Download Thermochemistry Chem 2/H

Thermochemistry:
Chemical Energy
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Energy and Its Conservation
Energy: The capacity to supply heat or do work.
Kinetic Energy (EK): The energy of motion.
Potential Energy (EP): Stored energy.
Law of Conservation of Energy: Energy cannot be
created or destroyed; it can only be converted from one
form into another. First Law of Thermodynamics
∆E = q + w
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Energy and Its Conservation
Thermal Energy: The kinetic energy of molecular motion,
measured by finding the temperature of an object.
Heat: The amount of thermal energy transferred from one
object to another as the result of a temperature difference
between the two.
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Calorimetry
The energy released or absorbed
during a chemical reaction.
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Some definitions
Exothermic: energy is given off
(temperature increases)
Endothermic: energy is required
(temperature decreases)
Enthalpy, DH: Internal energy in kJ/mol
- DH: exothermic + DH: endothermic
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Types of Calorimeters
Coffee cup
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Bomb
Coffee Cup Calorimetry
Measure the heat flow at constant pressure (∆H).
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Bomb Calorimetry
Measure the heat flow at constant volume (∆E).
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Coffee-cup calorimetry
q = -mcDT
q: quantity of heat in joules, J
m: mass of liquid, g
c: specific heat capacity, 4.18J/goC
DT: change in temperature in oC, Tfinal –
Tinitial
q = nΔH or ΔH = q/n
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There are two kinds of coffeecup calorimetry problems.
1st kind: Grams of a chemical are added to a
given mass or volume of water.
What to do:
Grams of water will be plugged into m.
DT, Last temperature minus first temperature.
Solve for q and change joules to kJ.
Grams of the other chemical convert to moles.
Solve for DH by putting kJ over moles and
divide.
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Example #1
In a coffee-cup calorimeter, 5g of sodium
are dropped in 300g of water at 20oC
the temperature rises to 35oC.
Calculate the enthalpy change for this
reaction.
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Example #2
A 1.5g sample of calcium is dropped into
275mL of water at 20oC in a coffee-cup
calorimeter. The temperature rises to
23oC. The density of water is 1g/mL.
What is the energy change for this
reaction in kJ/mol?
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There are two kinds of coffeecup calorimetry problems.
2nd kind: Two volumes of a solution are given.
What to do:
Add them and plug into m (the density of most
solutions = 1g/mL, so the mL = the grams)
DT, Final temperature minus Initial temperature.
Solve for q and change joules to kJ.
Pick one of the chemicals and multiply (L)(M) to
find moles.
Solve for DH by putting kJ over moles and
divide.
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Example #3
450mL of 2M HCl are mixed with 450mL
of 2M NaOH in a coffee-cup
calorimeter. The temperature rises from
23oC to 25oC. Calculate the enthalpy
change for this reaction.
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Example #4
In a coffee-cup calorimeter, 275mL of
0.5M Ca(OH)2 react with 275mL of 0.5M
H2SO4. The temperature increases
from 25oC to 30oC. Calculate the
energy in kJ/mol.
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Worked Example Calculating the Amount of Heat Released in a
Reaction
How much heat in kilojoules is evolved when 5.00 g of aluminum reacts with a stoichiometric amount of Fe 2O3?
Strategy
According to the balanced equation, 852 kJ of heat is evolved from the reaction of 2 mol of Al. To find out how
much heat is evolved from the reaction of 5.00 g of Al, we have to find out how many moles of aluminum are
in 5.00 g.
Solution
The molar mass of Al is 26.98 g/mol, so 5.00 g of Al equals 0.185 mol:
Because 2 mol of Al releases 852 kJ of heat, 0.185 mol of Al releases 78.8 kJ of heat:
✔ Ballpark Check
Since the molar mass of Al is about 27 g, 5 g of aluminum is roughly 0.2 mol, and the heat evolved is about
(852 kJ/2 mol)(0.2 mol), or approximately 85 kJ.
