SPECIFIC HEAT CAPACITY AND CALORIMETRY
| |
|[pic] |
|UNIT 4: CHEMICAL REACTIONS: THE MOLE, STOICHIOMETRY AND THERMODYNAMICS |
|Part D: Thermodynamics |
|Big Picture Ideas: |
|Enthalpy changes occur during a chemical reaction. |
|Entropy is the measure of disorder. |
|Chemical reactions, though they occur at different rates, are affected by common factors such as temperature, presence of a |
|catalyst/inhibitor, etc |
|Big Picture Questions: |
|o What factors determine the change in entropy of a system? |
|How does a catalyst or inhibitor affect reaction rate and what are some real-life applications of each? |
| Suggested Resources… |Key Terms: | |
|Homework Assignments |Thermochemistry |calorimetry |
|Classwork Assignments |exothermic reaction |heat capacity |
|Laboratory Activities |endothermic reaction |specific heat |
|Formative Assessments |enthalpy |kinetic theory |
|Textbook pages: Chapter 12 and 22-3 |standard enthalpy change | |
|Websites: | | |
|Directions: Use this information as a general reference tool to guide you through this unit. Don’t hesitate to ask your teacher for help! |
|By the conclusion of this unit, you should know the following: |By the conclusion of this unit, you should be able to do the following: |
| |Interpret energy diagrams for both endothermic and exothermic reactions. |
|Energy can neither be created nor destroyed but can be transformed from|Using the collision theory, explain how altering the temperature, surface area, |
|one type to another. |concentration and the addition of a catalyst affects the rate of a reaction. |
|Thermochemical equations show energy changes accompany chemical |Given an equation, determine whether entropy increases or decreases. |
|reactions. |Distinguish between endothermic and exothermic reactions. |
|Collision theory describes characteristics for a successful reaction. |Explain what is meant by enthalpy and enthalpy change |
|Factors can be altered to affect the rate of a reaction (temperature, |Define standard enthalpy change and explain how it is used. |
|surface area, catalyst, concentration). |Compare heat and temperature. |
|In any spontaneous process, the entropy of the universe always |Use stoichiometric relationships to calculate heat. |
|increases. | |
|In a calorimeter, there is an even transfer of heat between water and | |
|another substance. | |
|Energy is required during bond breakage; it is released during bond | |
|formation. | |
Practice Problems - Directions:
Identify whether the problem is : a) heat and stoichiometry b) q = m c ΔT c) m c ΔT = m c ΔT
Write all the given values in an orderly fashion on a separate sheet of paper and solve.
1
1. Nitrogen reacts with hydrogen to produce ammonia (NH3). The ΔH = +46.2 kJ.
How many kJ of heat is absorbed when 97grams of NH3 is produced?
2. The temperature of a metal bar with a mass of 87 grams is raised from 31° C to 132 °C. In the process, 790 joules of heat were absorbed. Find the specific heat of the metal.
3. A 23.6g sample of metal, originally at 88 °C was placed in 62 mL of water, originally at 17 °C. After some time, the system reached an equilibrium and the final temperature of the system was 21.5 °C. Find the specific heat of the metal.
4. A piece of clay at 94 °C was placed into 29g of water at 31 °C. After some time, the temperature stabilized at 35.6 °C. If the specific heat of the clay is 1.2 J/g° C, find the mass of the clay that was dropped in the water.
5. How much heat is transferred when 9.22 g of glucose (C6H12O6) in your body reacts with O2 to produce CO2 and H2O?
6. An 18.7 g sample of platinum metal increases in temperature by 2.3 °C when 5.7 joules of heat are added. What is the specific heat of platinum?
