Limiting and exess reactants.ppt

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6-1
Chapter 6:
Quantities in
Chemical
Reactions

6-2
Questions for Consideration
1. What do the coefficients in balanced equations
represent?
2. How can we use a balanced equation to relate the
number of moles of reactants and products in a chemical
reaction?
3. How can we use a balanced equation to relate the mass
of reactants and products in a chemical reaction?
4. How do we determine which reactant limits the amount
of product that can form?
5. How can we compare the amount of product we actually
obtain to the amount we expect to obtain?
6. How can we describe and measure energy changes?
7. How are heat changes involved in chemical reactions?

6-3
Chapter 6 Topics:
1. The Meaning of a Balanced Equation
2. Mole-Mole Conversions
3. Mass-Mass Conversions
4. Limiting Reactants
5. Percent Yield
6. Energy Changes
7. Heat Changes in Chemical Reactions

6-4
Introduction
 How can we predict
amounts of reactants
and products in a
reaction, such as that in
an internal combustion
engine?
 How can we predict the
amount of heat
generated or absorbed
during a reaction?
Figure 6.2

6-5
Internal Combustion Engine
 In an ICE, octane
burns with oxygen to
produce hot gases that
push against a piston
to do work.
 The amount of oxygen
that reacts is
dependent upon the
amount of octane that
burns.
 The amount of energy
produced also depends
on the amount of
octane that reacts.
Figure 6.3 (A)

6-6
Hydrogen Fuel Cell
 In a hydrogen fuel
cell, hydrogen reacts
with oxygen to water
and electrical energy.
 The amount of oxygen
that reacts is
dependent upon the
amount of hydrogen
available.
 The amount of energy
produced also depends
on the amount of
hydrogen that reacts.
Figure 6.3 (B)

6-7
6.1 The Meaning of a Balanced
Equation
 What does a balanced equation tell us about the
relative amounts of reactants and products? Let’s
consider the combustion of propane:
Figure 6.4

6-8
What do the coefficients in a balanced
chemical equation mean?
 Balanced, the equation is:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
 The coefficients in the balanced equation tell us
about the relative numbers of reactants that combine
and products that form. There is an implied
coefficient of 1 in front of C3H8(g).

6-9
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
 For every 1 molecule of propane that reacts with 5
molecules of oxygen gas, 3 molecules of carbon
dioxide and 4 molecules of water are produced.
Figure 6.4

6-10
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
 The coefficients in the balanced equation also tell us
about the relative numbers of moles of reactants and
products.
 For every 1 mole of C3H8(g) that reacts with 5 moles
of O2(g), 3 moles of CO2(g) and 4 moles of H2O(g) are
formed.

6-11
Summary – The Meaning of the Coefficients

6-12
Amounts of
Reactants and Products
 The process of determining the amount of a reactant
or product from another reactant or product in a
reaction is called stoichiometry.
 We use the mole relationships provided in a
balanced equation to calculate amounts of reactants
and products in a reactions.

6-13
6.2 Mole-Mole Conversions
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
 A mole ratio is used to relate the number of
moles of one reactant or product to another.
 Mole ratios are obtained from the coefficients in
the balanced equation.
 For example, the mole ratio of O2 to C3H8 is 5:1
or:
2
3 8
5 mol O
1 mol C H

6-14
Mole Ratios
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
 The mole ratio of O2 to C3H8 allows us to
calculate the amount of O2 that will react with
any amount of C3H8 that reacts. The mole ratio
is used as a conversion factor in a dimensional
analysis equation.
 If 0.40 mol C3H8 reacts:
3 8
0.40 mol C H 2
3 8
5 mol O
1 mol C H
 2
= 2.0 mol O

6-15
Activity: Mole-Mole Conversions
 Benzene (C6H6) burns in air according to the
following equation:
2C6H6(g) + 15O2(g)  12CO2(g) + 6H2O(g)
1. What is the mole ratio of O2 to C6H6?
2. How many moles of O2 are required to react
with each mole C6H6?
3. How many moles of O2 are required to react
with 0.38 mole of C6H6?

