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Chemistry
Quarter 3
Assessment
Review
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Thermodynamics
Determine the Specific Heat of a Metal
1. A group of students conducted an
experiment to determine the
specific heat of a metal. Their goal
was to use the specific heat to
identify the metal.
Mass of metal
Mass of water
Initial temperature of metal
Initial temperature of water
Final temperature of water and metal
∆Twater
∆Tmetal
a. Add an arrow to the diagram that
shows the direction of heat transfer
when the hot metal is added to the
water in the calorimeter.
20.0 g
100.0 g
100.0 °C
23.1 °C
26.3 °C
Heat flows from warmer to cooler.
Determine the Specific Heat of a Metal
b. Calculate the heat gained by the
water. The specific heat of water is
4.18 J/g°C.
1. A group of students conducted an
experiment to determine the
specific heat of a metal. Their goal
was to use the specific heat to
identify the metal.
Mass of metal
Mass of water
Initial temperature of metal
Initial temperature of water
Final temperature of water and metal
∆Twater
∆Tmetal
20.0 g
100.0 g
100.0 °C
23.1 °C
26.3 °C
3.2 °C
𝑞 = 𝑚 × 𝑐 × ∆𝑇
4.18 𝐽
𝑞 = (100.0 𝑔) ×
× (3.2 ℃)
𝑔∙℃
∆𝑇 = 𝑇2 − 𝑇1 = 26.3 ℃ − 23.1 ℃ = 3.2 ℃
𝑞 = 1337.6 𝐽
𝑞 = 1300 𝐽
Determine the Specific Heat of a Metal
1. A group of students conducted an
experiment to determine the
specific heat of a metal. Their goal
was to use the specific heat to
identify the metal.
Mass of metal
Mass of water
Initial temperature of metal
Initial temperature of water
Final temperature of water and metal
∆Twater
∆Tmetal
20.0 g
100.0 g
100.0 °C
23.1 °C
26.3 °C
3.2 °C
c. How does this calculated heat relate
to the metal?
– What ever heat is gained by the water
(endothermic) is exactly equal to the heat
lost by the metal (exothermic).
– This gives us a value for q to use when
solving for the specific heat of the metal.
Determine the Specific Heat of a Metal
1. A group of students conducted an
experiment to determine the
specific heat of a metal. Their goal
was to use the specific heat to
identify the metal.
Mass of metal
Mass of water
Initial temperature of metal
Initial temperature of water
Final temperature of water and metal
∆Twater
∆Tmetal
20.0 g
100.0 g
100.0 °C
23.1 °C
26.3 °C
3.2 °C
- 73.7 °C
d. Find the specific heat of the metal.
𝑞 = 𝑚 × 𝑐 × ∆𝑇
(𝑚 × ∆𝑇) (𝑚 × ∆𝑇)
𝑞
𝑐=
𝑚 × ∆𝑇
−1337.6 𝐽
𝑐=
(20.0 𝑔) × (−73.7 ℃)
∆𝑇 = 𝑇2 − 𝑇1 = 26.3 ℃ − 100.0 ℃ = −73.7 ℃
0.907 𝐽
𝑐=
𝑔℃
0.91 𝐽
𝑐=
𝑔℃
Determine the Specific Heat of a Metal
1. A group of students conducted an
experiment to determine the
specific heat of a metal. Their goal
was to use the specific heat to
identify the metal.
Mass of metal
Mass of water
Initial temperature of metal
Initial temperature of water
Final temperature of water and metal
∆Twater
∆Tmetal
20.0 g
100.0 g
100.0 °C
23.1 °C
26.3 °C
3.2 °C
-73.7 °C
d. Which metal did the students have?
– Based on the calculated value for specific
heat (0.91 J/g°C), the sample was most
likely aluminum.
Specific Heat Values, J/g°C
Al
0.900
Au
0.129
Cu
0.385
Enthalpy Diagrams
2. Consider the enthalpy diagram
below.
a. What is the heat of reaction, ΔH, for
this reaction?
– Difference between products and reactants
– 50 kJ - 150 kJ = -100 kJ
b. Make a claim about if the reaction is
endothermic or exothermic. Support
your claim with evidence.
– Exothermic: the reactants are higher than
the products, meaning heat must be given
off
∆H= -100 kJ
Enthalpy Diagrams
2. Consider the enthalpy diagram
below.
c. How much energy is required
(activation energy Ea) to break
reactant bonds?
– Reactants to the top of the hill
– 200 kJ – 150 kJ = 50 kJ
d. Compare the stabilities of the
products and the reactants.
– Higher energy = less stable
– Reactants are less stable than the products
because they are higher energy.
Ea= 50 kJ
Specific Heat of a Substance
3. A sample that weighs 234 grams is heated by 18°C when 1460 calories of
heat is added. What is the specific heat of the substance?
