a. energy changes energy can be however the energy is the capacity to and/or thermochemistry with...
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A. Energy Changes
energy can be however the
energy is the capacity to and/or
Thermochemistry
with every energy conversion, energy is always
do work,
converted to other forms total energy
lost as heat
generate electricitygenerate heat
of the system is conserved
(1st Law of Thermodynamics)
(2nd Law of Thermodynamics)
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the primary sources of energy are:
1. chemical: fossil fuels, plants
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2. nuclear: uranium, hydrogen
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3. solar: radiant energy, wind, hydroelectric
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4. geothermal: geysers, hotsprings
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there are 4 types of energy changes:
1. temperature change:
2. phase change:
3. chemical change:
4. nuclear change:
10’s kJ
10’s kJ
100-1000’s kJ
millions kJ
TYPICAL DIPLOMA QUESTION
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temperature or kinetic energy (EK) is the energy of
heat is the transfer of
B. Temperature Changes
ΔEK is the change in
motion of particles…
thermal energy
temperature)
increase in temperature means an
in EKincrease
kinetic energy
(results in a change in
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EK can be classified as:
1. vibrational motion:
2. rotational motion:
3. translational motion:
rapid back and forth movement of bundled atoms with no change of location -- solid, liquid, gas
molecular rotation, no change in position --liquid, gas
motion from one point to another --liquid, gas
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heat capacity is the heat required to change the temperature of a "unit mass" of a substance by 1C
ΔEK = q = mcΔt
where: ΔEK =
q =m =Δt =c =
change in kinetic energy in Jheat energy in Jmass in g
change in temperature in Cspecific heat capacity in J/gC
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ExampleFind the heat required to change 2.50 g of water from 10.0C to 27.0C .
q = mcΔt = (2.50 g)(4.19 J/gC) (27.0C - 10.0 C) = 178.075 J = 178 J
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Your Assignment: pg 1
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phase change is a change of
phase changes always involve but do not involve
C. Phase Changes1. The Basics
energy from the surroundings
state
energy changes a change in temperature
are associated with and not
separates the bonded molecules (intermolecular forces)
potential energykinetic energy
thereby increasing their potential energy, or Ep
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Types of Phase Changes
liquid
gassolid
melting
freezing
evaporation
condensation
sublimation
deposition
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exothermic =
melting, evaporation, sublimation
freezing, condensation, deposition
endothermic =
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2. Enthalpy and Molar Enthalpy
enthalpy is
unfortunately, the enthalpy of individual substances cannot be measured directly (EK can with a thermometer
but how do you measure EP?)
the total kinetic and potential energy
changes in enthalpy occur whenever heat is
change in enthalpy is measured in
released or absorbed in a physical or chemical change…
J or kJ (H)
of a chemical system under constant pressure and temperature
fortunately, this can be measured
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during a phase change
endothermic enthalpy changes are
exothermic enthalpy changes are
positive values
negative values
EK remains constant
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molar enthalpy can be used to calculate the enthalpy change of a phase change:
ΔH = nH
where: ΔH = n =H =
enthalpy change in J or kJ number of moles in mol molar enthalpy in J/mol or kJ/mol
molar enthalpy is the enthalpy change
molar enthalpy is measured in
per mole of a substance
J/mol or kJ/mol (H)
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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
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Your Assignment: pg 4 & 5 in workbook
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D. Total Energy Calculations
a heating curve is a showing the and changes as heat is added to a
substance over time
the total energy change that a substance goes through can be determined using a and the formulas and
during temperature changes the of the molecules change so you calculate heat using
during phase changes there only a change in so you calculate heat using
graph phasetemperature
heating curveq=mcΔt ΔH=nH
EKq = mct
EPH = nH
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Heating Curve
H2O(l) H2O(g)
H2O(s) H2O(l)
H2O(l)
H2O(s)
H2O(g)
100C
0C
Temperature(C)
Time (min)
BP
MP
For Water
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Cooling Curve
H2O(l) H2O(g)
H2O(s) H2O(l)
H2O(l)
H2O(s)
H2O(g)
100C
0C
Temperature(C)
Time (min)
BP
MP
For Water
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Steps:
1. Always
2. On the curve, put a point where and a point
3. Determine which formulas are needed and which part of the curve they apply to.
you begin
draw the heating curve first!!!!
where you end (temperatures).
