Transcript Stoichiometry

Thermochemistry
A. Energy Changes
energy is the capacity to do work, generate heat
and/or generate electricity
energy can be converted to other forms however the
total energy of the system is conserved
(1st Law of Thermodynamics)
with every energy conversion,
energy is always lost as heat
(2nd Law of Thermodynamics)
the primary sources of energy are:
1. chemical: fossil fuels, plants
2. nuclear: uranium, hydrogen
3. solar: radiant energy, wind, hydroelectric
4. geothermal: geysers, hotsprings
there are 4 types of energy changes:
1. temperature change: 10’s kJ
2. phase change: 10’s kJ
3. chemical change: 100-1000’s kJ
4. nuclear change:
millions kJ
TYPICAL DIPLOMA QUESTION
B. Temperature Changes
temperature or kinetic energy (EK) is the energy of
motion of particles… increase in temperature means an
increase in EK
heat is the transfer of thermal energy
ΔEK is the change in kinetic energy
(results in a change in temperature)
 EK can be classified as:
1. vibrational motion: rapid
back and forth movement of
bundled atoms with no
change of location -- solid,
liquid, gas
2. rotational motion:
molecular rotation, no
change in position --liquid,
gas
3. translational motion:
motion from one point to
another --liquid, gas
heat capacity is the heat required to change the
temperature of a "unit mass" of a substance by 1°C
ΔEK = q = mcΔt
where: ΔEK = change in kinetic energy in J
q = heat energy in J
m = mass in g
Δt = change in temperature in °C
c = specific heat capacity in J/gC
Example
Find the heat required to change 2.50 g of water from
10.0C to 27.0C .
q = mcΔt
= (2.50 g)(4.19 J/gC) (27.0C - 10.0 C)
= 178.075 J
= 178 J
Your Assignment: pg 1
C. Phase Changes
1. The Basics
phase change is a change of state
phase changes always involve energy changes but do
not involve a change in temperature
are associated with potential energy and not
kinetic energy
energy from the surroundings separates the bonded
molecules (intermolecular forces) thereby
increasing their potential energy, or Ep
Types of Phase Changes
liquid
evaporation
melting
condensation
freezing
sublimation
solid
deposition
gas
endothermic = melting, evaporation, sublimation
exothermic = freezing, condensation, deposition
2. Enthalpy and Molar Enthalpy
enthalpy is the total kinetic and potential energy
of a chemical system under constant pressure and
temperature
unfortunately, the enthalpy of individual substances
cannot be measured directly (EK can with a thermometer
but how do you measure EP?)
changes in enthalpy occur whenever
heat is released or absorbed in a
physical or chemical change…
fortunately, this can be measured
change in enthalpy is measured in
J or kJ (H)
during a phase change EK remains constant
endothermic enthalpy changes
are positive values
exothermic enthalpy
changes are
negative values
molar enthalpy is the enthalpy change per mole of a
substance
molar enthalpy is measured in J/mol or kJ/mol (H)
molar enthalpy can be used to calculate the enthalpy
change of a phase change:
ΔH =
nH
where: ΔH = enthalpy change in J or kJ
n = number of moles in mol
H = molar enthalpy in J/mol or kJ/mol
Example
Find the energy required to melt 2.50 g of ice.
H = nH
= m H
M
=
2.50 g
(6.01 kJ/mol)
18.02 g/mol
= 0.8337957825 kJ
= +0.834 kJ
Your Assignment: pg 4 & 5 in workbook
D. Total Energy Calculations
a heating curve is a graph showing the phase and
temperature changes as heat is added to a
substance over time
the total energy change that a substance goes
through can be determined using a heating curve
and the formulas q=mcΔt and ΔH=nH
during temperature changes the EK of the molecules
change so you calculate heat using q = mct
during phase changes there only a change in EP
so you calculate heat using H = nH
Heating Curve For Water
H2O(l)  H2O(g)
100C BP
Temperature
(C)
0C MP
H2O(l)
H2O(s)  H2O(l)
H2O(s)
Time (min)
H2O(g)
Cooling Curve For Water
H2O(g)
100C BP
Temperature
(C)
H2O(l)  H2O(g)
H2O(l)
H2O(s)  H2O(l)
0C MP
H2O(s)
Time (min)
Steps:
1. Always draw the heating curve first!!!!
2. On the curve, put a point where you begin and a
point where you end (temperatures).
3. Determine which formulas are needed and which
part of the curve they apply to.
each new line segment, you have a new formula
diagonal lines represent a change in
temperature  q=mcΔt
horizontal lines represent a phase change
 ΔH=nH (vap or fus)
4. Perform the calculation.
Example
Find the total energy required to change 1.0 g of ice
at -20C to steam at 110C.
Heating Curve For Water