𝑞 = 𝑚 × 𝑐 × ∆𝑇
(𝑚 × ∆𝑇) (𝑚 × ∆𝑇)
𝑞
𝑐=
𝑚 × ∆𝑇
1460 𝑐𝑎𝑙
𝑐=
(234 𝑔) × (18 ℃)
0.346 𝑐𝑎𝑙
𝑐=
𝑔℃
0.35 𝑐𝑎𝑙
𝑐=
𝑔℃
Endothermic and Exothermic Phase Change
4. Provide an example of a phase change that is endothermic and one that is
exothermic.
– Endothermic: any phase change that requires the substance to absorb heat
– Melting
– Boiling
– Exothermic: any phase change that releases heat/ cools the substance
– Freezing
– Condensing
Heat of Reaction (Stoichiometry)
5. Consider the reaction below.
67.8 𝑘𝐽 + 𝑁2 𝑔 + 𝑂2 𝑔 → 2𝑁𝑂2 (𝑔)
a. What is the ∆H of the reaction?
∆𝐻 = +67.8 𝑘𝐽
b. Is the reaction endothermic or exothermic? How do you know?
– The reaction is endothermic because heat is a reactant: heat must be absorbed for the reaction to
occur. This means that ∆H is positive.
c. How many grams of N2 are required to absorb 156 KJ of energy?
156 𝑘𝐽 1 𝑚𝑜𝑙 𝑁2 28.01 𝑔 𝑁2
×
×
= 64.46 𝑔 𝑁2 = 64.5 𝑔 𝑁2
1
67.8 𝑘𝐽
1 𝑚𝑜𝑙 𝑁2
𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑁2 = 2 × 14.007 = 28.014 𝑔/𝑚𝑜𝑙
Entropy and Disorder
6. Entropy is a measure of the disorder of a system. For each of the following,
determine if the disorder of the system is increasing or decreasing.
a. Ice melts into a liquid
–
Increase
b. Methane crystallizes into a solid
– Decrease
c. Solid iodine sublimes into a gas
– Increase
d. Water condenses on a cold can
– Decrease
Bond Breaking and Forming
7. In the reaction O2 + energy  O + O, bonds are being (circle one: broken/
formed) and energy is (circle one: released/ absorbed).
O O + energy  O + O
Standard Enthalpy of Formation
8. What is the standard enthalpy of formation of AgCl if 28.36 kJ of heat are
given off when 32 g of AgCl are formed?
– Enthalpy of formation has units of kJ/mol.
32 𝑔 𝐴𝑔𝐶𝑙
1 𝑚𝑜𝑙 𝐴𝑔𝐶𝑙
×
= 0.2233 𝑚𝑜𝑙 𝐴𝑔𝐶𝑙 = 0.22 𝑚𝑜𝑙 𝐴𝑔𝐶𝑙
1
143.32 𝑔 𝐴𝑔𝐶𝑙
𝑘𝐽
28.36 𝑘𝐽
𝑘𝐽
𝑘𝐽
∆𝐻 =
=
= 127.01
= 127.0
𝑚𝑜𝑙 0.2233 𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚𝑜𝑙
Heat Changes During Solution Formation
9. A 2.0 g sample of a solid dissolves in 50 g of liquid water at 25.0°C. Once
the solid is completely dissolve, the water temperature drops to 15.2°C.
a. Is this process endothermic or exothermic? Explain.
– The temperature of the water decreases, so the dissolving process must have absorbed that heat,
making it endothermic.
b. Calculate the amount of heat exchanged for this process.
𝑞 = 𝑚 × 𝑐 × ∆𝑇
4.18 𝐽
𝑞 = (50 𝑔) ×
× (−9.8 ℃)
𝑔∙℃
∆𝑇 = 𝑇2 − 𝑇1
= 15.2 ℃ − 25.0 ℃
= −9.8 ℃
𝑞 = 2048.2 𝐽
𝑞 = 2.0 × 103 𝐽
Heating and Cooling Curves
10. Using the word bank, label the heating curve below.
Boiling Point/
Condensation Point
Gas
Liquid
Melting Point/
Freezing Point
Solid
Word Bank
 Solid
 Liquid
 Gas
 Melting Point
 Freezing Point
 Boiling Point
 Condensation Point
 Phase Change
Phase change occurs
where graph is flat!
Time (Min) Temperature (°C)
0
-10
1
-5
2
0
10. Heating Curve and Data.
3
5
4
5
a. Based on the data, what is the melting point? The boiling point?
5
5
6
10
– Melting Point: 1st flat part, 5°C
7
15
nd
– Boiling Point: 2 flat part, 60°C
8
20
9
25
b. Which segment of the graph has the most kinetic energy? The least?
10
30
11
35
– Most: Gas; particles are moving the fastest and have the highest temperature
12
40
– Least: Solid; particles are vibrating in place and have the lowest temperature
13
45
14
50
c. Explain why the temperature remains constant during a phase
15
55
change.
