Download Unit 6 Interactive Reading Packet File - District 196 e

Chapter 5: Thermochemistry
Section
in Book
Book
problems
associated
with this
target
1. I can give examples of different forms of energy. I can also state common units of
energy and convert between these units.
5.1 & 5.2
13, 14 &
16
2. I can define the first law of thermodynamics and write the associated equation.
5.2
25 & 26
3. I can describe how the change in internal energy of a system is related to the
changes of heat and work between the system and its surroundings.
5.2
27 & 28
4. I can define enthalpy and relate the enthalpy change in a process to the heat
added (endothermic) to or lost (exothermic) by the system during the process.
5.2 & 5.3
33 & 34
5. I can define the term state function and be provide examples of state functions.
5.2
31 & 32
6. I can sketch an exothermic and endothermic energy diagram and label the
various parts.
5.2
39 & 40
7. I can make stoichiometric calculations based upon a thermochemical equation.
5.4
43, 45 &
47
8. I can differentiate between the terms heat capacity and specific heat.
5.5
49 & 50
9. I can solve problems using q = m•Cp•∆T; calculate any one of the quantities
given the other three.
5.5
53, 54 &
56
10. I can describe the difference between a coffe cup calorimeter and a bomb
calorimeter. I will also be able to solve simple bomb calorimetry problems.
5.5
57-60
11. I can state Hess’s law and use Hess’s Law to solve problems.
5.6
61, 63, 65
& 66
12. I can define the terms standard state and standard heat of formation. I will also
be to write a chemical reaction associated with the standard heat of formation.
5.7
67, 68 &
70
13. I can calculate the enthalpy change in a reaction when given the standard
enthalpies of formations of each reactant and product.
5.7
73, 75, 78
& 80
Target
Chapter 5 - Thermochemistry
1
Thermochemistry is the study of the relationships between chemical reactions and energy changes.
Target #1: I can give examples of different forms of energy. I can also state
common units of energy and convert between these units.
Energy – the capacity to do work or to transfer heat
2
Kinetic Energy - energy of motion; KE = 1/2 mv ; since all atoms have mass and are in motion, all atoms
have a certain amount of kinetic energy; the energy that a substance has because of its temperature is
related to its KE
Potential Energy - the energy that an object possess because of its composition or its position with
respect to another object
Electrostatic potential energy – a form of potential energy due to the interactions between charged
particles; this energy increases with increasing charge and decreases with increasing distance between the
charged particles
Internal energy of a system (E) - sum of all KE and PE in a system; the difference in changes of
energy from before the change to after the change; E = Efinal – Einitial;
Relevance of the signs of ∆E:

positive E = system has gained energy from the surroundings
negative E = system has lost energy to its surroundings
It is nearly impossible to determine the energy of a system, but it is relatively easy to calculate the changes of
energy in a system!
Common Units of Energy:
A- joule; 1 J = 1 kg-m2/s2
B- calorie - used in biochemistry; 1 cal = 4.184 J
C- kilocalorie (Calorie); 1000 cal = 1 kilocalorie = 1 Calorie
Other terms related to energy changes . . .
System: portion of the universe that we are studying with respect to thermochemical changes
Surroundings: everything that lies outside of the system
example: If we were studying the combustion of gasoline inside of a
cylinder, the gasoline and
oxygen would constitute the system and the cylinder and piston (and everything
that) would constitute the surroundings
the
outside of
Closed System: can exchange energy but not matter with the surroundings
Open System: can exchange both energy and matter with the surroundings
Heat: energy transferred from a hotter object to a colder object
Energy: capacity of an object to do work or to transfer heat (extensive property)
Target #2: I can define the first law of thermodynamics and write the
associated equation.
First Law of Thermodynamics - energy is conserved; any energy lost by a system must
be gained by its surroundings (and visa-versa)
2
Equation representing the First law of Thermodynamics:
E = q + w, where . . .
E = change in internal energy
q = heat gained or lost by the system
w = work done on or by the system
Target #3: I can describe how the change in internal energy of a system is
related to the changes of heat and work between the system and its
surroundings.
A positive “q” means the system has gained heat
A positive “w” means work is done on the system by its surroundings.
When
When
When
When
q
q
q
q
>
<
<
>
0
0
0
0
and
and
and
and
w
w
w
w
>
<
>
<
0,
0,
0,
0,
then
then
then
then
∆E
∆E
∆E
∆E
> 0
< 0
depends upon the magnitude of q and w
depends upon the magnitude of q and w
Target #4: I can define enthalpy and relate the enthalpy change in a process to
the heat added (endothermic) to or lost (exothermic) by the system during the
process.
Enthalpy (H) - heat absorbed or released in a chemical reaction (under constant pressure); the abbreviation
qp indicates that pressure is constant; note that ∆H=qp under constant pressure conditions . . . as in coffe cup
calorimeters!
Endothermic process: (+∆H) system absorbs heat

