a. energy – crash course video
DESCRIPTION
Thermochemistry. A. Energy – Crash course video. we use hydrocarbons in many areas of our lives. eg). glucose for cellular respiration. gasoline and natural gas for fuel. the energy stored in the chemical bonds of hydrocarbons originally comes from the. Sun***. - PowerPoint PPT PresentationTRANSCRIPT
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A. Energy – Crash course videowe use hydrocarbons in many areas of our
lives
Thermochemistry
eg)glucose for cellular respiration
gasoline and natural gas for fuel
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the energy stored in the chemical bonds of hydrocarbons originally comes from the
ancient plants captured the energy of the sun during ancient animals ate the plants, then death and trapping in rock formations as oil, gas, oil sand etc.
Sun***
photosynthesis,
They ask this question almost every year!!
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there are two fundamental types of energy:
1. the energy of (of particles)
2.
energy that is (in chemical bonds)
kinetic energy (EK):motion
potential energy (EP):
stored
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the First Law of Thermodynamics states that: the total energy of the universe is
energy can be however the of any system is (it be created or destroyed)
constant
converted to other formstotal energy
conserved cannot
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the Second Law of Thermodynamics states that: in the absence of energy input, a
system becomes more
heat will always transfer from until is reached
disordered
hot objects to cooler objects thermal equilibrium
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is a measure of the of the particles of a substance
the the particles are moving, the the temperature
B. Temperature Change
the amount of energy needed to heat a substance depends on three factors:
1. the of the substance (m) 2. the
3. the of substance (heat capacity) (c)
temperature average kinetic energy
fasterhigher
mass
change in temperature (Δt )
type
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heat capacity is heat required to change the temperature of of a substance by
Q = mcΔt
where: Q = heat energy in Jm =Δt =c =
mass in g change in temperature in °Cspecific heat capacity in J/gC
***You can find the heat capacities of common substances in your DATA BOOKLET
one gram1°C
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Example***Remember Δt = tf - ti
Find 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
*** Practice work: p.1, questions #1 - 6
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C. Energy and Enthalpy - video
is the sum of all forms of kinetic and potential energy in a system (EK + EP) 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, chemical or nuclear change…fortunately, this can be measured
enthalpy
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is the enthalpy change of a substance
molar enthalpy is also designated as H, although we will be using just for our formulas
***this can be confusing so pay close attention to the context…wording and units!!!
molar enthalpy per mole
J/mol or kJ/mol
H
molar enthalpy is measured in
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Summarizing Enthalpy
Word Symbol Meaning
Unit Formula
Heat Energy
Q General term for
heat
J or KJ Q = mc∆t
Enthalpy - Sum of energy
- -
Enthalpy change
∆H Change in
energy
J or KJ ∆H = mc∆t
Molar Enthalpy
H Change in
energy per mole
J/molOr
KJ/mol
nH = mc∆t
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is measured inJ or kJ
subscripts are sometimes used to denote the type of process, for example “r” for “reaction”
= enthalpy of reaction
the “” symbol is used to denote changes taking place at conditions
= standard enthalpy of reaction (at SATP 100 kPa and 25C)
enthalpy change, H,
rH
standard
rH
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D. Energy Change in Chemical Reactions changes in take
place during chemical reactions when one substance is
chemical bonds are sources of potential energy
bonds energy bonds energy
potential energy
converted into another substance stored
breaking requires
forming releases
Must know**
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if the energy added to break the bonds is than the energy released when the new bonds are formed then the reaction is
greater
endothermic
eg) photosynthesis
6 CO2(g) 6 O2(g)C6H12O6(s) ++ energy+ 6 H2O(l)
monster flower
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endothermic changes are designated as since energy is being to the system
positive values added
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if the energy added to break the bonds is than the energy released when the new bonds are formed then the reaction is
less
exothermic
eg) cellular respiration
6 CO2(g)6 O2(g)C6H12O6(s) + + energy+ 6 H2O(l)
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CO2(g) O2(g)hydrocarbon + + energy+ H2O(g)
hydrocarbon combustion eg)