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Bomb Calorimetry
q = -CDT
q: quantity of heat, kJ
C: heat capacity, kJ/oC
DT: change in temperature (increases
positive, decreases negative)
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Two kinds of Bomb also:
1st kind: Enthalpy (kJ/mol) is given
What to do:
Convert grams to moles
Multiply the moles by the enthalpy. This
answer plugs in for q
Find C or DT (whichever it asks for)
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Example #5
A 5g sample of hydrogen is burned in a
bomb calorimeter. The enthalpy of
combustion for hydrogen is -242kJ/mol.
What is the heat capacity of the
calorimeter if the temperature rises
5oC?
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Example #6
30g of magnesium are burned in a bomb
calorimeter that has a heat capacity of
18kJ/oC. If the enthalpy of combustion for
magnesium is -602kJ/mol, how much will
the
temperature rise?
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Still Bomb
2nd kind: Enthalpy is asked for (or energy
change in kJ/mol)
What to do:
Convert grams to moles.
Plug in C and DT and solve for q.
Solve for DH by putting kJ over moles and
divide.
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Example #7
If 10g of Ca are burned in a bomb
calorimeter that has a heat capacity of
12.8kJ/oC, the temperature increases by
12.38oC. What is the enthalpy of
combustion of calcium?
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Example #8
An 18g sample of sulfur is burned in a
bomb
calorimeter. The heat capacity of the
calorimeter is 8.9kJ/oC and the
temperature
increases 18.74oC. What is the enthalpy
of
combustion for sulfur?
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What is “Thermodynamics”?
The study of energy and its
interconversions.
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Some terms:
Enthalpy: Internal energy (energy stored in
the bonds) DH kJ/mol (+ endothermic, exothermic)
Entropy: disorder of a substance DS
J/mol.K
gases  aqueous  liquids  solids
most disordered to least disordered
Gibbs Free Energy: stored energy less the
degree of disorder DG kJ/mol
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How to calculate DH, DS & DG
DHo = S DHfo for products – SDHfo for
reactants
DSo = S So of products – SSo for reactants
DGo = S DGfo of products – SDGfo for
reactants
O superscripted means standard states
(1atm, 273K, etc)
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Calculating Enthalpy Changes Using
Standard Heats of Formation
Standard Heat of Formation (∆H°f): The enthalpy
change for the formation of 1 mol of a substance in its
standard state from its constituent elements in their
standard states.
Standard states
C(s) + 2 H2(g)
CH4(g)
∆H°f = -74.8 kJ
1 mol of 1 substance
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Calculating Enthalpy Changes Using
Standard Heats of Formation
∆H° = ∆H°f (Products) - ∆H°f (Reactants)
aA+bB
cC+dD
∆H°= [c ∆H°f (C) + d ∆H°f (D)] - [a ∆H°f (A) + b ∆H°f (B)]
Products
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Reactants
Calculating Enthalpy Changes Using
Standard Heats of Formation
Using standard heats of formation, calculate the standard
enthalpy of reaction for the photosynthesis of glucose
(C6H12O6) and O2 from CO2 and liquid H2O.
6 CO2(g) + 6 H2O(l)
C6H12O6(s) + 6 O2(g) ∆H° = ?
H°= [∆H°f (C6H12O6(s))] - [6 ∆H°f (CO2(g)) + 6 ∆H°f (H2O(l))]
∆H°= [(1 mol)(-1273.3 kJ/mol)] -
[(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)]
= 2802.5 kJ
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How to calculate DH, DS & DG
DSo and DGo can be calculated the same
way as DHo
DSo = S So of products – SSo for reactants
DGo = S DGfo of products – SDGfo for
reactants
O superscripted means standard states
(1atm, 273K, etc)
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Example #1
Calculate DHo, DSo and DGo for the
reaction below.
C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)
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Predicting Entropy
To decide which substance has more
entropy, first look at its state (solid,
liquid, etc.). The more disordered state
will have more disorder.
If they have the same state, the larger
molecule will have more entropy.
Both gaseous, higher temperature, less
pressure, more disorder.