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.
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Example
Find the total energy required to change 1.0 g of ice at -20C to steam at 110C.
Heating Curve For Water
Temperature(C)
Time (min)
0C
100C
-20C
110C
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= mcΔt + nHfus + mcΔt + nHvap + mcΔt
= (1.0g)(2.00J/gC)(20C) + (1.0g/18.02g/mol) (6010J/mol) + (1.0g)(4.19J/gC)(100C) + (1.0g/18.02g/mol) (40650J/mol) + (1.0g)(2.02J/gC)(10C)
ΔEtotal = + + + +
= 3068.545172 J = 3.1 103 J
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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.
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E. Calorimetrycalorimetry is a technological process of
the isolated system used to determine the heat involved in a phase change or in a chemical reaction is called a
Bomb Calorimeter
ENERGY
insulation water
enclosed system(bomb)
measuring energy changes using an isolated system
calorimeter
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here’s how it works:
reacting substances are placed in the bomb is placed in the
bomb
calorimeter
is recorded
reaction is initiated
is
recorded
initial temp of water
final (maximum) temp of water
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it is assumed that no energy is by the system except for the energy required or released by the
calculations are based on the Principle of Heat Transfer:
HEAT LOST = HEAT GAINED
gained or lost
reaction or phase change
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Example 1
A chemical reaction in a bomb calorimeter causes the temperature of 500 g of water to increase in temperature from 10.0C to 52.0C. Calculate the heat released by this reaction. Give your answer in kJ.
HL (rxn) = HG (water) q = mct q = (500 g)(4.19 J/gC)(52.0C – 10.0C ) q = 87 990 J
= 87.990 kJ = 88.0 kJ
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Example 2
150 g of unknown metal X is at 100C. It is placed in a calorimeter with 200 mL of water at 23.0C. If the equilibrium temperature reached is 25.0C, what is the specific heat capacity of metal X?
HL (metal) = HG (water) mct = mct
(150g)c(100C – 25.0C) = (200g)(4.19 J/gC)(25.0C – 23.0C) 11250 c = 1676
c = 0.148977777 J/gC c = 0.149 J/gC
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Your Assignment: pg 6 in workbook
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Example 3
When 80.0 g of NaOH is added to 850 mL of water at 23.0C, the temperature of the water rose to 28.5C after the NaOH had dissolved. Calculate the molar enthalpy of dissolving.
HL (dissolving) = HG (water) (m/M)H = mct
(80.0g/40.00g/mol)H = (850g)(4.19J/gC)(28.5C – 23.0C)
(2 mol)H = 19588.25 J H = 9794.125 J/mol H = 9.79 x 103 J/mol
exothermic
–
or – 9.79 kJ/mol
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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 18.0C and after the LiNO3 the temperature was 16.5C.
Calculate the molar enthalpy of dissolving.
HL (water) = HG (dissolving) mct = (m/M)H
(150g)(4.19J/gC)(18.0C – 16.5C) = (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
or +1.24 kJ/mol
+endothermic
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Your Assignment: Molar Enthalpy Worksheet
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Example 5
15 g of ice at 5.0C is placed in a beaker of water at 30C. Calculate the mass of the water in the beaker if the final temperature at equilibrium is 10C.
30C
5.0C
temperature
time
10C
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HL (water) = HG (ice)mct = mct + (m/M)Hfus + mct
m(4.19 J/gC)(30C – 10C ) = (15 g)(2.00 J/gC)(0C – (–5.0C)) +(15 g/18.02g/mol)(6010 J/mol) + (15 g)(4.19J/gC)(10C – 0C)
(83.8 J/g)m = 150 J + 5002.77 J… + 628.5 J
m = 68.966……g = 69 g
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Example 6
If 10 g of ice at 15C is placed in a calorimeter with 200 mL of water at 25C and stirred so that an equilibrium is reached, what is the final temperature of the mixture?