2 H2O(g)
2 H2(g)
2 H2O(g) + 483.6 kJ

+
O2(g) ;
2 H2(g)
H = +483.6 kJ (pos. sign)
+
O2(g)
Exothermic process: (-∆H) system releases heat
Hrxn =
2 H2(g)
+
O2(g)
2 H2(g)
+
O2(g)
H (products) -


2 H2O(g) ;
H = -483.6 kJ
(neg. sign)
2 H2O(g) + 483.6 kJ
H (reactants)
Target #5: I can define the term state function and be provide examples of
state functions.
State function - property of a system which is determined by its state (or conditions, such as temperature &
pressure) (see figure 5.10 on page 176); the value of a state function depends only upon the present state of
3
the system, not the path the system took to reach that state
State functions can be recognized by the abbreviation of the property.
If the abbreviation is a capitol
letter, the property is a state function. If the abbreviation is a lower case letter, then the prperty is not a state
function.
Practice: Which of the following are state functions?
a) potential energy
e) temperature
b) kinetic energy
f) pressure
c) heat
g) velocity
d) work
h) mass
Target #6: I can sketch an exothermic and endothermic energy diagram and
label the various parts.
Draw an energy diagram the following reactions. Label the ∆H, energy of reactants and
the energy of products.
CaO(s) + CO2(g)