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Example #2
Which in each pair has more entropy?
A. H2O(g)
H2O(l)
B. C6H12O6 (s) C12H22O11(s)
C. HCl(aq)
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HCl(g)
Predicting DS change in a
reaction.
Look at both sides of the reaction, identify
the state of the chemicals, how many
moles of chemicals are present (more
particles, more disorder)and how big the
particles are. These are in order of
importance.
If the right has more disorder, DS goes
up.
If the right has less disorder, DS goes
down.
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Example #3
Predict the sign for DS for each reaction.
A. 12CO2(g) + 11H2O(l)  C12H22O11(s) +
12O2(g)
B. 2H2O(g)  2H2(g) + O2(g)
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Worked Example Predicting the Sign of ΔS for a Reaction
Predict whether ΔS°is likely to be positive or negative for each of the following reactions:
a.
b.
Strategy
Look at each reaction, and try to decide whether molecular randomness increases or decreases. Reactions that
increase the number of gaseous molecules generally have a positive ΔS, while reactions that decrease the number
of gaseous molecules have a negative ΔS.
Solution
a. The amount of molecular randomness in the system decreases when 2 mol of gaseous reactants combine to
give 1 mol of liquid product, so the reaction has a negative ΔS° .
b. The amount of molecular randomness in the system increases when 9 mol of gaseous reactants give 10 mol of
gaseous products, so the reaction has a positive ΔS° .
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Spontaneity
A spontaneous reaction occurs without
the need of outside energy.
Do not confuse spontaneous with
instantaneous.
Spontaneous happens on its own,
instantaneous happens right away.
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How can you tell?
DS + tends to be spontaneous
DH – tends to be spontaneous
The only way to know for sure is if DG is -,
then it is guaranteed to be spontaneous.
DGo = DHo - T DSo
DG = DH - T DS
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Example #4
Predicting the signs of ΔH, ΔS and ΔG
A. When solid ammonium chloride is
dissolved in water, it forms an aqueous
solution and the temperature drops.
B. When hydrogen gas is mixed with
oxygen gas and a spark ignited, a fire
ball to the ceiling occurs. 2H2(g) + O2 (g)
 2H2O(g)
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Worked Conceptual Example Predicting the Signs of ΔH, ΔS, and ΔG
for a Reaction
What are the signs of ΔH, ΔS, and ΔG for the following nonspontaneous transformation?
Strategy
First, decide what kind of process is represented in the drawing. Then decide whether the process increases or
decreases the entropy (molecular randomness) of the system and whether it is exothermic or endothermic.
Solution
The drawing shows ordered particles in a solid subliming to give a gas. Formation of a gas from a solid increases
molecular randomness, so ΔS is positive. Furthermore, because we’re told that the process is nonspontaneous, ΔG
is also positive. Because the process is favored by ΔS (positive) yet still nonspontaneous, ΔH must be unfavorable
(positive). This makes sense, because conversion of a solid to a liquid or gas requires energy and is always
endothermic.
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Calculating DG using DH & DS
Be sure to change DS into kJ from J.
Change temperature to Kelvin by adding
273 to celsius.
Plug in and solve for the unknown.
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Example #5
For the reaction:
H2O(g)  H2O(l)
DHo = -44kJ/mol DSo = 119J/mol.K
Calculate DGo at 10oC, 0oC and -10oC.
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More info about DG
If DG is negative the reaction is
spontaneous (runs forwards)
If DG is positive it is not spontaneous
forwards (can be spontaneous
backwards)
If DG = 0 it is at equilibrium (no net
forward or backward movement)
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Temperature and spontaneity
You can change the temperature of
reactions and they will become
spontaneous at their new temperature.
Phase changes:
The boiling point and freezing (melting)
points are temperatures where DG =0.
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Free Energy
Reversible
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Example #6
For the reaction:
H2O(g)  H2O(l)
DHo = -44kJ/mol DSo = 119J/mol.K
At what temperature does this reaction
become spontaneous?