6-16
Activity Solutions: Mole-Mole Conversions
2C6H6(g) + 15O2(g)  12CO2(g) + 6H2O(g)
1. What is the mole ratio of O2 to C6H6?
Based on the coefficients in the balanced chemical equation,
the mole ratio can be written two ways:
15 moles O2 OR 2 moles C6H6
2 moles C6H6 15 moles O2
2. How many moles of O2 are required to react with each
mole C6H6?
15/2 or 7.5 moles of O2 are required to react with each mole
of C6H6
3. How many moles of O2 are required to react with 0.38
mole of C6H6?
6 6
0.38 mol C H 2
6 6
15 mol O
2 mol C H
 2
= 2.9 mol O

6-17
Activity: Mole Ratios
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
 What is the mole ratio for determining the moles
of CO2 that will be produced when 2.3 mol O2
reacts?
 How many moles of CO2 will be produced?
2 2
2.3 mol O = ? mol CO

2 2
moles CO = 2.3 mol O 2
2
3 mol CO
5 mol O
 2
= 1.4 mol CO
2
2
3 mol CO
5 mol O

6-18
Activity: Mole Ratios
2Al(s) + 3Cl2(g)  2AlCl3(g)
 How many moles of Cl2 are required to prepare
0.62 mol AlCl3?
2 3
moles Cl = 0.62 mol AlCl 2
3
3 mol Cl
2 mol AlCl
 2
= 0.93 mol Cl

6-19
6.3 Mass-Mass Conversions
 2Na(s) + Cl2(g)  2NaCl(s)
 What mass of chlorine gas is required to
react with 9.20 grams of sodium?
 We don’t measure reactants and products in
moles, but we commonly measure their mass.
 The balanced equation does not tell us a mass
relationship.
 How do we convert grams of reactant or
product to moles?

6-20
Mass-Mass Conversions
2Na(s) + Cl2(g)  2NaCl(s)
(9.20 g Na)
1. We convert grams to moles using molar
mass:
9.20 g Na
1 mol Na
22.99 g Na
 0.400 mol Na

For a
review, see
Section 4.2
or Math
Toolbox 4.1.
Figure from p. 220

6-21
(9.20 g Na)
(0.400 mol)
2. Next we relate moles of Na to moles of Cl2
using the mole ratio:
0.400 mol Na 2
1 mol Cl
2 mol Na
 2
0.200 mol Cl

Figure from p. 220

6-22
(9.20 g Na)
(0.400 mol Na) (0.200 mol Cl2)
3. The last step is to convert moles of Cl2 to
grams using the molar mass of Cl2:
2
0.200 mol Cl 2
2
70.90 g Cl
1 mol Cl
 2
14.2 g Cl

Figure from p. 220

6-23
(9.20 g Na) (14.2 g Cl2)
(0.400 mol Na) (0.200 mol Cl2)
3. The last step is to convert moles of Cl2 to
grams using the molar mass of Cl2:
2
0.200 mol Cl 2
2
70.90 g Cl
1 mol Cl
 2
14.2 g Cl


6-24
Steps for Mass-Mass Conversions
From page 221
9.20 g Na
1 mol Na

22.99 g Na
2
1 mol Cl

2 mol Na
2
2
70.90 g Cl
1 mol Cl
 2
14.2 g Cl


6-25
 What mass of NaCl should be produced when
9.20 g Na reacts with 14.2 g Cl2?
9.20 g Na
1 mol Na

22.99 g Na
2 mol NaCl

2 mol Na
58.44 g NaCl
1 mol NaCl
 23.4 g NaCl

2
14.2 g Cl
2
1 mol Cl

2
70.90 g Cl
2 mol NaCl

2
1 mol Cl
58.44 g NaCl
1 mol NaCl
 23.4 g NaCl

2
Or, from the law of conservation of mass:
9.20 g Na + 14.2 g Cl = 23.4 g NaCl

6-26
 What mass of NaCl should be produced when
9.20 g Na reacts with 14.2 g Cl2?
Figure 6.6

6-27
Activity: Mass-Mass Conversions
 Given:
2 H2(g) + O2(g)  2 H2O(g)
If 1.8 × 108 g of hydrogen was used in the liftoff
during a shuttle launch, then:
1. What mass of oxygen gas was consumed?
2. What mass of water vapor was produced?
3. Does the mass of the water vapor equal the
masses of the reactants? If so, then what law
describes this observation?