110C

100C
Temperature
(C)
0C 
-20C


Time (min)
ΔEtotal =  +  +  +  + 
= mcΔt + nHfus + mcΔt + nHvap + mcΔt
= (1.0g)(2.00J/gC)(20C) + (1.0g/18.02g/mol)
(6010J/mol) + (1.0g)(4.19J/gC)(100C) +
(1.0g/18.02g/mol) (40650J/mol) +
(1.0g)(2.02J/gC)(10C)
= 3068.545172 J
= 3.1  103 J
Total Energy Calculation Practice Question
Find the total energy required to
changed 4.4 g of methanol from a solid
at -102 ˚C to a gas at 89.0 ˚C. Fusion
of methanol occurs at -98.0 ˚C while
vaporization occurs at 65.0 ˚C.
E. Calorimetry
calorimetry is a technological process of measuring
energy changes using an isolated system
the isolated system used to determine the heat
involved in a phase change or in a chemical
reaction is called a calorimeter
Bomb Calorimeter
insulation
water
ENERGY
enclosed system
(bomb)
here’s how it works:

reacting substances are placed in the bomb

bomb is placed in the
calorimeter

initial temp of water
is recorded

reaction is initiated

final (maximum)
temp of water is
recorded
it is assumed that no energy is gained or lost by
the system except for the energy required or
released by the reaction or phase change
calculations are based on the Principle of Heat
Transfer:
HEAT LOST = HEAT GAINED
Example 1
A chemical reaction in a bomb calorimeter causes the
temperature of 500 g of water to increase in temperature
from 10.0C to 52.0C. Calculate the heat released by
this reaction. Give your answer in kJ.
HL (rxn) = HG (water)
q = mct
q = (500 g)(4.19 J/gC)(52.0C – 10.0C )
q = 87 990 J
= 87.990 kJ
= 88.0 kJ
Example 2
150 g of unknown metal X is at 100C. It is placed in a
calorimeter with 200 mL of water at 23.0C. If the
equilibrium temperature reached is 25.0C, what is the
specific heat capacity of metal X?
HL (metal) = HG (water)
mct = mct
(150g)c(100C – 25.0C) = (200g)(4.19 J/gC)(25.0C – 23.0C)
11250 c = 1676
c = 0.148977777 J/gC
c = 0.149 J/gC
Your Assignment: pg 6 in workbook
Example 3
When 80.0 g of NaOH is added to 850 mL of water at
23.0C, the temperature of the water rose to 28.5C after
the NaOH had dissolved. Calculate the molar enthalpy
of dissolving.
HL (dissolving) = HG (water)
(m/M)H = mct
(80.0g/40.00g/mol)H = (850g)(4.19J/gC)(28.5C – 23.0C)
(2 mol)H = 19588.25 J
H = 9794.125 J/mol
H = – 9.79 x 103 J/mol
or – 9.79 kJ/mol
exothermic
Example 4
When 52.5 g of LiNO3 is added to 150 mL of water in a
calorimeter the initial temperature of the water was
18.0C and after the LiNO3 the temperature was 16.5C.
Calculate the molar enthalpy of dissolving.
HL (water) = HG (dissolving)
mct = (m/M)H
(150g)(4.19J/gC)(18.0C – 16.5C) = (52.5 g/68.95 g/mol)H
942.75 J = (0.761… mol)H
1238.145 J/mol = H
+1.24 x 103 J/mol = H
endothermic
or +1.24 kJ/mol
Your Assignment: Molar Enthalpy Worksheet
Example 5
15 g of ice at 5.0C is placed in a beaker of water at
30C. Calculate the mass of the water in the beaker if
the final temperature at equilibrium is 10C.
30C
temperature

