Download Chem161 Chapter 6

Ch 6. Energy and Chemical
Change
Brady & Senese, 5th Ed.
Energy Is The Ability To Do Work
• Energy is the ability to do work (move mass
over a distance) or transfer heat
• Types: kinetic and potential
• kinetic: the energy of motion
• potential: the stored energy in matter
• Internal energy (E): the sum of the kinetic
and potential energy for each particle in the
system
2
Kinetic Energy: The Energy Of Motion
• KE=½mv2
• Energy can be transferred by moving
particles
• Collision of fast particles with slower
particles causes the slow particle to speed
up while the fast molecule slows
• this is why hot water cools in contact with cool
air
3
Potential Energy Depends on Position
• Potential energy increases when: objects that
attract move apart, or objects that repel move
toward each other
• Stored energy that can be converted to
kinetic energy
4
Your Turn!
Which of the following is not a form of kinetic
energy?
A. A pencil rolls across a desk
B. A pencil is sharpened
C. A pencil is heated
D. All are forms of kinetic energy
E. None are forms of kinetic energy
5
Law Of Conservation Of Energy
• Energy cannot be
created or destroyed
but can be transformed
from one form of energy
to another
• Also known as the first
law of thermodynamics
• How does water falling
over a waterfall
demonstrate this law?
6
Problems: 3, 5, 9
7
Heat And Temperature Are Not The Same
• The temperature of an object is proportional
to the average kinetic energy of its particles
—the higher the average kinetic energy, the
higher the temperature
• Heat is energy (also called thermal energy)
transferred between objects caused by
differences in their temperatures until they
reach thermal equilibrium
8
Units Of Energy
• SI unit is the Joule, J
• J = kg·m/s2
• If the calculated value is greater than 1000 J, use
the kJ
• British unit is the calorie, cal
• cal = 4.184 J (exact)
• Nutritional unit is the Calorie (note capital c),
which is one kilocalorie
• 1 Cal = 1 kcal = 4.184 kJ
9
Internal Energy is Conserved
• 1st Law of Thermodynamics: For an isolated
system the internal energy (E) is constant:
Δ E = Ef - E i = 0
Δ E = Eproduct - Ereactant = 0
• We canʼt measure the internal energy of
anything, so we measure the changes in energy
• E is a state function
• E = work + heat
10
What Is Temperature?
• Temperature (T) is proportional to the average
kinetic energy of all particle units: °C, °F, K
• Avg KE= ½ mvavg2
• At a high temperature, most molecules are
moving at higher average
11
State Function
• A property whose value depends only on the
present state of the system, not on the method
or mechanism used to arrive at that state
• Position is a state function: both train and car
travel to the same locations although their paths
vary
• The actual distance traveled does vary with path
New
York
Los
Angeles
12
Problems: 41, 43
13
Heat Transfer, q
• Heat (q) = the transfer of energy from regions
of high temperature to regions of lower
temperature
• units: J, cal, kg·m2/s2
• A calorie is the amount of energy needed to raise
the temperature of 1.00 g water from 14.5 to 15.5
°C
• A metal spoon at 25°C is placed in boiling
water. What happens?
14
Surroundings / System / Universe
• System: the reaction or area under study
• Surroundings: the rest of the universe
• Open systems can gain or lose mass and
energy across their boundaries
• i.e. the human body
• Closed systems can absorb or release
energy, but not mass, across the boundary
• i.e. a light bulb
• Isolated systems cannot exchange matter or
energy with their surroundings Adiabatic
• i.e. a stoppered Thermos bottle
15
The Sign Convention
• Endothermic systems require energy to be
added to the system, thus the q is (+)
• Exothermic reactions release energy to the
surroundings. Their q is (-)
• Energy changes are measured from the point
of view of the system
16
Your Turn!
A cast iron skillet is moved from a hot oven to a
sink full of water. Which of the following is
not true?
A. The water heats
B. The skillet cools
C. The heat transfer for the skillet has a (-) sign
D. The heat transfer for the skillet is the same
as the heat transfer for the water
E. None of these are untrue
17
Heat Capacity and Transfer
• Heat capacity (C)- the (intensive) ability of an
object with constant mass to absorb heat.