6-28
Activity Solutions: Mass-Mass Conversions
2 H2(g) + O2(g)  2 H2O(g)
1. What mass of oxygen gas was consumed?
Start with the number that was given in the problem –
1.8×108 g of H2(g). Convert to moles via the MM(H2), then
use a mole ratio to relate H2 to O2. Finally, convert from
moles of O2 to grams of O2 using the MM(O2).
Mass H2
(Given in
Problem)
Moles H2 Moles O2 Mass O2
MM(H2) MM(O2)
Mole
Ratio
8
2
1.8×10 g H
2
1 mol H
×
2
2.016 g H
2
1 mol O
×
2
2 mol H
2
2
32.00 g O
×
1 mol O
9
2
= 1.4×10 g O

6-29
2 H2(g) + O2(g) → 2 H2O(g)
2. What mass of water vapor was produced? Start again
with the number that was given in the problem: 1.8×108 of
H2(g). Convert to moles using the MM(H2), then use a mole
ratio to relate H2 to H2O. Finally, convert from moles of H2O
to grams of H2O using the MM(H2O).
Mass H2
(Given in
Problem)
Moles H2
Moles
H2O
Mass H2O
MM(H2) MM(H2O)
Mole
Ratio
8
2
1.8×10 g H
2
1 mol H
×
2
2.016 g H
2
2 mol H O
×
2
2 mol H
2
2
18.02 g H O
×
1 mol H O
9
2
= 1.6×10 g H O

6-30
 Given:
2 H2(g) + O2(g)  2 H2O(g)
1.8 × 108 g H2 + 1.4 × 109 g O2 = 1.6 × 109 g H2O
3. Does the mass of the water vapor equal the masses of
the reactants?
Yes or close to it – hydrogen and oxygen added
together equals 1.58 × 109 g H2O.
If so, then what law describes this observation?
The Law of Conservation of Mass

6-31
The Law of Conservation of Mass
 The law of conservation of mass states that the masses
of the reactants that are consumed must equal the
masses of the products that are formed.
Mass consumed reactants = Mass formed products
Figure 6.6

6-32
Activity: Mass-Mass Conversions
 When aluminum metal is exposed to oxygen gas, a
coating of aluminum oxide forms on the surface of
the aluminum. The balanced equation for the
reaction of aluminum metal with oxygen gas is:
4Al(s) + 3O2(g)  2Al2O3(s)
Suppose a sheet of pure aluminum gains 0.0900 g of
mass when exposed to air. Assume that this gain can
be attributed to its reaction with oxygen.
1. What mass of O2 reacted with the Al?
2. What mass of Al is reacted?
3. What mass of Al2O3 is formed?

6-33
Activity Solutions: Mass-Mass
Conversions
4Al(s) + 3O2(g)  2Al2O3(s)
Suppose a sheet of pure aluminum gains 0.0900 g of mass
when exposed to air. Assume that this gain can be
attributed to its reaction with oxygen.
1. What mass of O2 reacted with the Al?
0.0900 g O2
2. What mass of Al is reacted?
3. What mass of Al2O3 is formed?
2
0.0900 g O
2
1 mol O
×
2
32.00 g O
4 mol Al
×
2
3 mol O
26.98 g Al
×
1 mol Al
= 0.101g Al
2
0.0900 g O
2
1 mol O
×
2
32.00 g O
2 3
2 mol Al O
×
2
3 mol O
2 3
101.96 g Al O
×
1 mol Al
2 3
= 0.191g Al O

6-34
6.4 Limiting Reactants
 When two reactants are mixed, one usually does not react
completely because there is too much of it.
 In this reaction solid magnesium metal reacts with aqueous
hydrochloric acid (A). One reactant is in excess in B and the
other reactant is in excess in C.
 Can you identify the excess reactant in each case?
Figure 6.7

6-35
Limiting Reactant
 The limiting reactant is the reactant that reacts
completely and is therefore not present when the
reaction is complete.
 Since the limiting reactant reacts completely, its
amount determines the amount of product that
can form.