7. How much heat is released if 1.0 g of hydrogen peroxide (H2O2) decomposes?
2 H2O2 ⋄ 2H2O + O2 ΔH = -190kJ
SPECIFIC HEAT CAPACITY AND CALORIMETRY
1. Units of heat:
Joule: metric unit for heat/energy
calorie: amount of energy needed to change 1 gram of water 1°C
kilocalorie (Calorie or food calorie)
CONVERSION FACTOR: 4.18J = 1 cal
1000 cal = 1 Cal or kcal (food calorie)
1000 J = 1 kJ
TRY: 2500J = ? cal 2500 J 1 cal = 598 cal
4.18 J
4.25 Cal = ? J 4.25 Cal 1000 cal 4.18 J = 17,765 J 1 Cal 1 cal
11.2 kJ = ? Cal 11.2 kJ 1000 J 1 Cal = 2.7 Cal 1 Cal 4.18 J
1. Specific Heat Capacity how much energy it takes to change 1 gram of something’s temp. by 1°C
low specific heat capacity: it takes less energy to raise the temp (ex: metal)
high specific heat capacity: it takes more energy to raise the temp (ex: water)
2. Problems Involving Specific Heat Capacity
H = m x (sh) x ΔT
The specific heat of lead is 0.129 J/g°C. How much heat is necessary to raise the temperature of 45.0g of lead from 25°C to 38°C?
H = 45g X 0.129 J/g°C X 13 °C
H = 75.5 J (it takes 75.5 J of heat to raise 45g of lead, 13°C
b. What mass of lead can be heated from 23°C to 44°C using 125J of heat? The specific heat capacity of lead is 0.129 J/g°C.
125 = m (.129)(21)
46.1g = m
c. When 32.0g of a substance cools from 85°C to room temperature (25°C), 2400J of heat are released. Find the specific heat capacity of the substance.
2400 = (320)(sh)(60)
1.25 J/g°C = sh
d. A 16-g piece of iron absorbs 1090 joules of heat energy, and its temperature changes from 25ºC to 175ºC. Calculate the heat capacity of iron.
1090 = (16)(sh)(150)
.454 J/g°C = sh
e. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22ºC to 55ºC, if the specific heat of aluminum is 0.90 J/gºC?
H = 10(0.90)(33)
H = 297 J
f. To what temperature will a 50.0 g piece of granite raise if it absorbs 5275 joules of heat and its heat capacity is 0.50 J/gºC? The initial temperature of the granite is 20.0ºC.
5275 = (50)(.5)(ΔT)
211 = ΔT
211 + 20 = Ti
231 ºC = Ti
g. Calculate the heat capacity of a piece of silver if 1500 g of the silver absorbs 6.80kJ of heat, and its temperature changes from 32ºC to 57ºC.
6.80 kj = 6800 J 6800 = (1500)(sh)(25)
0.18 J/gºC = sh
h. 10.0 mL of 4ºC water is heated until its temperature is 37ºC. If the specific heat of water is 4.18 J/gºC, calculate the amount of heat energy needed to cause this rise in temperature. The density of water is 1.0g/mL.
H = 10(4.18)(33)
H = 1379.4 J
Calorimetry: heat gained by water = heat lost by process
m(sh)(ΔT) = m(sh)(ΔT)
CALORIMETER
[pic]
1. 55.0g of a “mystery metal” at 93°C is placed in a calorimeter containing 100.0g of water at 25°C. The metal sits in the water until the temperature levels off at 29°C. At this point, both the metal and the water are at 29°C. The specific heat capacity of water is 4.18 J/g°C.
| |METAL |WATER |
|mass |55g |100 g |
|specific heat capacity |? |4.18 J/g°C |
|Tf |29 °C |29 °C |
|Ti |93 °C |25 °C |
Use H = m x (sh) x ΔT to find the amount of heat gained by the water.
H = 100(4.18)(4)
H = 1672 J
How much heat was lost by the metal?
1672 J (same amount of energy gained by the water!!)
Use H = m x (sh) x ΔT to find the specific heat capacity of the metal.