5.0C
time
10C
HL (water)
= HG (ice)
mct = mct + (m/M)Hfus + mct
m(4.19 J/gC)(30C – 10C ) = (15 g)(2.00 J/gC)(0C – (–5.0C))
+(15 g/18.02g/mol)(6010 J/mol)
+ (15 g)(4.19J/gC)(10C – 0C)
(83.8 J/g)m = 150 J + 5002.77 J… + 628.5 J
m = 68.966……g
= 69 g
Example 6
If 10 g of ice at 15C is placed in a calorimeter with
200 mL of water at 25C and stirred so that an
equilibrium is reached, what is the final temperature of
the mixture?
25C
temperature

15C
time
tf
HL (water)
= HG (ice)
mct = mct + (m/M)Hfus + mct
(200 g)(4.19 J/gC)(25C – tf)= (10 g)(2.00 J/gC)(0C – (–15.0C))
+(10 g/18.02g/mol)(6010 J/mol)
+ (10 g)(4.19J/gC)(tf – 0C)
20 950 J – (838 J/C) tf = 300 J + 3335.18313 J + (41.9 J/C) tf
17314.81687 J = (879.9 J/C) tf
19.678… C = tf
20C = tf
Assignment: Enthalpy and Phase Change
G. Chemical Change
a chemical change is a transformation
involving an energy change in which one substance
is converted into another substance
uses of chemical energy (exothermic):
1. steam generators
from burning
fossil fuels
2. motor vehicles
where fuel is
burned
3. natural gas, propane,
coal, wood burned for
heating
4. batteries
5. living organisms, cellular respiration
a calorimeter can be used to quantify the amount of
heat lost or gained by a chemical reaction (still
sticking to the heat lost = heat gained principle!!!!)
Example 1
A 2.65 g sample of methanol (CH3OH) was burned in a
calorimeter which contained 500 mL of water at 25.0C.
If the final temperature of the water is 50.0C, what is
the molar heat of combustion for methanol?
heat lost (combustion)= heat gained (water)
(m/M)H = mct
(2.65g/32.05g/mol)H = (500g)(4.19J/gC)(50.0C – 25.0C)
(0.0826… mol )H = 52375 J
H = 633441.038 J/mol
H = – 6.33 x 105 J/mol
or –633 kJ/mol
Example 2
An 8.40 g sample of N2(g) is reacted with pure oxygen in a
bomb calorimeter containing 1.00 kg of water to produce
N2O. The temperature of the water dropped by 5.82C.
What is the molar heat of reaction of N2(g)?
heat lost (water) = heat gained (formation)
mct = (m/M)Hf
(1000 g)(4.19 J/gC)(5.82C) = (8.40 g/28.02 g/mol)H
24385.8 J = (0.299… mol )H
H = 81344.06143 J/mol
H = + 8.13 x 104 J/mol
or +81.3 kJ/mol
Your Assignment: pg 7 in workbook
H. Industrial Bomb Calorimeters
industrial calorimeters are used in research to measure
the heat of combustion of food, fuel, oil, crops, and
explosives
modern calorimeters have fixed components
eg) volume of water used, container (bomb) material,
stirrer and thermometer
in calculating the energy of combustion, you take all
components of the calorimeter into account:
Etotal = mct (H2O) + mct (stirrer) + mct (bomb) +
mct (thermometer)
all of the “mc” parts are constant so they are replaced
by one constant C, the heat capacity of the entire
system in kJ/C
Example 1
A 1.50 g sample of methane is completely burned in a
calorimeter with a heat capacity of 11.3 kJ/C. The
temperature increased from 20.15C to 27.45C.
Calculate the molar enthalpy of combustion for methane.
heat lost (combustion) = heat gained (calorimeter)
(m/M)H = Ct
(1.50 g/16.05 g/mol)H = (11.3 kJ/C)(27.45C – 20.15C)
(0.0934 … mol )H= 82.49 kJ
H = 882.6430002 kJ/mol
H = – 883 kJ/mol
Example 2
When 3.00 g of butter is burned in a bomb calorimeter