• a/k/a calorimeter constant
• Varies with the sample mass and the identity of
the substance
• Units: J/°C
• q=C×Δt
• q= heat transferred
• C= heat capacity of object
• Δt= Change in Temperature (tfinal-tinitial)
18
Learning Check:
A cup of water is used in an experiment. Its heat
capacity is known to be 720 J/ °C. How much
heat will it absorb if the experimental
temperature changed from 19.2 °C to 23.5 °C?
19
Heat Transfer and Specific Heat
• Specific heat (s)- The (extensive) ability of a
substance to store heat.
• C = m×s
• Units : J/g·°C; J/g · K; J/mol · K
• q=m×Δt×s
• q= heat transferred
• m= mass of object
• Δt= Change in Temperature (tfinal-tinitial)
20
Specific Heats
• Substances with
high specific heats
resist temperature
changes
• Note that water has
a very high specific
heat
• (this is why coastal
temperatures are
different from inland
temperatures)
Substance
Specific Heat,
J/ g °C
(25 °C)
Carbon (graphite)
0.711
Copper
0.387
Ethyl alcohol
2.45
Gold
0.129
Granite
0.803
Iron
0.4498
Lead
0.128
Olive oil
2.0
Silver
0.235
Water (liquid)
4.18
21
Learning Check:
Calculate the specific heat of a metal if it takes
235 J to raise the temperature of a 32.91 g
sample by 2.53°C.
22
Problems: 45, 47, 49
23
The First Law Of Thermodynamics
Explains Heat Transfer
• If we monitor the heat transfers (q) of all
materials involved and all work processes,
we can predict that their sum will be zero
• By monitoring the surroundings, we can
predict what is happening to our system
• Heat transfers until thermal equilibrium, thus
the final temperature is the same for all
materials
24
Learning Check:
A 43.29 g sample of solid is transferred from
boiling water (T=99.8°C) to 152 g water at
22.5°C in a coffee cup. The Twater rose to
24.3°C. Calculate the specific heat of the
solid.
qsample+ qwater + qcup=0
qcup is neglected in problem
qsample = - qwater
25
Your Turn!
What is the heat capacity of the container if
100.g of water (s=4.184 J/g·°C) at 100.°C
are added to 100. g of water at 25°C in the
container and the final temperature is
61°C?
A. 870 J/°C
B. 35 J/°C
C. -35 J/°C
D. -870 J/°C
E. None of these
26
Chemical Potential Energy
• Chemical bond: net attractive forces that
bind atomic nuclei and electrons together
• Exothermic reactions form stronger bonds in
the product than in the reactant and release
energy (-E)
• Endothermic reactions break stronger bonds
than they make and require energy (+E)
27
Work and Pistons
• Pressure = force/
area
• If the container
volume changes, the
pressure changes
• Work = -P×ΔV
• units: L • atm
• 1 L • atm = 101 J
• In expansion, ΔV>0,
and is exothermic
• Work is done by the
system in expansion
28
How does work relate to reactions?
• Work = Force · Distance
• is most often due to the expansion or contraction of
a system due to changing moles of gas.
• Gases push against the atmospheric pressure , so
Psystem = -Patm
• w = -Patm×ΔV
• The deployment of an airbag is one example of this
process.
1 C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
6 moles of gas → 7 moles of gas
29
Learning Check: P-V work
• Ethyl chloride is prepared by reaction of ethylene
with HCl. How much PV work (in J) is done if 89.5
g ethylene and 125 g of HCl are allowed to react at
atmospheric pressure and the volume change is
-71.5 L?
w = -1atm × -71.5L =71.5 L·atm
w = 0.708 J
• Calculate the work (in kilojoules) done during a
synthesis of ammonia in which the volume
contracts from 8.6 L to 4.3L at a constant external
pressure of 44 atm
w = -44atm × (4.3-8.6)L =189.2L·atm
w = 0.0019 kJ
30
Energy Can Be Transferred as Heat and
Work
• ΔE= q + w
• internal
energy
changes are
state
functions
31
Your Turn!