1672 = 55(sh)(64)
0.475 J/g°C = sh
2. 55.0g of a metal are heated to 112ºC, and then placed in a coffee cup calorimeter containing 60.0g of water at 32ºC. The final temperature in the calorimeter is 42ºC. What is the specific heat of the metal?
| |METAL |WATER |
|mass |55g |60g |
|specific heat capacity |? |4.18 J/g°C |
|Tf |42 ºC |42 ºC |
|Ti |112 ºC |32 ºC |
m(sh)(ΔT) = m(sh)(ΔT)
55(sh)(70) = 60(4.18)(10)
sh = 0.65 J/g°C
__________________
3. The specific heat of aluminum is 0.902 J/gºC. 15.0g of aluminum are heated to 115ºC, and added to a calorimeter containing water at 25ºC. The final temperature in the calorimeter is 45ºC. What mass of water was in the calorimeter?
| |METAL |WATER |
|mass |15g |? |
|specific heat capacity |0.902 J/g°C |4.18 J/g°C |
|Tf |45 ºC |45 ºC |
|Ti |115 ºC |25 ºC |
m(sh)(ΔT) = m(sh)(ΔT)
15(0.902)(70) = x(4.18)(20)
11.33g = x
___________________
THE ENERGY OF CHEMICAL REACTIONS
1. Define the following terms:
a. endothermic – feels cold, absorbs heat, ΔH = positive, heat on reactant side (right) of rxn
b. exothermic – feels hot, releases heat, ΔH =negative, heat on product side (left) of rxn
c. ΔH – change in heat or “enthalpy”
activation energy – amount of energy needed to start a reaction
e. catalyst – speeds up reaction by lowering the activation energy
2. Energy Diagrams
ENDOTHERMIC REACTION EXOTHERMIC REACTION
[pic]
***Chemical bonds break on the way “up” (require energy), and form on the way “down” (release energy)***
1. Label the following on each diagram above: reactants, products, activation energy, ΔH, breaking chemical bonds, bond formation.
2. A student burns a food sample in a calorimeter containing 100.0g of water at 25ºC. The temperature of the water in the calorimeter raises to 51ºC.
Calculate ∆H for the reaction.
H = (100)(4.18)(26)
H = 10868 J
is the reaction endothermic or exothermic?
Exothermic (burning releases energy!) so ∆H = -10868J
sketch an energy diagram for the reaction on the axis below. Label the reactants, products, ∆H, activation energy and the activated complex on the diagram.
****see previous page of notes for exothermic reaction diagram!
ENERGY AND STOICHIOMETRY
1. Consider the following reaction:
2 S + 3 O2 ⋄ 2 SO3 + 791.4kJ
Does the reaction absorb or release heat? _________________
Is the reaction endothermic or exothermic? ________________
c. What is the value of ΔH? _________-791.4 kJ________
d. Draw an enthalpy diagram for the reaction.
See notes page
e. How much heat will be given off if 10.0g of sulfur burn in excess oxygen?
10 g S 1 mole S -791.4 kJ = -123.3 kJ
32.1 g S 2 moles S
What mass of sulfur must burn in excess oxygen in order to release 250kJ of heat?
250 kJ 2 mole S 32.1 g = 20.2 g S
791.4 g S 1 moles S
2. Consider the following reaction:
H2 + Br2 ⋄ 2 HBr ΔH = 72.80kJ
Does the reaction absorb or release heat? _____________________
Is the reaction endothermic or exothermic? ____________________
c. On which side of the reaction would heat appear? ____72.80 kJ___________
d. Draw an enthalpy diagram for the reaction.
See notes
e. How much heat is required to form 125g of HBr?
125 g HBr 1 mole HBr 72.8 kJ = 56.2 kJ
80.9 g S 2 moles HBr
What mass of HBr can be formed if 525kJ of heat are absorbed by the reaction?