with a heat capacity of 9.22 kJ/C the temperature
changes from 19.62C to 31.89C. Calculate the specific
enthalpy of combustion in kJ/g.
***note that in this question we are asked for enthalpy of
combustion in kJ/g not kJ/mol. We substitute mass in for
moles in the formula Hcomb = nHcomb
heat lost (combustion) = heat gained (calorimeter)
*** mH = Ct
(3.00g)H = (9.22 kJ/C)(31.89C – 19.62C)
(3.00 g)H= 113.1294 kJ
H = 37.7098 kJ/g
H = – 37.7 kJ/g
Your Assignment: pg 8 in workbook
I. Reaction Enthalpies
the enthalpy change of a reaction refers to changes in
EP and is called the heat of reaction
the heat of reaction, H , can be expressed in 4 ways
(vocabulary):
1. Outside Equation
the heat of reaction can be given as a H value
outside the equation
2 SO2(g) + O2(g)  2 SO3(g)
H = –197.8 kJ
Example
Calculate the molar enthalpy of reaction (H) for sulphur
dioxide using the following information:
2 SO2(g) + O2(g)  2 SO3(g)
H = –197.8 kJ
n = 2 mol
H = –197.8 kJ
H = nH
H = H
n
= –197.8 kJ
2 mol
= –98.9 kJ/mol
2. Inside Equation
the heat of reaction can be written in the equation
endothermic reaction… heat is on reactant side
eg) H2O(l) + 285.8 kJ  H2(g) + ½ O2(g)
exothermic reaction… heat is on product side
eg) Mg(s) + ½ O2(g)  MgO(s) + 601.6 kJ
Example
Calculate the molar enthalpy (H) for oxygen in the
decomposition of water using the following information:
H2O(l) + 285.8 kJ  H2(g) + ½ O2(g)
H = +285.8 kJ
n = ½ mol
H = nH
H = H
n
= +285.8 kJ
½ mol
= +571.6 kJ/mol
3. Molar Enthalpy, H
use the formula H = (m/M)H to calculate H
Example
Find the molar enthalpy when 5.0 g of butane
produces 850 kJ of energy.
H = –850 kJ
m = 5.0 g
M = 58.14 g/mol
H = (m/M)H
H = H
(m/M)
=
–850 kJ
(5.0 g/58.14 g/mol)
= –9883.8 kJ/mol
= –9.9 x 103 kJ/mol
4. Potential Energy Diagram
 reactants are separated from the products
 shape indicates whether the reaction is endothermic or
exothermic
Endothermic
Exothermic
products
EP
(kJ)
H > 0
positive
reactants
Reaction Progress
reactants
EP
(kJ)
H < 0
negative
products
Reaction Progress
Your Assignment: 1. pg 2 in workbook
2. pg 3 in workbook
J. Bond Energy
bond energy is the energy required to break a
chemical bond or the energy released when a bond
is formed
in both endothermic and exothermic reactions, the
energy required to “pull apart” the atoms in the
reactants is called the activation energy
the activation energy is always higher than the energy
contained in the reactants and the products
Endothermic
2 H2O + energy  2 H2 + O2
HHHHOO
Energy
(kJ)
activation
energy needed
to break bonds
energy released when
bonds form
2 H2 + O 2
net energy for
reaction
2 H2O
Reaction Progress
Exothermic
H2 + Cl2  2 HCl + energy
activation
energy needed
to break bonds
Energy
(kJ)
H2 + Cl2
H H Cl Cl
energy released when
bonds form
net energy
for reaction
Reaction Progress
2 HCl
 if a catalyst is used, it acts to lower the activation
energy for the reaction (***typical diploma
question)
Endothermic
2 H2O + energy  2 H2 + O2
= catalyzed reaction
HHHHOO
Energy
(kJ)
activation
energy needed
to break bonds
energy released when
bonds form
2 H2 + O2
net energy for
reaction
2 H2 O
Reaction Progress
Exothermic
H2 + Cl2  2 HCl + energy