When TNT is combusted in air, it is according to the
following reaction:
4C6H2(NO2)3CH3(s) + 17 O2(g) →24 CO2(g) + 10H2O(l) + 6N2(g)
The reaction will do work for all of these reasons except:
A. The moles of gas increase
B. The volume of gas increases
C. The pressure of the gas increases
D. None of these
32
Problems: 23, 27
33
Calorimetry Is Used To Measure Heats Of
Reaction
• Heat of reaction: the amount of heat absorbed
or released in a chemical reaction
• Calorimeter: an apparatus used to measure
temperature changes in materials surrounding a
reaction that result from a chemical reaction
• From the temperature changes we can calculate
the heat of the reaction, q
• qv; heat measured under constant volume conditions
• qp: heat measured under constant pressure
conditions
34
Internal Energy Is Measured With A Bomb Calorimeter
• Used for reactions
in which there is
change in the
number of moles
of gas present
• Measures qv
• Immovable walls
mean that work is
zero
• ΔE = qv
35
Learning Check: Bomb Calorimeter
500. mg naphthalene (C10H8) is combusted in a
bomb calorimeter containing 1000. g of water.
The temperature of the water increases from
20.00°C to 24.37°C. The calorimeter constant
is 420 J/°C. What is the change in internal
energy for the reaction? swater = 4.184 J/g°C
qv reaction + qwater + qcal = 0 by the first law
qwater= 1000. g ×(24.37-20.00)°C× 4.184 J/g°C
qcal= 420 J/g ×(24.37-20.00)°C
qv = ΔE = -2.0×104 J
36
Enthalpy Of Combustion
• When one mole of a fuel substance is
reacted with elemental oxygen, a
combustion reaction can be written whose
enthalpy is ΔΗc˚
• Is always negative
• Learning Check: What is the equation
associated with the enthalpy of combustion
of C6H12O6(s)?
C6H12O6(s) + 9O2(g) →6CO2(g) + 6H2O(l)
37
Your Turn!
252 mg of benzoic acid, C6H5CO2H, is combusted in a
bomb calorimeter containing 814 g water at
20.00ºC. The reaction increases the temperature of
the water to 21.70. What is the internal energy
released by the process?
A.
B.
C.
D.
E.
-711 J
-2.85 J
+711 J
+2.85 J
None of these
swater = 4.184 J/g°C
38
Enthalpy Change (ΔH)
• enthalpy is the heat transferred at constant
pressure
• ΔH = qp
• ΔE = qp - P ΔV = ΔH – PΔV
• ΔH = ΔHfinal - ΔHinitial
• ΔH= ΔHproduct- ΔHreactant
39
Enthalpy Measured in a Coffee Cup
Calorimeter
• when no change in moles of gas is
expected, we may use a coffee
cup calorimeter
• the open system allows the
pressure to remain constant
• thus we measure qp
• ΔE=q + w
• since there is no change in the
moles of gas present, there is no
work
• thus we also are measuring ΔE
40
Learning Check: Coffee Cup Calorimetery
When 50.0 mL of .987 M H2SO4 is added to 50.0 mL of
1.00 M NaOH at 25.0 °C in a calorimeter, the
temperature of the aqueous solution increases to 31.7
°C. Calculate heat for the reaction per mole of limiting
reactant.
Assume that the specific heat of the solution is 4.18 J/g°C, the density is 1.00
g/mL, and that the calorimeter itself absorbs a negligible amount of heat
moles H2SO4 = 0.04935
moles NaOH = 0.0500, is limiting
qv rxn + qcal + qsoln = 0
qsoln = 100.gsoln×(31.7-25.0)°C ×4.18J/g°C
qv rxn = -2.8×103 J
qv rxn = -5.6×104 J
H2SO4(aq) + 2 NaOH(aq) → 2H2O(l) + Na2SO4(aq)
41
Your Turn!
A sample of 50.00mL of 0.125M HCl at 22.36 ºC
is added to a 50.00mL of 0.125M Ca(OH)2 at
22.36 ºC. The calorimeter constant was 72
J/g ºC. The temperature of the solution
(s=4.184 J/g ºC, d=1.00 g/mL) climbed to
23.30 ºC. Which of the following is not
true?
A. qczl=67.9 J
B. qsolution= 393.3J
C. qrxn = 461.0 J
D. qrxn = -461.0 J
E. None of these
42
Calorimetry Overview
• The equipment used depends on the reaction
type.
• If there will be no change in the moles of gas ,
we may use a coffee-cup calorimeter or a
closed system. Under these circumstances,
we measure qp.