525 kJ 2 mole HBr 80.9 g = 1167 g HBr
72.8 g S 1 moles HBr
THERMODYNAMICS: ENERGY & ENTROPY IN CHEMICAL REACTIONS
Enthalpy of a Reaction (ΔH): change in heat (Hproducts – Hreactants)
ΔH = “+” = endothermic = H absorbed
ΔH = “-“ = exothermic = H released
Entropy: (ΔS) – change in order/how disorganized a substance is
ΔS = “+” = system is more disorganized
ΔS = “-“ = system is less disorganized
Which process(es) below would have a positive change in entropy (ΔS)?
cleaning your room
- ΔS
a bottle breaking
+ ΔS
leaves falling off of a tree
+ ΔS
water freezing
- ΔS
States of matter, solutions and entropy:
Most ordered
Solid
Liquid
Gas
Less ordered
Solution
Which reactions below would have a positive change in entropy (ΔS)?
Al2(CO3)3(s) ⋄ Al2O3(s) + 3CO2(g)
Positive ΔS
KCl(aq) + Br2(l) ⋄ KBr(aq) + Cl2(g)
Positive ΔS
FeCl2(aq) + H2SO4(aq) ⋄ FeSO4(s) + HCl(aq)
Negative ΔS
How can you tell if a reaction will be spontaneous? (able to occur?)
|ΔS change in entropy |ΔH change in heat |Spontaneous??? |
|+ more disorder | exothermic releases heat |Yes – it can occur at any temp. |
|+ more disorder | + endothermic |Spontaneous at high temperature |
| |requires heat | |
|- less disorder/more order | - exothermic |Spontaneous at low temperature |
| |releases heat | |
|- less disorder/more order | + endothermic |No – it cannot occur |
| |requires heat | |
Determine whether each reaction will have a positive or negative change in entropy (ΔS).
Label each reaction as endothermic or exothermic.
Determine which reaction(s) below will naturally occur (be spontaneous)? Which will require added energy? Which will only occur naturally at low temperatures?
a. Na2CO3 (s) ⋄ Na2O (s) + CO2 (g) ΔS = __+___ ΔH = 321kJ
Spontaneous? ___@ high temp__
b. Mg (s) + 2HCl (aq) ⋄ MgCl2 (aq) + H2 (g) ΔS = __+__ ΔH = - 457.6kJ
Spontaneous? _yes - always_____
c. N2 (g) + 3H2 (g) ⋄ 2NH3 (g) ΔS = __-___ ΔH= -91.8kJ
Spontaneous? __@ low temp____
d. 2KCl (s) + 3O2 (g) ⋄ 2KClO3 (s) ΔS =__-___ ΔH = 90.0kJ
Spontaneous? ___never______
Specific Heat Worksheet
Show all work and proper units.
1. A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25°C to 175°C. Calculate the specific heat capacity of iron.
1086.75 = 15.75(sh)(150)
0.46 J/g°C = sh
2. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C?
H = 10(0.90)(33)
H = 297 J
3. To what temperature will a 50.0 g piece of glass raise if it absorbs 5275 joules of heat and its specific heat capacity is 0.50 J/g°C? The initial temperature of the glass is 20.0°C.
5275 = (50)(.5)(ΔT)
211 = ΔT
211 + 20 = Ti
231 ºC = Ti
4. Calculate the heat capacity of a piece of wood if 1500.0 g of the wood absorbs 6.75×104 joules of heat, and its temperature changes from 32°C to 57°C.
6.75 x 104= (1500)(sh)(25)
1.8 J/gºC = sh
5. 100.0 mL of 4.0°C water is heated until its temperature is 37°C. If the specific heat of water is 4.18 J/g°C, and the density of water is 1.0g/mL. Calculate the amount of heat energy needed to cause this rise in temperature.
H = 10(4.18)(33)
H = 1379.4 J
6. 25.0 g of mercury is heated from 25°C to 155°C, and absorbs 455 joules of heat in the process. Calculate the specific heat capacity of mercury.