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exothermic changes are designated as since energy is being from the system
negative values removed
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the amount of energy lost or gained is to the amount of substances that react ie) if 100 g of a substance burning will
release energy compared to when 50 g of that same substance burns
directly proportional
twice as much
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E. Communicating Enthalpy Changes
1. rH Notation
the heat of reaction can be given as a rH value of the equation
Exothermic Reactions
the sign on rH is since the enthalpy of the system is
outside
negativedecreasing
2 SO2(g) + O2(g) 2 SO3(g) rH = 197.8 kJ
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2. Inside the Equation
in an exothermic reaction enthalpy is included as aproduct
Mg(s) + ½ O2(g) MgO(s)+ 601.6 kJ
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3. Potential Energy Diagrams shows the potential energy of the
and the of a chemical reaction
reactants have potential energy than the products in an exothermic reaction
the difference between the reactants and products is the
reactantsproducts
more
rH
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Exothermic
EP
(kJ)
Reaction Progress
reactants
products
rH = -300kJ
700 kJ
400 kJ
Products have EP than the reactants energy has been to the surroundings
Reactants = Products + rH
less
lost
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1. rH Notation
the sign on rH is since the enthalpy of the system is
2 SO3(g) 2 SO2(g) + O2(g)
Endothermic Reactions
positiveincreasing
rH = +197.8 kJ
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2. Inside the Equation
in an endothermic reaction enthalpy is included as a
H2O(l)
reactant
+ 285.8 kJ H2(g) + ½ O2(g)
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3. Potential Energy Diagrams reactants have potential
energy than the products in an exothermic reaction
the difference between the reactants and products is the
less
rH
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PRACTICE QUESTION
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Endothermic
EP
(kJ)
Reaction Progress
reactants
products
H = +350kJ
500 kJ
150 kJ
Products have EP than the reactants energy has been by the system
Reactants + rH = Products
more
gained
Answer: 3,1,2,4
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PRACTICE QUESTIONS
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PRACTICE QUESTIONAnswer: C
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PRACTICE QUESTIONAnswer: A
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ENDOTHERMIC VS EXOTHERMIC SUMMARY
Endothermic Exothermic
- Products get cold (Reactants take in heat)
- Products get hot (Reactants release heat)
- (In the system) Reactants + energy
Products (Outside the initial system)
- (In the system) Reactants Products (Outside the initial system)
- Energy is a positive value because energy is added to the initial system
- Energy is negative because energy is released outside the initial system
Ex: reactants → products ; ∆H = + 250 KJ
Ex: reactants → products ; ∆H = - 250 KJ
- Reactants Ep ↓ than the Ep of products
- Reactants Ep ↑ than the Ep of products
Practice questions: page 1 and 2
Answer: C
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cellular respiration is a low temperature combustion of the hydrocarbon glucose, producing
C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l)
cH = 2802.5 kJ
CO2(g) and H2O(l)
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ΔrH = nrH
where: ΔrH = n =
rH =
enthalpy change in kJ (or J)number of moles in mol molar enthalpy in kJ/mol or J/mol
F. Calculating Enthalpy Changes molar enthalpy, in kJ/mol, and the number
of moles of a substance can be used to calculate the enthalpy change of a chemical change:
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the number of moles can either come from the in the chemical reaction or from the formulabalancing
n = m/M
always have a sign on H or H
o for (energy absorbed)
o for (energy released)
positive endothermic
negative exothermic
ΔrH = Q = nH = mc∆t = H
m
M
m
M
Summarize the options
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PRACTICE QUESTIONS
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Example 1 Calculate the molar enthalpy of combustion for oxygen given the following information:
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g) cH =
2856.8 kJ
cH = ncH
cH = 408.11 kJ/mol2856.8 kJ = (7 mol)cH
Answer: B
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Example 2 Find the enthalpy change when 5.50 g of pentane burns.
cH = 3244.8 kJ/mol of pentane.
cH = ncH
= 247 kJ= 247.2828…kJ
= 5.50 g 3244.8kJ/mol 72.17 g/mol
= m cH M
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Example 3 When methane is burned, oxygen is consumed. Determine the mass of oxygen consumed if the change in enthalpy is 250 kJ and the molar enthalpy of reaction for oxygen is 401.3 kJ/mol.