• If there is a large change in the moles of gas,
we use a bomb calorimeter to measure qv.
43
Problems: 51, 53, 55
44
Thermochemical Equations
• Relate the energy of a reaction to the
quantities involved
• Must be balanced, but may use fractional
coefficients
• quantities are presumed to be in moles
45
Learning Check:
• 2C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)
ΔE = -2511 kJ
• The reactants (acetylene and oxygen) have 2511
kJ more energy than the products
• How many KJ are released for 1 mol C2H2?
1,256 kJ
46
Learning Check:
• 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g)
Δ H = 2816 kJ
• How many kJ are required for 44 g CO2
(MM=44.00986 g/mol)?
470 kJ
• If 100. kJ are provided, what mass of CO2 can
be converted to glucose?
9.38 g
47
Learning Check: Calorimetry of Chemical
Reactions
The meals-ready-to-eat (MRE) in the military can be
heated on a flameless heater. Assume the reaction in
the heater is Mg(s) + 2H2O(l) → Mg(OH)2(s) + H2(g)
ΔH = -353kJ
What quantity of magnesium is needed to supply the
heat required to warm 25 mL of water from 25 to 85
°C? Specific heat of water = 4.179 J/g °C. Assume the
density of the solution is the same as for water at
25°C, 1.00 g/mL
masssoln = 25 mL×1.00g/mL= 25 g
qsoln = 25 g ×(85-25) °C ×4.179 J/g °C= 6.269 ×103 J
massMg =0.43 g
48
Your Turn!
Consider the thermite reaction. The reaction is
initiated by the heat released from a fuse or
reaction The enthalpy change is –852 kJ/mol
Fe2O3 at 298K.
2 Al(s) + Fe2O3(s) →2Fe(s) + Al2O3(s)
What mass of Fe (MM=55.847 g/mol) is made
when 500. kJ are released?
A. 65.5 g
B. 0.587 g
C. 32.8 g
D. None of these
49
Learning Check: Ethyl Chloride Reaction
Revisited
• Ethyl chloride is prepared by reaction of
ethylene with HCl:
• C2H4(g) + HCl(g) → C2H5Cl(g) ΔH° = -72.3kJ
• What is the value of ΔE if 89.5 g ethylene and
125 g of HCl are allowed to react at
atmospheric pressure and the volume change
is -71.5 L?
mol HCl: 3.4283 mol C2H4: 3.1903, is Limiting
ΔHrxn =-230.66
w = -1atm × -71.5L =71.5
kJ
ΔE= -230 L·atm
kJ
MM ethylene=28.054 g/mol; MM HCl=36.4611 g/mol
50
Enthalpy Diagram
51
Problems: 57, 59
52
Hess’s Law
• The overall enthalpy change for a reaction is equal
to the sum of the enthalpy changes for individual
steps in the reaction
• For example:
ΔH= -822.2 kJ
• 2 Fe + 3/2 O2 →Fe2O3(s)&
• Fe2O3(s) + 2Al(s) →Al2O3(s) + 2Fe ΔH= -852 kJ
• 3/2 O2 + 2Al(s) →Al2O3(s ΔH= -822.2 kJ + -852 kJ
-1674 kJ
53
Rules for Adding Thermochemical
Reactions
1. When an equation is reversed—written in
the opposite direction—the sign of H must
also be reversed.
2. Formulas canceled from both sides of an
equation must be for the substance in
identical physical states.
3. If all the coefficients of an equation are
multiplied or divided by the same factor, the
value of H must likewise be multiplied or
divided by that factor.
54
Strategy for Adding Reactions Together:
1. Choose the most complex compound in the
equation (1)
2. Choose the equation (2 or 3 or…) that
contains the compound
3. Write this equation down so that the
compound is on the appropriate side of the
equation and has an appropriate coefficient
for our reaction
4. Look for the next most complex compound
55
Hess’s Law (Cont.)
5. Choose an equation that allows you to
cancel intermediates and multiply by an
appropriate coefficient
6. Add the reactions together and cancel
like terms
7. Add the energies together, modifying the
enthalpy values in the same way that
you modified the equation
•
•
if you reversed an equation, change the
sign on the enthalpy
if you doubled an equation, double the
energy
56
Learning Check:
• How can we calculate the enthalpy change for
the reaction 2 H2(g) + N2(g) → N2H4(g) using these
equations ?