= catalyzed reaction
activation
energy needed
to break bonds
Energy
(kJ)
H2 + Cl2
H H Cl Cl
energy released when
bonds form
net energy
for reaction
Reaction Progress
2 HCl
K. Predicting Enthalpy (Hr) Changes
1. Using Hess’s Law
because of the law of conservation of energy, the heat
of reaction is the same whether the reactants are
converted to the products in a single reaction or in
a series of reactions
G.H. Hess (1840) suggested
that if two or more
thermochemical equations
are added to give a final
equation then the enthalpies
can be added to give the
enthalpy for the final equation
 sometimes the heat of reaction for a chemical change
is not easily measured due to time of reaction, cost,
rarity of reactants etc. so we use Hess’s Law to
calculate Hr
Steps:
1. Write the net reaction, if it is not given.
2. Manipulate the given equations so they will add to
yield the net equation.
 if you multiply or divide an equation, multiply or
divide the H by the same factor
 if you flip an equation, flip the sign on H
3. Cancel the reactants and products where possible
to simplify (you should end up with your net equation!)
4. Add the component enthalpy changes to get the
net enthalpy change.
Example 1
Find the heat of reaction for C(s, di)  C(s, gr) using the
following reactions:
flip C(s, gr) + O2(g)  CO2(g)
H = –393.5 kJ
H = –395.4 kJ
C(s, di) + O2(g)  CO2(g)
CO2(g)  C(s, gr) + O2(g)
H = –395.4 kJ
H = +393.5 kJ
C(s, di) 
H = –1.9 kJ
C(s, di) + O2(g)  CO2(g)
C(s, gr)
Example 2
Find the heat of reaction for H2O2(l)  H2O(l) + ½ O2(g)
using the following reactions:
flip H2(g) + O2(g)  H2O2(l)
H2(g) + ½ O2(g)  H2O(l)
H2(g) + ½ O2(g)  H2O(l)
H = –187.8 kJ
H = –285.8 kJ
H2O2(l)  H2(g) + ½ O2(g)
H = –285.8 kJ
H = +187.8 kJ
H2O2(l)  H2O(l) + ½ O2(g)
H = –98.0 kJ
Example 3
Find the heat of reaction for C(s) + 2 H2(g)  CH4(g)
using the following reactions:
C(s) + O2(g)  CO2(g)
 2 H2(g) + ½ O2(g)  H2O(l)
flip CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
C(s) + O2(g)  CO2(g)
H = –393.5 kJ
H = –285.8 kJ
H = –890.5 kJ
H = –393.5 kJ
CO2(g) + 2 H2O(l)  CH4(s) + 2 O2(g) H = +890.5 kJ
2 H2(g) + 1 O2(g)  2 H2O(l)
C(s) + 2 H2(g)  CH4(g)
H = –571.6 kJ
H = –74.6 kJ
Your Assignment: pgs 9-10 in workbook
pg 11 in workbook
2. Using Standard Heats of Formation Hf
sometimes it is not easy to measure the heat change for
a reaction (too slow/expensive)
in this case, H can be determined using
heats of formation
heats of formation (Hf ) are the changes in EP that
occur when compounds are formed from their
elements
Hf for elements cannot be directly measured therefore
they are designated as zero…all other Hf values are
in reference to this
Hf for common compounds are listed on pages 6-7
in data booklet
the Hf is an indirect measure of the stability
of a compound
the more exothermic the formation , the more
more stable the compound
(this means you have to add that energy to
decompose it)
eg) List the following compounds in order from
most stable to least stable.
3
5
4
2
1
H2O(l)
C2H4(g)
N2O4(g)
PCl3(l)
Al2O3(s)
Hf =
Hf =
Hf =
Hf =
Hf =
–285.8 kJ/mol
+52.4 kJ/mol
+11.1 kJ/mol
–319.7 kJ /mol
–1675.7 kJ /mol
Hess’s Law formula states that the Hr is the