cH = m cH M
250 kJ = m 401.3 kJ/mol
32.00 g/molm = 19.935…g = 19.9 g
Practice p.2-3, #1 - 7
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PRACTICE QUESTION
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G. Calorimetry – calorimetry 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
Calorimeter
ENERGY
insulation water
enclosed chemical system
measuring energy changes using an isolated system
calorimeter
Answer:
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Steps for Using a Simple Calorimeter
1. Measure the of the water in the calorimeter.
2. Add the to the calorimeter. 3. Allow reaction to proceed, the solution to ensure even temperature.
4. Measure the of the water in the calorimeter (maximum temperature for exothermic reactions, minimum temperature for endothermic reactions)
Video on calorimetry
Video #2 on calorimetry
initial temperature
reactants
stirring
final temperature
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calorimetry is based on the First and Second Law of Thermodynamics…energy is conserved and energy is transferred from hotter objects to cooler objects until thermal equilibrium is reached
Lab #4 – Calorimetry (Virtual lab link)Video – Don’t do this!!!Video – Do this
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PRACTICE QUESTION
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it is assumed that is by the system except for the energy required or released by the
calculations are based on the Principle of Heat Transfer: Video on calculation
HEAT LOST = HEAT GAINED
gained or lost
chemical change
no energy
remember, you must use a with your H values (either enthalpy change or molar enthalpy)
endothermic = value
exothermic = value
sign
positive
negative
Answer: B
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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)
= 88.0 kJ= 87.990 kJ
= 87 990 J
= (500 g)(4.19 J/gC)(52.0C – 10.0C )= mct= Q
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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) in kJ/mol? heat lost (water) = heat gained
(formation)Q = H
mct = (m/M) H(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 = +81.3 kJ/mol
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Example 4 A student mixes 100.0 mL of 0.500 mol/L HBr(aq) with 100.0 mL of 0.500 mol/L KOH(aq). The initial temperature of both solutions is 21.00C and the highest temperature reached after mixing is 24.40C. Calculate the molar enthalpy of neutralization in kJ/mol for the HBr(aq). Assume both solutions have the density and heat capacity of pure water. heat lost (neut HBr(aq) ) = heat gained (water)
H = Q cv H = mct
(0.1000L)(0.500 mol/L) H= (200g)(4.19J/gC)(24.40C-21.00 C)
(0.0500 mol ) H = 2849.2 J H = – 56984 J/mol H = – 57.0 kJ/mol
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Example 5 A student built a simple calorimeter with a 25.0 g tin can and 150 mL of water. Calculate the molar enthalpy of combustion of ethanol in kJ/mol if 0.166 g of this fuel increased the temperature of the calorimeter by 7.00C. Remember to include not only the heat gained by the water but also by the calorimeter. heat lost (combustion) = heat gained (water + tin can)
H = Q (m/M) H = mct + mct
(0.166/46.08g/mol) H = (150g)(4.19J/gC)(7.00C)+ (25.0g)(0.227J/gC)(7.00C)
(0.00360… mol ) H = 4439.225 J H = – 1232286.057 J/mol H = – 1232.2… KJ/mol H = –1.23 103 kJ/mol
Practice questions : page 3 , #1-7
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PRACTICE QUESTION
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H. Industrial Bomb Calorimeters industrial calorimeters are used in
to measure the heat of combustion of food, fuel, oil, crops, and explosives (Video)
modern calorimeters have eg) volume of water used, container
(bomb) material, stirrer and thermometer
research
fixed components
Answer: 2.81
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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 the of the in
entire system
one constant C, heat capacitykJ/C
you can also be asked to calculate instead of kJ/mol in calorimetry questions
kJ/g
you use the formula instead of nH to give kJ/g (video)
mH
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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.
heat lost (combustion) = heat gained (calorimeter)
m H = 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/gPractice questions : page 4 , #1-3
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PRACTICE QUESTIONS
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I. Predicting Enthalpy (rH) Changes1. 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
same
single reaction series of reactions
thermochemical equationsadded
enthalpies
enthalpy for the final equation
Answer: D
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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 rHVideo: Hess’ Law
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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
Manipulateadd net equation.