• N2H4(g) + H2(g) → 2 NH3(g)
ΔH° = -187.6 kJ
• 3H2(g) + N2(g) → 2NH3(g)
ΔH° = -92.2 kJ
• Reverse the first reaction (and change sign)
• 2 NH3(g) → N2H4(g) + H2(g)
ΔH° = +187.6 kJ
• Add the second reaction (and add the enthalpy)
N2(g) → 2NH3(g)
• 3H2(g) + 2
ΔH° = -92.2 kJ
• 2 NH3(g)+ 3H2(g) + N2(g) → N2H4(g) + H2(g)+ 2NH3(g)
• 2 H2(g) + N2(g) → N2H4(g) (187.6-92.2)= +95.4 kJ
57
Learning Check:
Calculate ΔH for 2C(s) + H2(g) → C2H2(g) using:
ΔH° = -2599.2
• 2 C2H2(g) + 5O2(g) → 4 CO2(g) + 2H2O(l)
ΔH° = -393.5 kJ
• C(s) + O2(g) → CO2(g)
ΔH° = +285.9 kJ
• H2O(l) → H2(g) + ½ O2(g)
58
Your Turn!
What is the energy of the following process:
6A + 9B + 3D + 1 F→2 G
Given that:
• C → A + 2B
∆H=20.2 kJ/mol
• 2C + D →E + B ∆H=30.1 kJ/mol
• 3E + F →2G
∆H=-80.1 kJ/mol
A. 70.6 kJ
(20.2×-6)+(30.1×3)+(-80.1)
B. -29.8 kJ
C. -111.0 kJ
=-113.4
D. None of these
kJ
59
State matters!
• C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
• ΔH= -2043 kJ
• C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
• ΔH= -2219 kJ
• note that there is a difference in energy
because the states do not match
• If H2O(l) → H2O(g) ΔH = 44 kJ/mol
• 4H2O(l) → 4H2O(g) ΔH = 176 kJ/mol
• -2219 +176 kJ = -2043 kJ
60
Standard State
Most stable form of the pure substance at
• 1 atm pressure
• Stated temperature. If temperature is not
specified, assume 25 °C
• Solutions are 1M in concentration.
• Measurements made under standard state
conditions have the ° mark: ΔH°
• Most ΔH values are given for the most stable
form of the compound or element.
61
Determining the Most Stable State
• The most stable form of a substance:
• below the melting point is solid
• above the boiling point is gas
• between these temperatures is liquid
•
•
•
•
What is the standard state of GeH4?
mp -165 °C
bp -88.5 °C gas
What is the standard state of GeCl4?
mp -49.5 °C
bp 84 °C liquid
62
Problems: 61, 63, 65, 67, 69
63
Allotropes
• Are substances that have more than one
form in the same physical state
• You should know which form is the most
stable
• C, P, O and S all have multiple allotropes.
Which is the standard state for each?
64
Enthalpy Of Formation
• Is the enthalpy change ΔH°f for the formation
of 1 mole of a substance in its standard state
from elements in their standard states
• Note: ΔH°f = 0 for an element in its standard
state
• Learning Check:
• What is the equation that describes the formation
of CaCO3(s)?
Ca(s) + C(gr) +3/2 O2(g) →CaCO3(s)
65
Calculating ΔH For Reactions Using ΔH°f
ΔH°rxn = [sum of ΔH°f of all products]
– [sum of ΔH°f of all reactants]
• 2Fe(s) + 6H2O(l)
0
-285.8
→ 2Fe(OH)3(s) + 3H2(g)
-696.5
• CO2(g) + 2H2O(l) → 2O2(g) + CH4(g)
-393.5 -285.8
0
-74.8
0
ΔH°f
ΔH°f
66
Your Turn!
What is the enthalpy for the following reaction?
2 HCH3CO2(aq)
∆Hfº
-487
kJ/mol
A.
B.
C.
D.
E.
-756.28 kJ
756.28 kJ
-1545.74 kJ
1545.74 kJ
None of these
+ 2 OH-(aq)
→ 2H2O(l)
+ 2 C2H3O2(aq)
-229.94
-285.84
-542.96
kJ/mol
kJ/mol
kJ/mol
Problems: 71, 73, 75, 83, 85
68