455 = 25 (sh)(130)
0.14 J/g°C = sh
CALORIMETRY PRACTICE
| |METAL |WATER |
|mass | | |
|specific heat capacity | |4.18 J/g°C |
|Tf | | |
|Ti | | |
1. A piece of metal with a mass of 38.0 g and a temperature of 115°C is placed into 75.0 mL of water at 15°C. The final temperature of the system is 20°C. The density of water is 1.0 g/mL. What is the specific heat of the metal?
38g (sh) (95°C) = 75 g (4.18 J/g°C) (5°C)
sh = 0.43 J/g°C
2. A piece of gold (C= 0.129 J/g°C) is heated to 100°C and placed into 60 mL of water at 20.5°C. The final temperature of the system is 22°C. What was the mass of the metal?
M ( 0.129 J/g°C) (78°C) = (60g) (4.18 J/g°C) (1.5°C)
M = 37.4 g
3. 25.0 g of copper (C=.389 J/g°C) at 95°C is placed into water at 25°C. The final temperature of the system is 28.5°C. What was the mass of the water?
25g (0.389 J/g°C) (66.5°C) = M (4.18 J/g°C) (3.5°C)
44.2 g = M
4. A piece of aluminum (C=0.900 J/g°C) with a mass of 55.0 g is heated and placed into 45.0 mL of water at 18°C. The final temperature of the system is 22°C. What was the starting temperature of the metal?
55g (0.9 J/g°C) (ΔT) = 45g (4.18 J/g°C) (4°C)
ΔT = 15.2
15.2 + 22 = 37.2°C
[pic]
HOMEWORK: ENTROPY AND SPONTANEITY OF CHEMICAL REACTIONS
Determine whether ΔS will be positive (disorder increases) or negative (disorder decreases) for each process below:
a. burning paper _____+________________
b. organizing your binder _____-________________
c. water freezing _____-________________
d. dry ice sublimating _____+________________
2. Determine the sign of ΔS for each reaction below. Then determine under which conditions (if at all) the reactions will be spontaneous.
CO2 (g) → C(graphite) + O2 (g) ΔH = +393.5 kJ/mol
ΔS = ___-____ Spontaneous? ____no_______
SKIP b!
2HCl(g) → H2 (g) + Cl2 (g) ΔH = +184.6 kJ/mol
ΔS = _________ Spontaneous?_____________
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ΔH = -890.4 kJ/mol
ΔS = __-___ Spontaneous? __@ low temp___
2 Na(s) + 2 H2O(l) ⋄ 2 NaOH (aq) + H2(g) ΔH = -446 kJ/mol
ΔS = __+____ Spontaneous? _yes_______
3. A student tries to dissolve a salt (ionic compound) in water. He is having a hard time, so he decides to use the bunsen burner to heat up the water as the salt dissolves. It works! Determine the signs for ΔH and ΔS for this process.
ΔH ___positive___________ ΔS ___positive____________
4. When you did the reaction in your “ziploc bag” lab, the bag became cold. Explain in terms of entropy and enthalpy why the reaction was able to occur.
Energy Content of Foods
All human activity requires “burning” food for energy. In this experiment, you will determine the energy released (in kJ/g) as various foods, such as cashews, marshmallows, peanuts, and popcorn, burn. You will look for patterns in the amounts of energy released during burning of the different foods.
[pic]
Figure 1
MATERIALS
| |UTILITY CLAMP |
| |2 STIRRING RODS |
| |RING STAND AND 10-CM (4-INCH) RING |
|THERMOMETER |100-ML GRADUATED CYLINDER |
|2 FOOD SAMPLES |SMALL CAN |
|FOOD HOLDER |COLD WATER |
|WOODEN SPLINT |MATCHES |
PROCEDURE
1. OBTAIN A PIECE OF ONE OF THE TWO FOODS ASSIGNED TO YOU AND A FOOD HOLDER LIKE THE ONE SHOWN IN FIGURE 1. FIND AND RECORD THE INITIAL MASS OF THE FOOD SAMPLE AND FOOD HOLDER. CAUTION: DO NOT EAT OR DRINK IN THE LABORATORY.