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Example #7
For the phase change:
Br2(l)  Br2(g)
DHo = 31kJ/mol and DSo = 93J/mol.K
What is the normal boiling point of
bromine?
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Worked Example Using the Free-Energy Equation to Calculate
Equilibrium Temperature
Lime (CaO) is produced by heating limestone (CaCO3) to drive off CO2 gas, a reaction used to make Portland
cement. Is the reaction spontaneous under standard conditions at 25 °
°
°C? Calculate the temperature at which the
reaction becomes spontaneous.
Strategy
The spontaneity of the reaction at a given temperature can be found by determining whether ΔG is positive or
negative at that temperature. The changeover point between spontaneous and nonspontaneous can be found by
setting ΔG = 0 and solving for T.
Solution
At 25 °
C (298 K), we have
Because ΔG is positive at this temperature, the reaction is nonspontaneous. The changeover point between
spontaneous and nonspontaneous is approximately
The reaction becomes spontaneous above approximately 1120 K (847 °
°
°C).
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Calculating Enthalpy Changes Using
Hess’s Law
Hess’s Law: The overall enthalpy change for a reaction
is equal to the sum of the enthalpy changes for the
individual steps in the reaction.
Haber Process: 3 H2(g) + N2(g)
2 NH3(g) ∆H°= -92.2 kJ
Multiple-Step Process
2 H2(g) + N2(g)
N2H4(g)
∆H°1 = ?
N2H4(g) + H2(g)
2 NH3(g)
∆H°2 = -187.6 kJ
3 H2(g) + N2(g)
2 NH3(g)
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∆H°reaction = -92.2 kJ
Calculating Enthalpy Changes Using
Hess’s Law
∆H°1 + ∆H°2 = ∆H°reaction
∆H°1 = ∆H°reaction - ∆H°2
= -92.2 kJ - (-187.6 kJ) = +95.4 kJ
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Worked Example Using Hess’s Law to Calculate ΔH°
Methane, the main constituent of natural gas, burns in oxygen to yield carbon dioxide and water:
Use the following information to calculate ΔH°in kilojoules for the combustion of methane:
Strategy
It often takes some trial and error, but the idea is to combine the individual reactions so that their sum is the desired
reaction. The important points are that:
• All the reactants [CH4(g) and O2(g)] must appear on the left.
• All the products [CO2(g) and H2O(l)] must appear on the right.
• All intermediate products [CH2O(g) and H2O(g)] must occur on both the left and the right so that they cancel.
• A reaction written in the reverse of the direction given [
] must have the sign of its ΔH° reversed
(Section 8.7).
• If a reaction is multiplied by a coefficient [
is multiplied by 2], then ΔH° for the reaction must be
multiplied by that same coefficient.
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Worked Example Using Hess’s Law to Calculate ΔH°
Continued
Solution
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Worked Example Using Hess’s Law to Calculate ΔH°
Water gas is the name for the mixture of CO and H2 prepared by reaction of steam with carbon at 1000 °
°C:
The hydrogen is then purified and used as a starting material for preparing ammonia. Use the following information
to calculate ΔH°in kilojoules for the water-gas reaction:
Strategy
As in Worked Example 8.6, the idea is to find a combination of the individual reactions whose sum is the desired
reaction. In this instance, it’s necessary to reverse the second and third steps and to multiply both by 1/2 to make
the overall equation balance. In so doing, the signs of the enthalpy changes for those steps must be changed and
multiplied by 1/2. Note that CO2(g) and O2(g) cancel because they appear on both the right and left sides of
equations.
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Worked Example Using Hess’s Law to Calculate ΔH°
Continued
Solution
The water-gas reaction is endothermic by 131.3 kJ.