6-36
What is the limiting reactant?
Cu(s) + AgNO3(aq) 
Figure from
Example 6.3

6-37
Consider an Analogy:
The construction of a model solar car
Figure 6.8 (A)

6-38
If you have 4 frames, 5 solar cells, 6
motors, and 12 wheels, how many solar
cars can you make? Figure 6.8 (A and B)

6-39
What is the limiting part?
Figure 6.8 (B)
 The wheel is the limiting part because it will be
completely used up.
 The number of cars made depends on the number
of wheels available. If there were 4 more wheels,
then another car could have been made with the
remaining parts.

6-40
How many solar cars?
Figure 6.8 B( and C)

6-41
Limiting Reactants at a Molecular
Level
 In a chemical reaction, the balanced equation
tells us the relative number of molecules (or
moles) that combine in the reaction.
Figure 6.9

6-42
Limiting Reactants at a Molecular
Level
 If reactants are not present in this ratio, then
there will be a limiting reactant and excess of
the other reactant.
Figure 6.10

6-43
Activity: Limiting Reactant
2H2(g) + O2(g)  2H2O(g)
 Complete the after picture. How many H2O
molecules form?
 What is the limiting reactant? What is in excess?
Figure from p. 251

6-44
Activity Solution: Limiting Reactant
2H2(g) + O2(g)  2H2O(g)
What is the limiting reactant?
Using molecule ratios from the balanced chemical
equation:
The limiting reactant is O2 because there is only enough of
it to create 6 molecules of H2O.
Which reactant is left over?
H2 is left over at the end of the reaction.
2
3 molecules O 2
2
2 molecules H O
1 molecule O
 2
2
= 6 molecules H O
8 molecules H 2
2
2 molecules H O
2 molecules H
 2
= 8 molecules H O

6-45
2H2(g) + O2(g)  2H2O(g)
 You can also use “molecule” ratios to determine the limiting
reactant. The 8 H2 molecules need 4 O2 molecules to react
with them. There are only 3 O2 molecules, so all the H2 cannot
react.
 H2 is in excess, and O2 is the limiting reactant.
Figure from p. 251

6-46
Activity Solution: Limiting Reactant
 The molecular-level diagram shows a mixture of reactant
molecules (three O2 molecules and 8 H2 molecules) for the
following reaction:
2H2(g) + O2(g)  2H2O(g)
The after picture should have 6 H2O molecules and 2 H2
molecules:
Figure from
p. 251

6-47
Steps for Determining the
Limiting Reactant
1. Calculate the amount of one reactant (B) needed to
react with the other reactant (A).
2. Compare the calculated amount of B (amount
needed) to the actual amount of B that is given.
a. If calculated B = actual B, there is no limiting reactant.
Both A and B will react completely.
b. If calculated B > actual B, B is the limiting reactant. Only
B will react completely.
a. If calculated B < actual B, A is the limiting reactant. Only
A will react completely.

6-48
Activity: Limiting Reactants
(Mole Scale)
 The balanced equation for the reaction of
phosphorus and oxygen gas to form diphosphorus
pentoxide is:
P4(s) + 5O2(g)  2P2O5(s)
What is the limiting reactant when each of the
following sets of quantities of reactants is mixed?
a. 0.50 mol P4 and 5.0 mol O2
b. 0.20 mol P4 and 1.0 mol O2
c. 0.25 mol P4 and 0.75 mol O2