CaCO3(s)
CO2(g) + 2 H2O(l)
∆H = -178 kJ

CH4(g) + 2 O2(g)
∆H = 890 kJ
Target #7: I can make stoichiometric calculations based upon a
thermochemical equation.
Consider the following reaction:
2 KClO3(s)  2 KCl(aq) + 3 O2(g) ; ∆H = -89.4 kJ
For this reaction, calculate the ∆H for the formation of:
4
a) 1.24 moles of O2
b) 4.89 grams of KCl
Target #8: I can differentiate between the terms heat capacity and specific
heat.
Definition
Intensive or
Extensive?
Units
Heat capacity
Specific heat (or specific
heat capacity)
Molar heat capacity
Target #9: I can solve problems using q = m•Cp•∆T, calculate any one of the
quantities given the other three.
calorimetry - the measure of heat flow
calorimeter - instrument used to measure heat flow
bomb calorimeter - a special type of calorimeter used for combustion reactions
q = (grams) (specific heat) (∆T)
*When given three of these four terms, you should be able to solve for the fourth!
example 1: How much heat would it take to raise the temperature of 50.0 kg of water from 25.0oC to
75.0oC?
example 2: Calculate the enthalpy change (in kJ) upon converting 36.0 g of ice at -25.0oC to vapor at 125oC
under constant pressure at 1 atm. Use the following constants:
Cp (ice) = 2.09 J/g-K
Cp (H2O-liquid) = 4.18 J/g-K
Cp (water vapor) = 1.84 J/g-K
∆H fusion = 6.01 kJ/mol
∆H vaporization = 40.67 kJ/mol
5
AP Chemistry Practice
Chapter 5 – Thermochemistry Review
Targets #1-9
Name_________________________
Date ____________
Period____
Directions: Answer each of the following questions.
1. The KE of a 7.3 kg steel ball traveling at 18.0 m/s is _______ J.
2. One Joule equals . . . .
a) 1 kg m2/s2
b) 2 kg
c) 4.184 cal
6
d) 1 g cm/sec
e) none of these
3. An 35.0 mg object must be moving at a speed of ___ km/hr in order to posses a KE of 1.20 J.
4. A ______ system can exchange energy with the surroundings but cannot exchange matter
with the surroundings.
5. The change in energy (∆E) of a system that releases 12.4 J of heat and does 4.2 J of work
on the surroundings is ____ J.
6. q has a _____ (+/ -) value when heat is transferred from the surroundings to the system and
work has _____ (+/ -) value when work is done by the system on the surroundings.
7. The sign of ∆E is always negative when q ___ 0 and when w ___ 0.
Fill in the blanks with <, >, or =.
8. Circle all of the following which are state functions?
∆E
q
heat
work
temperature
∆PE
enthalpy
9. A ____ ∆H (+/ -) corresponds to an endothermic process.
10. The value for ∆H for the following reaction is -72 kJ. How many kJ of heat will be evolved
when 1.0 mol of HBr is formed?
H2(g) + Br2(g)  2 HBr(g)
o
11. The value of ∆H for the reaction below is -790 kJ. The enthalpy change accompanying the
reaction of 0.95 g of S is _____ kJ.
2 S(s) + 3 O2(g) -----> 2 SO3(g)
12. Draw an energy diagram for the following reaction:
4 Al(s) + 3 O2(g)  2 Al2O3(s) + 3351 kJ
7
13. ∆H = q when ______________________ is held constant. ∆H = ∆E when the
_____________ is held constant.
14. If 73.51 J of heat were added to 25.6 g of water at 25.3oC, what would be the final
temperature of the water?
15. Circle all of the following is/are endothermic processes?
dry ice subliming
water evaporating
water freezing
16. Under what conditions does ∆H = q?
17. Which of the following is/are true concerning enthalpy?
a) ∆H for a reaction depends upon the states of the reactants and products.
b) ∆H (forward reaction) = -∆H (reverse reaction)
c) The enthalpy of a system is independent of the way the system achieved its current
state.
d) Changes in enthalpy must be measured at a constant T, P, and V.
18. 2 Na2O2(s) + 2 H2O(l)  4 NaOH(s) + O2(g) ; ∆H = -126 kJ
How much heat is produced when 2.00 mol of NaOH is formed by the above reaction?
Circle true or false for #19-24.
T
T
T
T
F
F
F
F
T F
T F
19.
20.
21.
22.
Enthalpy is an intensive property.
H2O(g)  H2O(l) is an example of an endothermic process.
The enthalpy of a system can be measured.
The enthalpy change for a reaction depends upon the state of the reactants and
products.
23. ∆Hforward = 1/ ∆ Hreverse
24. The difference between ∆ E and ∆H is the amount of pressure volume work done
by the system.
8
Extra Calorimetry Problems
(Target 9)
q = m•Cp•∆T
Cp of H2O(l) = 4.184 J/g•oC
Cp of Hg(l) = 0.14 J/g•oC
Cp of Al(s) = 0.90 J/g•oC
Directions: Solve each of the following. Use the information above to help. Check your answers
on the WIKI page!
1. What is the heat capacity (J/oC) of 534 grams of liquid mercury?
2. How many kJ of heat are required to raise the temperature of 450. g of water
from 23.0oC to 85.0oC?
3. Assume you added 0.500 kJ of heat to a 100.0 gram piece of aluminum at 17.0oC.
What is the final temperature of the piece of aluminum?
4. When a 3.88 gram sample of solid NH4NO3 dissolves in 60.0 grams of water in a
coffee cup calorimeter, the temperature drops from 23.0oC to 18.4oC. Calculate
the ∆H (in kJ/mol) for NH4NO3 for the dissolving process. Assume that the
specific heat and density of the solution is the same as pure water.
5. A 75.0 gram piece of aluminum is heated to 95.0oC. The hot piece of aluminum is
placed in a coffee cup calorimeter which holds 100.0 mL of water at 22.0oC.
What will be the final temperature of the water in the calorimeter?
9
Target #10: I can describe the difference between a coffe cup calorimeter
and a bomb calorimeter. I will also be able to solve simple bomb calorimetry
problems.
What is constant?
Important Relationships
Coffe cup calorimeters
Bomb calorimeters
Bomb Calorimetry Problems
The following are examples of the various types of bomb problems that you may see on the test over this unit.
These problems are all based on the idea that a calorimeter has a certain heat capacity (usually expressed in
J/oC or kJ/oC). The heat capacity of the calorimeter is the amount of heat actually absorbed by the
calorimeter itself. The rigid steel container (the “bomb”) usually contains a known mass of the substance to
be ignited. The bomb is surrounded by water inside of the calorimeter. The energy change of the reaction
inside of the bomb is determined by measuring the increase in the temperature of the water and of the other
parts of the calorimeter.
Example #1: A 0.5269-gram sample of octane, C8H18, is placed in a bomb calorimeter known to have a
total heat capacity of 11.3 kJ/oC. The octane is ignited and the temperature increase of the calorimeter is
2.25oC. Calculate the amount of energy (in kJ/mol) released by this reaction.
Example #2: Camphor (C10H16O) has a heat of combustion of 5903.6 kJ/mol. When a 0.1204 g sample of
camphor is burned in a bomb calorimeter, the temperature increases by 2.28oC. Calculate the heat capacity
of the calorimeter.
Example #3: A 1.05 g sample of benzoic acid is combusted completely in a bomb calorimeter. Calculate the
change in temperature given that the heat capacity of the calorimeter is 1.80 kJ/K and the heat of combustion
of benzoic acid is 26.4 kJ/g.
10
Example #4: The combustion of methanol (CH3OH) is 239 kJ/mol. A 3.750-g sample of methanol is burned
in a bomb calorimeter. The temperature of the calorimeter increased from 20.00oC to 22.90oC. If the
calorimeter contained 1.500 kg of water, calculate the heat capacity of the dry calorimeter.
Example #5: 1.048 g of benzene (C6H6) is placed in the bomb compartment of a bomb calorimeter. The
water which surrounds the bomb compartment has a mass of 945 g. When the benzene is burned, the
temperature increases from 23.640oC to 32.692oC. The heat capacity of the dry calorimeter is 891 J/oC.
Find the heat of combustion of benzene (in kJ/g and kJ/mol).
Extra Bomb Calorimetry Problems
1.
A quantity of 1.435 g of naphthalene (C10H8) was burned in a
constant-volume bomb calorimeter. Consequently, the temperature of the water
rose from 20.17oC to 25.84oC. If the quantity of water surrounding the
calorimeter was exactly 2000 grams, and the heat capacity of the dry calorimeter
was 1.80 kJ/oC, calculate the molar heat of combustion of naphthalene.
11
2.
A 0.828 gram sample of methanol (wood alcohol, CH3OH) is placed in a bomb
calorimeter along with enough oxygen gas to ensure complete combustion. The
calorimeter contains 1.35 kg of water, and the heat capacity of the calorimeter
interior (without the water) is 1.06 kJ/oC. The molar heat of combustion for
methanol is 725.9 kJ/mol. Assuming the initial temperature of the water is
23.10oC, what will be the final temperature of the water after combustion?
Target #11: I can state Hess’s law and use Hess’s Law to solve problems.
Hess’s Law - if a reaction is carried out in a series of steps, ∆H for the reaction is equal to the sum of the
enthalpy changes for the individual steps
Example 1: Find the ∆H for the following reaction using Hess’s Law and the equations that below:
C(graphite) + 2H2(g)