16
60
17
60
– Heat is still being added, but the temperature remains constant. The added
18
60
heat is instead being used to break apart the intermolecular attractions that
19
60
are holding the substance together in that phase (i.e. as a solid)
20
60
21
65
22
70
Heating and Cooling Curves
Behavior of Gases
and Gas Laws
Ideal vs. Real Gases
11. Under which conditions of temperature and pressure will a real gas behave
most like an ideal gas?
– Ideal gases experience no attraction between gas particles.
– High temperature: gases are moving too fast (kinetic energy too high) to attract/
interact with each other
– Low pressure: particles are diffuse (far apart), and unlikely to collide and experience
attractions
Partial Pressures
12. A sample of hydrogen gas is
collected over water at a
temperature of 29 °C.
a.
What two gases are present above
the water?
– Hydrogen gas
– Water vapor
Partial Pressures
12. A sample of hydrogen gas is
collected over water at a
temperature of 29 °C.
b.
If the total pressure is 300 mmHg,
what is the pressure of the hydrogen
gas alone?
Total = Hydrogen Gas + Water Vapor
PTotal = PHydrogen Gas + PWater Vapor
300 mmHg = PHydrogen Gas + 30 mmHg
PHydrogen Gas = 270 mmHg
Gas Pressure and Manometers
13. A sample of gas is collected in
a manometer, as shown below.
a. Which exerts more pressure, the
gas or the atmosphere? Explain.
– The atmosphere is exerting more
pressure, as the mercury is forced up
the tube towards the gas.
b. If the pressure of the atmosphere is
755 mmHg, what pressure is exerted
by the gas in the flask?
𝑃𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑒 > 𝑃𝑔𝑎𝑠
Need to subtract the height difference
(which is already in mmHg)
755 𝑚𝑚𝐻𝑔 − 23 𝑚𝑚𝐻𝑔
= 732 𝑚𝑚𝐻𝑔
Behavior of Gases
14. Consider the two cylinders below.
a. How many moles of gas are present in
each cylinder?
– Both pistons have the same initial
conditions of P, T, and V
– Therefore as long as they are ideal
gases, they have the same number of
moles of each gas.
Behavior of Gases
14. Consider the two cylinders below.
b. If the conditions in cylinder A change
to a pressure of 5 atm and a
temperature of 300 K, what is the
new volume of the gas?
P2 = 5 atm
V2 = ? L
T2 = 300 K
P1 = 2 atm
V1 = 6.5 L
T1 = 200 K
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1
𝑇2
2 𝑎𝑡𝑚 × 6.5 𝐿 5 𝑎𝑡𝑚 × 𝑉2
=
200 𝐾
300 𝐾
2 atm × 6.5 𝐿 × 300𝐾 = (200𝐾 × 5 𝑎𝑡𝑚 × 𝑉2 )
(200 𝐾 × 5 𝑎𝑡𝑚)
(200 𝐾 × 5 𝑎𝑡𝑚)
𝑉2 = 3.9 𝐿
Gas Laws: Boyle, Charles, Gay-Lussac
Gas Law
Equation
Practice
5.6 L of gas at a pressure of 1.5 atm is compressed to a volume of
4.8 L. What is the new pressure of the gas?
𝑃1 × 𝑉1 = 𝑃2 × 𝑉2
Graph
𝑃1 × 𝑉1 = 𝑃2 × 𝑉2
𝑃1 × 𝑉1
= 𝑃2
𝑉2
1.5 𝑎𝑡𝑚 × 5.6 𝐿
= 𝑃2
4.8 𝐿
Pressure
Boyle
P1: 1.5 atm
V1: 5.6 L
P2: ?
V2: 4.8 L
Volume
𝑃2 = 1.75 𝑎𝑡𝑚
Gas Laws: Boyle, Charles, Gay-Lussac
Gas Law
Equation
𝑉1 𝑉2
=
𝑇1 𝑇2
Graph
Volume
Charles
Practice
A balloon contains 45 L of helium at 25°C. If the temperature is
increased to 55°C, what will the new volume of the balloon be?
V1: 45 L
T1: 25 °C + 273 = 298 K
V2: ?
T2: 55 °C + 273 = 328 K
𝑉1 𝑉2
=
𝑇1 𝑇2
𝑉1 × 𝑇2
= 𝑉2
𝑇1
45 𝐿 × 328 𝐾
= 𝑉2
298 𝐾
Temperature
𝑉2 = 49.5 𝐿
Gas Laws: Boyle, Charles, Gay-Lussac
Gas Law
Equation
𝑃1 𝑃2
=
𝑇1 𝑇2
Graph
Pressure
GayLussac
Practice
A container of gas is initially at 0.500 atm and 25 ˚C. What will the
pressure be at 125 ˚C?
P1: 0.500 atm
T1: 25 °C + 273 = 298 K
P2: ?
T2: 125 °C + 273 = 398 K
𝑃1 𝑃2
=
𝑇1 𝑇2
𝑃1 × 𝑇2
= 𝑃2
𝑇1
0.500 𝑎𝑡𝑚 × 398 𝐾
= 𝑃2
298 𝐾
Temperature
𝑃2 = 0.67 𝑎𝑡𝑚