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Example #8
Using the reactions:
C(s) + O2(g)  CO2(g) DH = -393.5kJ/mol
CO(g) + ½ O2(g)  CO2(g) DH = -283kJ/mol
Calculate the enthalpy for:
C(s) + ½ O2(g)  CO(g)
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Example #9
Calculate DH for the reaction:
2C(s) + H2(g)  C2H2(g)
Given the following:
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O (l)
DH = -2599.2kJ/mol
C(s) + O2(g)  CO2(g) DH = -393.5kJ/mol
2H2(g) + O2(g)  2H2O (l) DH = -571.6kJ/mol
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Example #10
Given the following data:
H2(g) + ½ O2(g)  H2O(l) DH = -286kJ
N2O5(g) + H2O (l)  2HNO3(l) DH = -77kJ
½ N2(g) + 3/2 O2(g) + ½ H2(g)  HNO3 (l)
DH = -174kJ
Calculate the DH for:
2 N2(g) + 5 O2(g)  2 N2O5(g)
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Hess’s Law and Gibb’s Free
Energy
Hess’s Law works the same way for
Gibb’s Free Energy, you just use DG
values instead of DH values.
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Example #11
Calculate DG for the reaction
2CO(g) + O2(g)  2CO2(g)
From the following:
2CH4(g) + 3O2(g)  2CO(g) + 4H2O(g)
DG = 1088kJ/mol
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
DG = -801kJ/mol
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Example #12
Calculate DG for the reaction
C2H4 (g) + H2O(l)  C2H5OH(l)
From the following:
4C(s) + 6H2(g) + O2(g)  2C2H5OH(l)
DG = 700kJ/mol
2H2(g) + O2(g)  2H2O(l) DG = 474kJ/mol
2C(s) + 2H2(g)  C2H4(g) DG =
136kJ/mol
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More Ways to find DG! 
DGo = - R T lnK
DG = DGo + RT lnQ
K: equilibrium constants (Kc, Kp, Ka,
Kb, Kw, Ksp….)
Q: reaction quotient
R: 0.008314kJ/mol.K
T: temperature in Kelvin
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Example #13
At 127oC the synthesis of ammonia has
the following equilibrium concentrations:
[NH3] = 0.031M
[N2] = 0.85M
[H2] = 0.0031M
N2(g) + 3H2(g)  2NH3(g)
Calculate DGo for this reaction using the
equilibrium concentrations.
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Example #14
Calculate DG for the synthesis of
methanol at 25oC where carbon
monoxide is present at 5atm and
hydrogen gas is present at 3atm.
CO(g) + 2H2(g)  CH3OH(l)
Is this reaction spontaneous with these
pressures at this temperature?
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Example #15
Determine K for the following reaction:
6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g)
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Example #16
For a reaction DHo = -600kJ/mol and DSo
= 150J/mol.K and 25oC.
Calculate K for this reaction.
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Calculating Enthalpy Changes Using
Standard Heats of Formation
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Calculating Enthalpy Changes Using
Standard Heats of Formation
Using standard heats of formation, calculate the standard
enthalpy of reaction for the photosynthesis of glucose
(C6H12O6) and O2 from CO2 and liquid H2O.
6 CO2(g) + 6 H2O(l)
C6H12O6(s) + 6 O2(g) ∆H° = ?
H°= [∆H°f (C6H12O6(s))] - [6 ∆H°f (CO2(g)) + 6 ∆H°f (H2O(l))]
∆H°= [(1 mol)(-1273.3 kJ/mol)] -
[(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)]
= 2802.5 kJ
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Calculating Enthalpy Changes Using
Standard Heats of Formation
6 CO2(g) + 6 H2O(l)
(1)
(2)
C6H12O6(s) + 6 O2(g) ∆H° = 2802.5 kJ
(3)
Why does the calculation “work”?
1. Reverse the “reaction” and reverse the sign on the
standard enthalpy change.
C(s) + O2(g)
CO2(g)
∆H° = -393.5 kJ
Becomes
CO2(g)
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C(s) + O2(g) ∆H° = 393.5 kJ
Calculating Enthalpy Changes Using
Standard Heats of Formation
6 CO2(g) + 6 H2O(l)
(1)
(2)
C6H12O6(s) + 6 O2(g) ∆H° = 2802.5 kJ
(3)
Why does the calculation “work”?