same factorflip
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 kJH2(g) + ½ O2(g) H2O(l) H = –285.8 kJCH4(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) +
Practice questions: page 4-6
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J. Using Standard Heats of Formation fH 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 (fH) are the changes in EP that occur when
heats of formation
compounds are formed from their elements
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f H for elements cannot be directly measured therefore they are designated as …all other f H values are in reference to this…see pages 4-5 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) fH = C2H4(g) fH =N2O4(g) fH =PCl3(l) fH =Al2O3(s) fH =
–285.8 kJ/mol+52.4 kJ/mol+11.1 kJ/mol–319.7 kJ /mol–1675.7 kJ /mol
1
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
rH = nfH(products) nfH(reactants)
rH
reactantsproducts
rH = fH(products) fH(reactants)
Take the energy released during the formation of the products
And subtract the initial energy of the system (energy released when reactants were
formed)
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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
cH = nfH (products) nfH (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 and draw the EP diagram for this reaction. 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
cH = nfH (products) nfH (reactants) = (-3024.8 kJ) – (-168.0 kJ)= -2856.8 kJ
+(6 mol)(-241.8 kJ/mol)
-3024.8 kJ
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EP
(kJ)
EP Diagram for 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 16 H2O(g)
2 C2H6(g) + 7 O2(g)
4 CO2(g) + 16 H2O(g)
-168.0
-3024.8
H = -2856.8 kJ
Reaction Progress
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Example 3Calculate the molar enthalpy of combustion for 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
cH = nfH (products) nfH (reactants) = (-3024.8 kJ) – (-168.0 kJ)= -2856.8 kJ
+(6 mol)(-241.8 kJ/mol)
-3024.8 kJ
cH = cH n= 2856.8 kJ 2 mol= 1428.4 kJ/mol
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Example 4Calculate the energy released when 25.0 g of methanol is burned. The products of combustion are gases.
2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g) (2mol)(-239.2kJ/mol) + 0 kJ (2 mol)(-393.5
kJ/mol) -478.4 kJ
cH = nfH (products) nfH (reactants) = (-1754.2 kJ) – (-478.4 kJ)= -1275.8 kJ
+(4 mol)(-241.8 kJ/mol)
-1754.2 kJ
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cH = cH n= 1275.8 kJ 2 mol= 637.9 kJ/mol
cH = m cH M
= 498 kJ
cH = 25.0 g 637.9 kJ/mol
32.05 g/mol
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Example 5Calculate the molar heat of formation for ethylene glycol given the following information:
(CH2OH)2(l) + 5/2 O2(g) 2 CO2(g) + 3 H2O(g) (1mol)x + 0 kJ (2 mol)(-393.5
kJ/mol)
H = nfH (products) nfH (reactants) 1178.0 kJ = (-1930.2 kJ) – (1 mol)x+ 1930.2 = +1930.2+752.2 kJ =– (1 mol)x(-1 mol) = (-1 mol)
x = 752.2 kJ/mol
+(3 mol)(-241.8 kJ/mol)
-1930.2 kJ (1mol)x
H = 1178.0 kJ
Complete p.7 of workbook
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PRACTICE QUESTION
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PRACTICE QUESTIONSAnswer: 1668.4 KJ
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PRACTICE QUESTIONSAnswer: 2,4,6,7
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K. Energy and Efficiency
most of Canada’s energy (electricity) comes from processes such as the combustion of
electricity is also generated through
processes
chemical
fossil fuels
nuclear
Answer: C
Video
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how does the energy from physical, chemical and nuclear processes compare:
both methods involve changing (a change) which turns turbines to generate electrical energy
water into steamphas
e
changes involve the breaking and forming of forces ( kJ/mol)
physicalintermolecular
1 – 100
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changes involve the breaking and forming of ( kJ/mol)
chemical chemical bonds
100 – 10000
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changes involve changes within the ( kJ/mol)
nuclear nuclei of atoms
millions to billions
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is the ratio of produced (energy ) to in its production (energy )
% Efficiency = Energy output 100
Energy input
efficiency useful energyoutput energy used
input
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it is important to be as as possible with appliances and vehicles
saving energy and it helps to (greenhouse effect and acid rain)
efficient
saves you moneysave the environment
we have developed many technologies that help us to solve practical problems
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in any process, the that take place, the the process because of in transfer
more energy conversionsless efficient
heat loss
gas furnace (natural gas) is about efficient since it is used to directly supply heat
90%
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natural gas power plant is only about efficient because there are several energy conversions that take place before electricity is generated (water to steam to kinetic energy to mechanical energy to electrical energy)
37%
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L. Fuelling Society
we must assess the of relying on any fuel source
when selecting an energy source, efficiency is not the only consideration
must also be considered
risks and benefits
environmental impact
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Advantages vs. Disadvantages of Fossil Fuels
Advantages Disadvantagesrelatively 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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processes produce