6-49
Activity Solutions: Limiting Reactants
(Mole Scale)
P4(s) + 5O2(g) → 2P2O5(s)
a. 0.50 mol P4 and 5.0 mol O2
Each mole of P4 requires 5 moles of O2, so 0.50
mol P4 requires:
The amount of O2 present is more than the amount
required, so P4 is the limiting reactant and O2 is
present in excess.
4
0.50 mol P 2
4
5 mol O
1 mol P
 2
= 2.5 mol O

6-50
(Mole Scale)
P4(s) + 5O2(g)  2P2O5(s)
b. 0.20 mol P4 and 1.0 mol O2
Each mole of P4 requires 5 moles of O2, so 0.20
mol P4 requires:
Since this is the amount of O2 present, there is no
limiting reactant. Both reactants are consumed
completely.
4
0.20 mol P 2
4
5 mol O
1 mol P
 2
= 1.0 mol O

6-51
(Mole Scale)
P4(s) + 5O2(g)  2P2O5(s)
c. 0.25 mol P4 and 0.75 mol O2
Each mole of P4 requires 5 moles of O2, so 0.25 mol
P4 requires:
Since the required amount of O2 is greater than that
present, the limiting reactant is O2 and P4 is present
in excess.
4
0.25 mol P 2
4
5 mol O
1 mol P
 2
= 1.25 mol O

6-52
Limiting Reactants – Mole Scale
 N2(g) + 3H2(g)  2NH3(g)
1. Identify the limiting reactant when the following
are mixed:
a) 2.0 mol N2 and 5.0 mole H2
b) 3.10 mole N2 and 10.2 mol H2
2. How many moles of NH3 can be produced in
each case?

6-53
Limiting Reactants – Mole Scale
 N2(g) + 3H2(g)  2NH3(g)
1. Identify the limiting reactant when the following
are mixed:
a) 2.0 mol N2 and 5.0 mole H2
b) 3.10 mole N2 and 10.2 mol H2
2. How many moles of NH3 can be produced in
each case?
a) 3.3 mole NH3
b) 6.2 mole NH3

6-54
Limiting Reactants (Mass Scale)
 Two approaches can be used when masses of the
reactants are given. In each approach, it is
necessary to convert masses of reactants to moles
of reactants as a first step.
 In the preferred approach, determine the limiting
reactant on a mole scale as before, then proceed to
calculate the mass.
 In an alternate approach, calculate the amount of
product predicted from each reactant. The
calculation that gives the least amount of product
identifies the limiting reactant.

6-55
Activity: Limiting Reactants
(Mass Scale)
 The balanced chemical equation for the reaction of
aluminum metal and chlorine gas is:
2Al(s) + 3Cl2(g)  2AlCl3(s)
Assume 0.40 g Al is mixed with 0.60 g Cl2.
a. What is the limiting reactant?
b. What is the maximum amount of AlCl3, in grams,
that can be produced?

6-56
(Mass Scale)
2Al(s) + 3Cl2(g)  2AlCl3(s)
a. What is the limiting reactant?
Calculate the moles of each reactant and determine which
is the limiting reactant as in earlier examples:
For every 2 moles of Al, 3 moles of Cl2 are required, so
0.148 moles of Al requires:
The limiting reactant is Cl2 because the required amount
is greater than the amount present.
0.40 g Al
1mol Al
×
26.98 g Al
2
= 0.148mol Al
0.60 g Cl 2
2
1mol Cl
×
70.90 g Cl
2
= 0.00846mol Cl
0.148 mol Al 2
3mol Cl
×
2 mol Al
2
= 0.222mol O

6-57
(Mass Scale)
2Al(s) + 3Cl2(g)  2AlCl3(s)
b. What is the maximum amount of AlCl3, in grams,
that can be produced?
All of the 0.00846 mol Cl2 present in the system
will react, so we use this amount to calculate the
mass of product:
2
0.00846 mol Cl
3
2 mol AlCl
×
2
3 mol Cl
3
3
133.3 g AlCl
×
1 mol AlCl
3
= 0.75 g AlCl