2. Determine and record the mass of an empty can. Add 50 mL of cold water to the can. Obtain the cold water from your teacher. Determine and record the mass of the can and water.
3. Use your thermometer to record the initial temperature of the water in the can. Read one place past the decimal. Be sure to record the temperature of the water, not the can. You will need to hold the thermometer in the water throughout the experiment.
4. Set up the apparatus as shown in Figure 1. Use a ring and stirring rod to suspend the can about 2.5 cm (1 inch) above the food sample. Remove the food sample from under the can and use a wooden splint to light it. Quickly place the burning food sample directly under the center of the can. Allow the water to be heated until the food sample stops burning. CAUTION: Keep hair and clothing away from open flames.
5. Stir the water with the thermometer. Continue stirring the water until the temperature stops rising. Record this maximum temperature, t2.
6. Determine and record the final mass of the food sample and food holder.
7. When you are done, place burned food, used matches, and partially-burned wooden splints in the container provided by the teacher. Repeat using the second food item.
Processing the data
1. FIND THE MASS OF WATER HEATED FOR EACH SAMPLE.
2. Find the change in temperature of the water, Δt, for each sample.
3. Calculate the heat absorbed by the water, q, using the equation
q = Cp•m•Δt
where q is heat, Cp is the specific heat capacity, m is the mass of water, and Δt is the change in temperature. For water, Cp is 4.18 J/g°C. Change your final answer to kJ.
4. Find the mass (in g) of each food sample burned.
5. Use the results of Steps 3 and 4 to calculate the energy content (in kJ/g) of each food sample.
6. 1 calorie = 4.18J. 1 Cal (food calorie) is 1 kilocalorie. Find the energy content in Cal/g.
DATA and calculations
|FOOD TYPE |––––––––––––––––––– |––––––––––––––––––– |
|INITIAL MASS OF FOOD AND HOLDER |G |G |
|FINAL MASS OF FOOD AND HOLDER |G |G |
|MASS OF FOOD BURNED |G |G |
|MASS OF CAN AND WATER |G |G |
|MASS OF EMPTY CAN |G |G |
|MASS OF WATER HEATED |G |G |
|FINAL TEMPERATURE, T2 |°C |°C |
|INITIAL TEMPERATURE, T1 |°C |°C |
|TEMPERATURE CHANGE, ΔT |°C |°C |
CALCULATIONS: SHOW ALL WORK IN THE SPACE PROVIDED!
|Food Type | |
|Heat, q | |
| | |
| | |
| | |
| |kJ |
|Energy content in kJ/g | |
| | |
| | |
| |kJ/g |
|Energy content in Cal/g |Cal/g |
|Manufacturer Reported Energy Content in Cal/g |Cal/g |
|% Error | |
|Food Type | |
|Heat, q | |
| | |
| | |
| | |
| |kJ |
|Energy content in kJ/g | |
| | |
| | |
| |kJ/g |
|Energy content in Cal/g |Cal/g |
|Manufacturer Reported Energy Content in Cal/g |Cal/g |
|% Error | |
List three sources of error and effect on results (Cal/g)
1.
2.
3.
-----------------------
ENERGY
TIME
TIME
95kJ
55kJ
20kJ
ΔH is
negative
ΔH is positive
ΔH is negative
ΔH is positive
ENERGY
30kJ
45kJ
55kJ
R
R
P
P
Ea (25kJ)
Ea (40kJ)
ΔH = P-R
= 45kJ – 30kJ = 15 kJ
ΔH = P-R
= 20kJ – 55kJ = -35 kJ
ΔS is -
ΔS is +
S
L
G
-----------------------
19
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