1. Multiply the coefficients by a factor and multiply the
standard enthalpy change by the same factor.
CO2(g)
C(s) + O2(g)
∆H°= 393.5 kJ
Becomes
6 CO2(g)
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6 C(s) + 6 O2(g) ∆H°= 6(393.5 kJ) = 2361.0 kJ
Calculating Enthalpy Changes Using
Standard Heats of Formation
6 CO2(g) + 6 H2O(l)
(1)
(2)
C6H12O6(s) + 6 O2(g) ∆H° = 2802.5 kJ
(3)
Why does the calculation “work”?
2. Reverse the “reaction” and reverse the sign on the
standard enthalpy change.
1
H2(g) + O2(g)
2
H2O(l)
∆H°= -285.8 kJ
Becomes
H2O(l)
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1
H2(g) + O2(g) ∆H°= 285.8 kJ
2
Calculating Enthalpy Changes Using
Standard Heats of Formation
6 CO2(g) + 6 H2O(l)
(1)
(2)
C6H12O6(s) + 6 O2(g) ∆H°= 2802.5 kJ
(3)
Why does the calculation “work”?
2. Multiply the coefficients by a factor and multiply the
standard enthalpy change by the same factor.
H2O(l)
1
H2(g) + O2(g) ∆H° = 285.8 kJ
2
Becomes
6 H2O(l)
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6 H2(g) + 3 O2(g) ∆H°= 6(285.8 kJ) = 1714.8 kJ
Calculating Enthalpy Changes Using
Standard Heats of Formation
6 CO2(g) + 6 H2O(l)
(1)
(2)
C6H12O6(s) + 6 O2(g) ∆H°= 2802.5 kJ
(3)
Why does the calculation “work”?
6 C(s) + 6 H2(g) + 3 O2(g)
C6H12O6(s)
∆H°= -1273.3 kJ
6 CO2(g)
6 C(s) + 6 O2(g) ∆H°= 2361.0 kJ
6 H2O(l)
6 H2(g) + 3 O2(g) ∆H°= 1714.8 kJ
6 CO2(g) + 6 H2O(l)
C6H12O6(s) + 6 O2(g)
∆H°= 2802.5 kJ
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Calculating Enthalpy Changes Using
Bond Dissociation Energies
Bond Dissociation Energy:
•
The amount of energy that must be supplied to break a
chemical bond in an isolated molecule in the gaseous
state and is thus the amount of energy released when
the bond forms.
or
•
Standard enthalpy changes for the corresponding
bond-breaking reactions.
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Calculating Enthalpy Changes Using
Bond Dissociation Energies
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Calculating Enthalpy Changes Using
Bond Dissociation Energies
H2(g) + Cl2(g)
2 HCl(g)
∆H°= D(Reactant bonds) - D(Product bonds)
∆H°= (DH-H + DCl-Cl) - (2 DH-Cl)
∆H°= [(1 mol)(436 kJ/mol) + (1mol)(243 kJ/mol)] (2mol)(432 kJ/mol)
= -185 kJ
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Worked Example Using Bond Dissociation Energies to Calculate
ΔH°
Use the data in Table 8.3 to find an approximate ΔH°in kilojoules for the industrial synthesis of chloroform by
reaction of methane with Cl2.
Strategy
Identify all the bonds in the reactants and products, and
look up the appropriate bond dissociation energies in
Table 8.3. Then subtract the sum of the bond dissociation
energies in the products from the sum of the bond
dissociation energies in the reactants to find the enthalpy
change for the reaction.
Solution
The reactants have four C–H bonds and three Cl–Cl bonds;
the products have one C–H bond, three C–Cl bonds, and
three H–Cl bonds. The approximate bond dissociation
energies from Table 8.3 are:
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Worked Example Using Bond Dissociation Energies to Calculate
ΔH°
Continued
Subtracting the product bond dissociation energies from the reactant bond dissociation energies gives the
approximate enthalpy change for the reaction:
The reaction is exothermic by approximately 330 kJ.
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