25C
15C
temperature
time
tf
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HL (water) = HG (ice)mct = mct + (m/M)Hfus + mct
(200 g)(4.19 J/gC)(25C – tf)= (10 g)(2.00 J/gC)(0C – (–15.0C))
+(10 g/18.02g/mol)(6010 J/mol) + (10 g)(4.19J/gC)(tf – 0C)
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
20C = tf
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Assignment: Enthalpy and Phase Change
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G. Chemical Changea is a transformation involving
an energy change in which one substance is converted into another substance
chemical change
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uses of chemical energy (exothermic):
1.
2.
steam generators from burning fossil fuels
motor vehicles where fuel is burned
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4.
3. natural gas, propane, coal, wood burned for heating
batteries
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5. living organisms, cellular respiration
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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!!!!)
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Example 1
A 2.65 g sample of methanol (CH3OH) was burned in a
calorimeter which contained 500 mL of water at 25.0C. If the final temperature of the water is 50.0C, what is the molar heat of combustion for methanol?
heat lost (combustion)= heat gained (water) (m/M)H = mct
(2.65g/32.05g/mol)H = (500g)(4.19J/gC)(50.0C – 25.0C)
(0.0826… mol )H = 52375 J H = 633441.038 J/mol H = – 6.33 x 105 J/mol
or –633 kJ/mol
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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 5.82C.
What is the molar heat of reaction of N2(g)?
heat lost (water) = heat gained (formation) mct = (m/M)Hf
(1000 g)(4.19 J/gC)(5.82C) = (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
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Your Assignment: pg 7 in workbook
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H. Industrial Bomb Calorimeters industrial calorimeters are used in to measure
modern calorimeters have
eg) volume of water used, container (bomb) material, stirrer and thermometer
the heat of combustion of food, fuel, oil, crops, and explosives
research
fixed components
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Etotal =
in calculating the energy of combustion, you take all components of the calorimeter into account:
mct (H2O) + mct (stirrer) + mct (bomb) +
mct (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
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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 20.15C to 27.45C. Calculate the molar enthalpy of combustion for methane.
heat lost (combustion) = heat gained (calorimeter) (m/M)H = Ct
(1.50 g/16.05 g/mol)H = (11.3 kJ/C)(27.45C – 20.15C)
(0.0934 … mol )H= 82.49 kJ H = 882.6430002 kJ/mol H = – 883 kJ/mol
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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 19.62C to 31.89C. 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 = Ct (3.00g)H = (9.22 kJ/C)(31.89C – 19.62C)
(3.00 g)H= 113.1294 kJ H = 37.7098 kJ/g H = – 37.7 kJ/g
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Your Assignment: pg 8 in workbook
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I. Reaction Enthalpies the enthalpy change of a refers to changes in
and is called the
the heat of reaction, , 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
reactionEP heat of reaction
H
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ExampleCalculate 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
H = –197.8 kJn = 2 mol
H = nH H = H n = –197.8 kJ
2 mol = –98.9 kJ/mol
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2. Inside Equation
the heat of reaction can be written in the equation
endothermic reaction…
eg)
exothermic reaction…
eg)
heat is on reactant side
heat is on product side
H2O(l) + 285.8 kJ H2(g) + ½ O2(g)
Mg(s) + ½ O2(g) MgO(s) + 601.6 kJ
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ExampleCalculate 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 kJn = ½ mol
H = nH H = H n = +285.8 kJ
½ mol = +571.6 kJ/mol
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3. Molar Enthalpy, H
use the formula H = (m/M)H to calculate H
ExampleFind the molar enthalpy when 5.0 g of butane produces 850 kJ of energy.