6-58
6.5 Percent Yield
 The amount of product
actually obtained in the
lab (actual yield) is
usually less than the
amount predicted by
calculations (theoretical
yield).
 Yields describe the
amount of product, and
can be in mass units,
moles, or number of
molecules.
Figure 6.11

6-59
Percent Yield
 The percent yield describes how much of a
product is actually obtained relative to the
amount that should form assuming
complete reaction of the limiting reactant.
actual yield
Percent Yield = 100%
theoretical yield


6-60
Activity: Percent Yield
 2Na(s) + Cl2(g)  2NaCl(s)
 If 0.20 mol chlorine reacts with excess sodium,
and 0.40 mol NaCl are produced, what is the
percent yield for the reaction?
0.40 mol
Percent Yield = ×100% =100%
0.40 mol
actual yield
Percent Yield = ×100%
theoretical yield
2
Theoretical yield = 0.20 mol Cl
2
2mol NaCl
×
1 mol Cl
= 0.40mol NaCl

6-61
 2Na(s) + Cl2(g)  2NaCl(s)
 If 0.450 mol chlorine reacts with excess
sodium, and 0.385 mol NaCl are produced,
what is the percent yield for the reaction?
0.385 mol
Percent Yield = ×100% = 42.8%
0.900 mol
actual yield
theoretical yield
2
Theoretical yield = 0.450 mol Cl
2
2mol NaCl
×
1 mol Cl
= 0.900mol NaCl

6-62
 A student was synthesizing aspirin in the laboratory. Using
the amount of limiting reactant, she calculated the amount of
aspirin that should form as 8.95 g. When she weighed her
aspirin product on the balance, its mass was 7.44 g.
a. What is the actual yield of the aspirin?
b. What is the theoretical yield of the aspirin?
c. Calculate the percent yield for this synthesis.

6-63
Activity Solutions: Percent Yield
 A student was synthesizing aspirin in the laboratory. Using
the amount of limiting reactant, she calculated the amount of
aspirin that should form as 8.95 g. When she weighed her
aspirin product on the balance, its mass was 7.44 g.
a. What is the actual yield of the aspirin?
7.44 g aspirin
b. What is the theoretical yield of the aspirin?
8.95 g aspirin
c. Calculate the percent yield for this synthesis.
actual yield
theoretical yield
7.44 g
= ×100% = 83.1%
8.95 g

6-64
6.6 Energy Changes
 When methane reacts with
oxygen when you use a gas
stove, it’s obvious that an
energy change is occurring.
Heat is released to the
surroundings and is used to
heat water and cook food.
 All chemical and physical
changes are accompanied
by energy changes. Figure 5.8F

6-65
Law of Conservation of
Energy
 When energy is released from the atoms of a
chemical reaction, it is transferred to the
surroundings. The amount released by the reaction
is equal to the amount absorbed by the surroundings.
 Energy can also be converted from one type to
another.
In an ICE,
32% of the
energy is
converted
to work.
Figure
6.12

6-66
Conversion Efficiencies
 Efficiency is
the amount of
useful work
that is
achieved from
an energy
conversion.
Figure 6.13

6-67
Efficiency
 No conversion is totally efficient – some amount of
energy is always lost to the surrounding as heat.
 The efficiency of a conversion is the percentage
that ends up in the form that we want.
Figure 6.14

6-68
Energy Changes that Accompany
Chemical Reactions
 Is energy absorbed or released by the reaction in each
case? What happens to the temperature of the
surroundings?
Figure
6.15

6-69
Endothermic and Exothermic Reactions
 Most chemical reactions are
exothermic – they release
energy to the surroundings.
 e.g. 2H2(g) + O2(g)  2H2O(g)
 Some reactions are endothermic,
absorbing energy from the
surroundings and cooling them.
Figure 5.5
Figure 5.8C
Figure from p. 250

6-70
Quantities of Heat
 Heat is energy that is transferred
between two objects because of a
difference in their temperatures.
 In chemistry, quantities of energy
(and heat) are usually expressed
with units of joules (J) or calories
(cal).
 1 cal = 4.184 J
 Nutritionists use the Calorie (Cal)
which is a kilocalorie (kcal), or
1000 cal.
1 Cal = 1000 cal = 1 kcal Figure 6.16

