -393.5 kJ
In thermochemical equation, it is important to indicate the physical state
H2(g) + 1/2O2 (g) H2O(g) H = -241.8 kJ
H2(g) + 1/2O2 (g) H2O(l) H = -285.8 kJ
34
Heat of Combustion The heat from the reaction that
completely burns 1 mole of a substance
Note Table 11.4, page 305
35
Section 11.3Heat in Changes of State
OBJECTIVES:
• Classify, by type, the heat changes that occur during melting, freezing, boiling, and condensing.
36
Section 11.3Heat in Changes of State
OBJECTIVES:
• Calculate heat changes that occur during melting, freezing, boiling, and condensing.
37
Heats of Fusion and Solidification
Molar Heat of Fusion (Hfus) - the heat absorbed by one mole of a substance in melting from a solid to a liquid
Molar Heat of Solidification (Hsolid) - heat lost when one mole of liquid solidifies
38
Heats of Vaporization and Condensation
Molar Heat of Vaporization (Hvap) - the amount of heat necessary to vaporize one mole of a given liquid.
Table 11.5, page 308
39
Heats of Vaporization and Condensation
Molar Heat of Condensation (Hcond) - amount of heat released when one mole of vapor condenses
Hvap = - Hcond
40
Heats of Vaporization and Condensation
Note Figure 11.5, page 310 The large values for Hvap and Hcond
are the reason hot vapors such as steam is very dangerous• You can receive a scalding burn
from steam when the heat of condensation is released!
41
Heats of Vaporization and Condensation
H20(g) H20(l) Hcond = - 40.7kJ/mol Sample Problem 11-5, page 311
42
Heat of Solution Heat changes can also occur when
a solute dissolves in a solvent. Molar Heat of Solution (Hsoln) -
heat change caused by dissolution of one mole of substance
43
Section 11.4Calculating Heat Changes
OBJECTIVES:
• Apply Hess’s law of heat summation to find heat changes for chemical and physical processes.
44
Section 11.4Calculating Heat Changes
OBJECTIVES:
• Calculate heat changes using standard heats of formation.
45
Hess’s Law If you add two or more
thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.
Called Hess’s law of heat summation Example shown on page 314 for
graphite and diamonds
46
Why Does It Work? If you turn an equation around, you change
the sign: If H2(g) + 1/2 O2(g) H2O(g) H=-285.5 kJ
then, H2O(g) H2(g) + 1/2 O2(g) H =+285.5 kJ
also, If you multiply the equation by a number,
you multiply the heat by that number: 2 H2O(g) H2(g) + O2(g) H =+571.0 kJ
47
Standard Heats of Formation The H for a reaction that produces 1
mol of a compound from its elements at standard conditions
Standard conditions: 25°C and 1 atm. Symbol is H f
0
The standard heat of formation of an element = 0
This includes the diatomics
48
What good are they? Table 11.6, page 316 has standard
heats of formation The heat of a reaction can be calculated
by:
• subtracting the heats of formation of the reactants from the products
Ho = (H f0 H f
0Products) - ( Reactants)
49
Examples CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
H f0
CH4 (g) = - 74.86 kJ/molH f
0O2(g) = 0 kJ/mol
H f0
CO2(g) = - 393.5 kJ/mol
H f0
H2O(g) = - 241.8 kJ/mol H= [-393.5 + 2(-241.8)] - [-74.68 +2 (0)]
H= - 802.4 kJ