H = –850 kJm = 5.0 gM = 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
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4. Potential Energy Diagram
reactants are separated from the products shape indicates whether the reaction is endothermic or
exothermic
Endothermic Exothermic
EP
(kJ)
EP
(kJ)
Reaction Progress Reaction Progress
reactants
products
H > 0positive
reactants
products
H < 0negative
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Your Assignment: 1. pg 2 in workbook 2. pg 3 in workbook
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J. Bond Energy
bond energy is the energy or the energy
in both endothermic and exothermic reactions, the energy required to the atoms in the
is called the
the activation energy is always than the energy contained in the reactants and the products
required to break a chemical bond
“pull apart” activation energy
released when a bond is formed
reactants
higher
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Endothermic 2 H2O + energy 2 H2 + O2
Energy(kJ)
Reaction Progress
2 H2O
H H H H O O
2 H2 + O2
net energy for reaction
activation energy needed to break bonds
energy released when bonds form
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Exothermic H2 + Cl2 2 HCl + energy
Energy(kJ)
Reaction Progress
H2 + Cl2
H H Cl Cl
2 HCl
net energy for reaction
activation energy needed to break bonds energy released when
bonds form
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if a is used, it acts to for the reaction (***typical diploma question)
catalyst lower the activation energy
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Endothermic 2 H2O + energy 2 H2 + O2
Energy(kJ)
Reaction Progress
2 H2O
H H H H O O
2 H2 + O2
net energy for reaction
activation energy needed to break bonds
energy released when bonds form
= catalyzed reaction
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Exothermic H2 + Cl2 2 HCl + energy
Energy(kJ)
Reaction Progress
H2 + Cl2
H H Cl Cl
2 HCl
net energy for reaction
activation energy needed to break bonds energy released when
bonds form
= catalyzed reaction
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K. Predicting Enthalpy (Hr) Changes
1. Using Hess’s Law
because of the law of conservation of energy, the heat of reaction is the whether the reactants are converted to the products in a or in a
G.H. Hess (1840) suggested that if two or more
are to give a final equation then the can be added to give the
samesingle reaction
series of reactions
thermochemical equationsadded
enthalpies
enthalpy for the final equation
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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
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Steps:1. Write the , if it is not given.
2. the given equations so they will to yield the if you multiply or divide an equation, multiply or
divide the H by theif you flip an equation, the sign on H
3. (you should end up with your net equation!)
the reactants and products where possible to
4. the component enthalpy changes to get the
net reaction
Manipulate addnet equation.
same factor
flip
Cancelsimplify
Addnet enthalpy change.
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Example 1Find the heat of reaction for C(s, di) C(s, gr) using the
following reactions:C(s, gr) + O2(g) CO2(g) H = –393.5 kJ
C(s, di) + O2(g) CO2(g) H = –395.4 kJ
C(s, di) + O2(g) CO2(g)
CO2(g) C(s, gr) + O2(g) H =+393.5 kJ
flip
C(s, di) C(s, gr) H = –1.9 kJ
H = –395.4 kJ
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Example 2Find the heat of reaction for H2O2(l) H2O(l) + ½ O2(g)
using the following reactions: H2(g) + O2(g) H2O2(l) H = –187.8 kJ
H2(g) + ½ O2(g) H2O(l) H = –285.8 kJ
H2(g) + ½ O2(g) H2O(l)
H2O2(l) H2(g) + O2(g) H =+187.8 kJ
flip
H2O2(l) H2O(l) H = –98.0 kJ
H = –285.8 kJ
½
+ ½ O2(g)
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Example 3Find the heat of reaction for C(s) + 2 H2(g) CH4(g) using
the following reactions: C(s) + O2(g) CO2(g) H = –393.5 kJ
H2(g) + ½ O2(g) H2O(l) H = –285.8 kJ
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) H = –890.5 kJ
C(s) + O2(g) CO2(g)
CO2(g) + 2 H2O(l) CH4(s) + 2 O2(g)
H =–571.6 kJ
flip
2 H2(g) CH4(g) H = –74.6 kJ
H = –393.5 kJ
H =+890.5 kJ
2
2 H2(g) + 1 O2(g) 2 H2O(l)
C(s) +
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Your Assignment: pgs 9-10 in workbook pg 11 in workbook
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2. Using Standard Heats of Formation Hf 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 (Hf ) are the changes in EP that
occur when
heats of formation
compounds are formed from their elements
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Hf for elements cannot be directly measured therefore
they are designated as …all other Hf values are
in reference to this
Hf for common compounds are listed on pages 6-7
in data booklet
the Hf is an indirect measure of the stability
zero
of a compound
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the more , the more
eg) List the following compounds in order from most stable to least stable.