6-71
 One serving of this cereal
has 190 Calories.
 What is the energy
content in units of joules?
1 Cal = 1000 cal = 1 kcal
4.184 J = 1 cal
Figure 6.16
Energy =190 Cal
1000 cal
×
1 Cal
4.184 J
×
1 cal
5
= 7.9×10 J
Activity: Units of Energy

6-72
Specific Heat
 When heat is added to a substance, the substance
increases in temperature.
 The amount of heat required to increase the
temperature of 1 gram of a substance is dependent
upon the identity of that substance, and is called
the specific heat of that substance.
 Specific heat units: J/(g ºC) or cal/(g ºC)
 The specific heat of water is 4.184 J/(g ºC). Other
values are given in Table 6.2.

6-73

6-74
Activity: Specific Heat
 How much heat is required to increase the
temperature of 1.00 g of water by 10.0 ºC?
 How much heat is required to increase the
temperature of 100.0 g of water by 1.00 ºC?
J
= 4.184
g
q o
C
1.00 g
 o
10.0 C
 = 41.8 J
J
= 4.184
g
q o
100.0 g
C
 o
1.00 C
 = 418 J

6-75
Activity: Example 6.11

6-76
 heat = mass × specific heat × temperature change
 q = m × C × ΔT
 How much heat must be added to 15.0 g of water to
increase its temperature from 25.0 to 75.0 ºC?
 What is the heat change when 15.0 grams of water
cools from 75.0 to 25.0 ºC?
= 15.0 g
q
J
4.184
g
 o
C
o
50.0 C
 = 3140 J
= 15.0 g
q
J
4.184
g
 o
C
o
-50.0 C
 = -3140 J

6-77
Meaning of the Sign of q
qsystem + qsurroundings = 0
Figure 6.17

6-78
 The specific heat of aluminum is 0.895 J/(g°C). If
156 g of aluminum at 75.0°C is cooled to 25.5°C, how
much heat is transferred? What is the sign of q, and
what is its significance?

6-79
Activity Solution: Specific Heat
 The specific heat of aluminum is 0.895 J/(g° C). If
156 g of aluminum at 75.0°C is cooled to 25.5°C,
how much heat is transferred? What is the sign of
q, and what is its significance?
q = mC∆T
∆T = Tf – Ti = 25.5°C – 75.0°C = -49.5°C
q = 156 g × 0.895 J/(g°C) × -49.5°C
q = -6.91 × 103 J
The sign of q is negative, which means that this process is
exothermic and therefore releases 6.91 × 103 J into its
surroundings.

6-80
 If a 1 kJ quantity of heat was added to each of the
following 50.0-gram substances, which would
increase in temperature by the greatest amount?
 Aluminum: 0.895 J/(g ºC)
 Copper: 0.377 J/(g ºC)
 Lead: 0.129 J/(g ºC)
The greatest
temperature change
would be for the
metal with the
smallest value of the
specific heat, lead.

6-81
Calorimetry
 Calorimetry is used to
determine the heat
change of a system by
measuring the heat
change of its
surroundings.
 In this calorimeter, an
insulated cup is used
so the surroundings is
limited to what is
inside the cup.
Figure 6.18
qsystem + qsurroundings = 0

6-82
Activity: Calorimetry
 A 92.0-g piece of
copper pipe is heated
and then placed into
this calorimeter with
100.0 g of water at
25.00C. The final
temperature of the
mixture is 29.45C.
 What is the heat
change (q) of the
pipe?
Figure 6.18

6-83
Activity Solution: Calorimetry
 q = m × C × ΔT
 qwater = 100.0 g × 4.184 J/(gºC) × (29.4525.00 ºC)
 qwater = 1860 J
 qpipe + qwater = 0
 qpipe=  qwater
 qpipe =  (1860 J)
 qpipe =  1860 J
 The copper pipe released 1860 J to the
surroundings.
Figure 6.18 Figure 6.17 (A)

6-84
Activity: Calorimetry
 A sample of a metal alloy is heated and then placed
in 125.0 g of water held in a calorimeter at 22.5°C.
The final temperature of the water is 29.0°C.
Assume heat exchange only occurs between the
water and the alloy.
a. Was the initial temperature of the alloy greater
than or less than the initial temperature of the
water?
b. What is the heat change of the alloy?