but they do produce which lasts for thousands of years
nuclear do notgreenhouse gasesradioactive waste
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turbines use a energy source (sun indirectly) but are not free of problems…they are
wind renewable
noisy and dangerous to birds
Video
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power is also but damming rivers on both sides of the dams
hydroelectric renewableaffects ecosystems
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the amount of released by a fuel determines how it is
fuels that use renewable energy sources ( etc.) and nuclear power are considered the
CO2(g)“clean”
solar, wind, geothermal
“cleanest”
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is the fossil fuel and is themethane “cleanest”coal least clean
regardless of which source of energy we use, we must think about the impact that our fuels have on the environment
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M. Bond Energy and Activation Energy bond energy is the energy
or the energy
required to break a chemical bond released when a bond is formed
the of a reaction represents the from the bonds in the reactant(s) and the bonds of the product(s) in reactions, bond breaking absorbs energy than the bond formation gives off, resulting in a
change in enthalpynet effect breaking
forming
exothermicless
H
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Exothermic H2 (g) + Cl2 (g) 2 HCl (g) + energy
Energy(kJ)
Reaction Progress
H2 + Cl2
H H Cl Cl
2 HCl H
bond breaking
bond formation
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in reactions, bond breaking absorbs energy than the bond formation gives off, resulting in a
endothermicmore
+H
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Endothermic2 H2O(l) + energy 2 H2 (g) + O2 (g)
Energy(kJ)
Reaction Progress
2 H2O
H H H H O O
2 H2 + O2
+H
bond breaking
bond formation
Activated complex
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ACTIVATION ENERGY ANALOGY – LEGO
You got a transformers Lego for Christmas but want a doll house.
Transformers = reactants ; Doll house = products
You need to take apart each Lego before starting to build; individual Legos = activated complex.
Once you’ve reached the point of no return = change over point
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the energy barrier that must be overcome for a chemical reaction to occur is called the
the atoms in the have to be in order for them to bond in a different configuration and become the products
the activation energy is always than the energy contained in the reactants and the products, however the amount of activation energy necessary is dependent on the reaction
activation energy
reactants“pulled apart”
higher
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the of the activation energy barrier on a potential energy diagram represents the of the reaction
in both endothermic and exothermic reactions, the molecules of the reactants are moving with a certain amount of
when the reactants with each other, the kinetic energy is transformed into
top
changeover point
kinetic energy
collide
potential energy
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this potential energy is then stored in bonds of the chemical species that exists at the top which is called the
this is a transitional species that is neither a reactant nor a product which has and is
when the partial bonds of the activated complex re-form as chemical bonds in the products, the stored potential energy is converted back into as the product molecules
activated complex
partial bondshighly unstable
kinetic energymove apart
Video – Activated complex
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EP (kJ)
Reaction Progress
Potential Energy Diagram: Exothermic
H
Ea
*
* activated complex
Reactants
Products
Change over point
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EP (kJ)
Reaction Progress
Potential Energy Diagram: Endothermic
Ea
+H
*
* activated complex
Reactants
Products
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N. Catalysts a catalyst is a substance that
of a chemical reaction by the reaction
catalysts provide for chemical reactions
they the required for a reaction to take place which results in the production of a of products in a given length of time (even at a lower temperature)
catalyzed reactions can be shown on EP diagrams:
increases the ratewithout being consumed
alternate pathways
lower activation energy
greater yield
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EP (kJ)
Reaction Progress
Potential Energy Diagram: Exothermic
uncatalyzed reaction
catalyzed reaction
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EP (kJ)
Reaction Progress
Potential Energy Diagram: Endothermic
uncatalyzed reaction
catalyzed reaction
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O. Catalysts in Industry catalysts are often used in industry to
speed up the reactions and obtain a reasonable reaction rate under reasonable conditions in cars, we have that use Pt(s), Pd(s) and Rh(s) to speed up the combustion of exhaust gases so that more of the products are ( instead of , instead of )
catalytic converters
harmless N2 NOx CO2
CO
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the oil and gas industry uses catalysts (Pt(s), HF(aq), H2SO4(aq) etc) in the of and to make more marketable fuels like gasoline
cracking and reformingcrude oil bitumen
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P. Enzymes compounds that act as catalysts in living
systems are called
called catalysts
chemical reactions in the body occur at very temperature (37C) and without catalysts many would be too
enzymes
biological
low
slow
eg) amylase, peptidase, lactase
Extra practice : p.8 from workbook