difference between the standard heats of formation of
the reactants and the products
Hr = nHf(products)  nHf(reactants)
Example 1
Calculate the standard heat of combustion for
2 CO(g) + O2(g)  2 CO2(g) and draw the EP diagram for this
reaction.
2 CO(g)
O2(g) 
+
(2 mol)(-110.5 kJ/mol) +
0 kJ
+
0 kJ
-221.0 kJ
2 CO2(g)
(2 mol)(-393.5 kJ/mol)
-787.0 kJ
Hc = nHf(products)  nHf(reactants)
= -787.0 kJ – (-221.0 kJ)
= -566.0 kJ
EP Diagram for 2 CO(g) + O2(g)  2 CO2(g)
-221.0 2 CO(g) + O2(g)
EP
(kJ)
-787.0
H = -566.0 kJ
2 CO2(g)
Reaction Progress
Example 2
Find the heat of combustion of ethane. The products of
combustion are gases.
2 C2H6(g)
(2mol)(-84.0kJ/mol)
-168.0 kJ
+ 7 O2(g)  4 CO2(g) + 6 H2O(g)
+ 0 kJ
+ 0 kJ
(4 mol)(-393.5 kJ/mol) + (6 mol)(-241.8
kJ/mol)
-1574.0 kJ
+ -1450.8 kJ
Hc = nHf(products)  nHf(reactants)
= (-1574.0 kJ + (-1450.8 kJ)) – (-168.0 kJ)
= -2856.8 kJ
Your Assignment: pgs 12-13 in workbook
L. Energy Systems in Biological Processes
Photosynthesis:
energy + CO2(g) + H2O(l)  C6H12O6(s) + O2(g)
Cellular Respiration:
C6H12O6(s) + O2(g)  CO2(g) + H2O(l) + energy
**Water vapor condenses into liquid
M. Nuclear Change
enthalpy changes in nuclear reactions are the result of
EP changes as rearrangements among the
subatomic particles (protons and neutrons) occur
ie) intranuclear forces
there are two types of nuclear reactions:
1. Fusion (Joining)
fusion of hydrogen to helium occurs on the sun and
other stars
these types of reaction produce the greatest amount
of energy and are necessary for life on Earth
require a great deal of heat and pressure
2
1
H + 31 H  42 He + 10 n + 1.70 x 109 kJ
Top number = mass number
Bottom number = number of protons
2. Fission (Splitting)
basis for nuclear power plants
uranium atoms can be split into two smaller nuclei
which produces large quantities of energy
was discovered in the late 1930's when uranium was
bombarded by neutrons causing it to split
235
U
92
141
1
10 kJ
+ 10 n  92
Kr
+
Ba
+
3
n
+
1.9
x
10
36
56
0
the neutrons produced by fission allow a chain
reaction to occur to keep the reaction self
sustaining
N. Society and Technological Connections
we must assess the risks and benefits of relying on
fossil fuels and nuclear energy as energy sources
we are limited by our scientific knowledge and by the
technology that has been developed to date
 fossils fuels are the most common source of energy
many aspects of our society are based on the price of
fuels like gasoline
Advantages vs. Disadvantages of Fossil Fuels
Advantages
relatively low cost
readily available (market)
plant set-up, vehicle
design, expertise
affordable
used all over the world
deposits are large
Disadvantages
release of gases that
contribute to the greenhouse
effect and acid rain when
burned
mining is detrimental to
the environment
non-renewable
Advantages vs. Disadvantages of Nuclear Power
Advantages
low cost of fuel
lots of energy from small
amount of fuel
Disadvantages
high cost of plant set-up,
expertise, decommission
cause thermal pollution
doesn’t add to greenhouse
effect, acid rain
difficult to dispose of
nuclear fuel wastes
Canada has lots of
uranium ore
possibility of
catastrophic accidents
preserves existing fossil
fuels
non-renewable