6-85
Activity Solutions: Calorimetry
a. Was the initial temperature of the alloy greater than or
less than the initial temperature of the water?
Since the temperature of the water increases, the initial
temperature of the alloy must be greater than the initial
temperature of the water.
b. What is the heat change of the alloy?
qalloy + qwater = 0
qalloy = -qwater
qwater = mwaterCwater∆Twater
∆Twater = Tf – Ti = 29.0°C – 22.5°C = 6.5°C
qwater = 125.0 g × 4.184 J/(g°C) × 6.5°C = 3.4 × 103 J
qalloy = -qwater = -3.4 × 103 J

6-86
6.7 Heat Changes in Chemical
Reactions
 The heat change for a chemical reaction can be
determined in the same way as the heat change
for the piece of copper pipe, if the reaction
takes place in solution.
qreaction+ qsurroundings = 0
 If the reaction does not take place in solution,
such as a combustion reaction, then a bomb
calorimeter is required.

6-87
Bomb Calorimeter
(For Non-Aqueous Reactions)
Figure 6.19

6-88
Bomb Calorimeter
(For Non-Aqueous Reactions)
 When an exothermic
reaction takes place,
the water in the
calorimeter absorbs
the heat from the
reaction and increases
in temperature.
 qreaction = qcalorimeter
Figure 6.19
qreaction + qcalorimeter = 0

6-89
Activity: Heat Changes in Chemical
Reactions
 The balanced equation for the combination reaction
of hydrogen gas and solid iodine is:
H2(g) + I2(s)  2HI(g)
The heat change q for this reaction is +53.00 kJ per
mole of I2 that reacts.
a. Is this reaction endothermic or exothermic?
b. What is the energy change when 2.50 mol of I2
reacts?

6-90
Activity Solutions: Heat Changes in
Chemical Reactions
H2(g) + I2(s)  2HI(g)
The heat change q for this reaction is +53.00
kJ per mole of I2 that reacts.
a. Is this reaction endothermic or exothermic?
Endothermic (indicated by the positive sign on q)
b. What is the energy change when 2.50 mol of I2
reacts?
2
2.50 mol I
2
+53.00 kJ
×
1 mol I
= +133 kJ

6-91
Activity: Bomb Calorimeter
 When 5.00 grams of methanol
is burned in a calorimeter,
the temperature of the
calorimeter increases from
20.0 to 35.0ºC. The heat
capacity of the calorimeter is
7.70 kJ/ ºC.
 What is the heat of
combustion per gram of
methanol? Figure 6.19

6-92
Activity Solution: Bomb Calorimeter
 When 5.00 grams of methanol is burned in a
calorimeter, the temperature of the calorimeter
increases from 20.0 to 35.0ºC. The heat capacity
of the calorimeter is 7.70 kJ/ ºC.
 What is the heat of combustion per gram of
methanol?
 First calculate the heat change for the sample:
q = 7.70 kJ/oC × (35.0 – 20.0)oC = 115.5 kJ
 Then calculate the heat change per gram:
115.5 kJ/5.00 g = 23.1 kJ/g

6-93
Combustion of Fuels

6-94
Activity: Combustion of H2
 The heat of combustion of H2 is -286 kJ/mol. How
much heat is released when a balloon filled with
0.0540 mol H2 is burned?
2
= 0.0540mol H
q
2
-286kJ
×
1 mol H
= -15.4kJ

6-95
Energy Content of Foods
as presented in most nutritional labels
Figure 6.16

Limiting and exess reactants.ppt

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