CH4(g)
(overall reaction)
We can calculate the enthalpy change indirectly using Hess’s Law. The following reactions involving C, H2,
and CH4 with O2 have been studied and the ∆Horxn values are accurately known:
12
(a)
C(graphite) + O2(g) ------>
CO2(g)
(b)
2 H2(g) + O2(g) --------> 2 H2O(g)
(c)
CH4(g) + 2 O2(g) -------->
CO2(g) + 2 H2O(g)
Horxn = -393.5 kJ
Horxn = -571.6 kJ
Horxn = -890.4 kJ
Since we want to obtain one equation containing only C and H2 as reactants and CH4 as product, we must
reverse (c) to get . . .
CO2(g) + 2 H2O(g) -------> CH4(g) + 2 O2(g)
Horxn = +890.4 kJ
The next step is to add the equations:
(a)
(b)
Horxn = -393.5 kJ
Horxn = -571.6 kJ
C(graphite) + O2(g) ------> CO2(g)
2 H2(g) + O2(g) --------> 2 H2O(g)
(c)
CO2(g) + 2 H2O(g) -------> CH4(g) + 2 O2(g)
----------------------------------------------------------------------------C(graphite) + 2H2(g) --------> CH4(g)
Example 2: Calculate the ∆H for the following reaction . . .
2 C(graphite) + H2(g) --------> C2H2(g)
Use the following equations . . .
(a)
C(graphite) + O2(g) --------> CO2(g)
(b)
H2(g)
+
1/2 O2(g) -------->
(c)
2 C2H2(g) +
H2O(l)
5 O2(g) --------> 4 CO2(g) + 2 H2O(l)
Horxn = +890.4 kJ
--------------------------Horxn = -74.7 kJ
Horxn = -393.5 kJ
Horxn = -285.8 kJ
Horxn = -2598.8 kJ
Solution . . . We want to eliminate O2, CO2, and H2O from the equation. First we reverse (c) in order to get
the C2H2(g) on the products side.
(a)
(b)
(c)
C(graphite) + O2(g) --------> CO2(g)
H2(g)
+
1/2 O2(g) -------->
4 CO2(g) + 2 H2O(l) ------->
H2O(l)
2 C2H2(g) +
5 O2(g)
Horxn = -393.5 kJ
Horxn = -285.8 kJ
Horxn = +2598.8 kJ
Next, we multiply (a) by 4 and (b) by 2 and carry out the addition:
(a)
4 C(graphite) + 4 O2(g) --------> 4 CO2(g)
(b)
2 H2(g)
+
O2(g) -------->
2 H2O(l)
13
Horxn = -1574.0 kJ
Horxn = -571.6 kJ
(c)
4 CO2(g) + 2 H2O(l) ------->
2 C2H2(g) +
5 O2(g)
---------------------------------------------------------------------4 C(graphite) + 2 H2(g) --------> 2 C2H2(g)
or
2 C(graphite) + H2(g) --------> C2H2(g)
Horxn = +2598.8 kJ
-----------------------------Horxn = 453.2 kJ
Horxn = 226.6 kJ
Example 3: Determine the heat of reaction for the synthesis of propane from its elements by using reactions
#1-3 below:
3C(s) + 4H2 (g) ------->
CH3CH2CH3(g)
rxn #1: CH3CH2CH3(g) + 5 O2(g) ----> 3CO2(g) + 4H2O(g)
rxn #2: 2H2 (g) + O2(g) ------> 2H2O(g)
rxn #3: C(s) + O2(g) -------> CO2(g)
Horxn = ?????
o
H 1 = -2,044 kJ
o
H 2 = -483.6 kJ
o
H 3 = -393.5 kJ
Target #12: I can define the terms standard state and standard heat of
formation. I will also be to write a chemical reaction associated with the
standard heat of formation.
Standard enthalpy of formation (heats of formation) =
Hof = heat lost or gained when forming one mole of a
substance from its elements at standard state conditions (25oC and 1 atm)
Example:
Na(s) + ½ Cl2(g)