exothermic the formation more stable the compound
(this means you have to add that energy to decompose it)
H2O(l) Hf =
C2H4(g) Hf =N2O4(g) Hf =PCl3(l) Hf =Al2O3(s) Hf =
–285.8 kJ/mol+52.4 kJ/mol+11.1 kJ/mol–319.7 kJ /mol–1675.7 kJ /mol1
4
3
2
5
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Hess’s Law formula states that the is the difference between the standard heats of formation of the and the
Hr = nHf(products) nHf(reactants)
Hr
reactants products
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Example 1Calculate 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 CO2(g)
(2 mol)(-110.5 kJ/mol) + 0 kJ (2 mol)(-393.5 kJ/mol)
-221.0 kJ + 0 kJ -787.0 kJ
Hc = nHf(products) nHf(reactants)
= -787.0 kJ – (-221.0 kJ)= -566.0 kJ
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EP
(kJ)
EP Diagram for 2 CO(g) + O2(g) 2 CO2(g)
2 CO(g) + O2(g)
2 CO2(g)
-221.0
-787.0
H = -566.0 kJ
Reaction Progress
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Example 2Find the heat of combustion of ethane. The products of combustion are gases.
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)
(2mol)(-84.0kJ/mol) + 0 kJ (4 mol)(-393.5 kJ/mol)
-168.0 kJ + 0 kJ -1574.0 kJ
Hc = nHf(products) nHf(reactants)
= (-1574.0 kJ + (-1450.8 kJ)) – (-168.0 kJ)= -2856.8 kJ
+ (6 mol)(-241.8 kJ/mol)
+ -1450.8 kJ
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Your Assignment: pgs 12-13 in workbook
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L. Energy Systems in Biological Processes
Photosynthesis:
Cellular Respiration:
energy + CO2(g) + H2O(l) C6H12O6(s) + O2(g)
C6H12O6(s) + O2(g) CO2(g) + H2O(l) + energy
**Water vapor condenses into liquid
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M. Nuclear Change
enthalpy changes in nuclear reactions are the result of
there are two types of nuclear reactions:
EP changes as rearrangements among the
subatomic particles (protons and neutrons) occur ie) intranuclear forces
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1. Fusion (Joining)
fusion of hydrogen to helium occurs on the sun and other stars
these types of reaction produce
require a great deal of
the greatest amount of energy and are necessary for life on Earth
heat and pressure
H + H He + n + 1.70 x 109 kJ01 11
2432
Top number =
Bottom number =
mass number
number of protons
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2. Fission (Splitting)
basis for nuclear power plants
uranium atoms can be
was discovered in the late 1930's when uranium was bombarded by neutrons causing it to split
the neutrons produced by fission allow
split into two smaller nuclei which produces large quantities of energy
U + n Kr + Ba + 3 n + 1.9 x 1010 kJ123592 0 0
1 9256
14136
a chain reaction to occur to keep the reaction self sustaining
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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
are the most common source of energy
many aspects of our society are based on the price of fuels like gasoline
fossils fuels
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Advantages vs. Disadvantages of Fossil Fuels
Advantages Disadvantages
relatively low cost
readily available (market)
plant set-up, vehicle design, expertise affordable
used all over the world
deposits are large
release of gases that contribute to the greenhouse effect and acid rain when burned
mining is detrimental to the environment
non-renewable
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Advantages vs. Disadvantages of Nuclear Power
Advantages Disadvantages
low cost of fuel
lots of energy from small amount of fuel
doesn’t add to greenhouse effect, acid rain
Canada has lots of uranium ore
preserves existing fossil fuels
high cost of plant set-up, expertise, decommission
cause thermal pollution
difficult to dispose of nuclear fuel wastes
possibility of catastrophic accidents
non-renewable