NaCl(s) ;
Hof = -410.9 kJ/mol
Table 5.3 and Appendix C in your text lists the standard enthalpies of formations of several compounds and
elements. A formation reaction is defined as one in which the reactants are elements in their standard states
and forming only ONE mole of product. Below are listed 2 examples of formation reactions:
2 Fe(s) + 3/2 O2(g)  Fe2O3(s)
Hof = -822.16 kJ/mol
2 C(graphite) + 3 H2(g) + 1/2 O2(g)  C2H5OH(l)
Hof = -277.7 kJ/mol
14
Note: The standard enthalpy of formation for the most stable form of any element is zero.
Practice #1: Which of the following reactions is the value for ∆Horxn equal to ∆Hof for the product?
a)
b)
c)
d)
 H2O(l)
2 H2 (g) + O2 (g)  2 H2O(l)
2 H2O(l) + O2 (g)  2 H2O2 (l)
H2O(l) + ½ O2 (g)  H2O2 (l)
H2(g) + ½ O2 (g)
Practice #2: Circle all of the following which have a ∆Hof equal to zero.
Ne(g)
Cl2(g)
H2O(g)
NH3(aq)
Fe(s)
H2O(l)
F2(l)
Practice #3: Write the formation reaction for butanol, C4H9OH.
Target #13: I can calculate the enthalpy change in a reaction when given the
standard enthalpies of formations of each reactant and product.
Use the following equation when solving enthalpy of formation problems:
Horxn =
Hoproducts Horeactants
Example #1: Consider the following chemical equation and answer the questions that follow:
2 Al(s) + Fe2O3(s)  2 Fe(s) + Al2O3(s)
A) Solve for the enthalpy change (∆Ho) for the above reaction using the thermochemical
data from Appendix C.
B) Solve for the enthalpy change (∆Ho) for the reaction given above by using the equations below:
4 Al(s) + 3 O2(g)

2 Al2O3(s)
15
; ∆Ho = -3339.65 kJ
2 Fe(s) + 3/2 O2(g)  Fe2O3(s)
; ∆Ho = -822.16 kJ
Example #2:
A) Write the formation reactions for C3H8(g), CO2(g), and H2O(l). Look up the ∆Hof for each.
B) Using ∆Horxn = ∑∆Hoproducts - ∑∆Horeactants, calculate the ∆Ho for the burning of 1 mole of
C3H8(g).
C) Now, calculate the ∆Ho for the combustion of propane using Hess’s Law.
16