Transcript CH 6: THERMOCHEMISTRY Vanessa Prasad-Permaul Valencia College CHM 1045

CH 6: THERMOCHEMISTRY
Vanessa Prasad-Permaul
Valencia College
CHM 1045
1
•INTRODUCTION TO THERMODYNAMICS
•KEY DEFINITIONS
•FIRST LAW OF THERMODYNAMICS
•HEAT, WORK & INTERNAL ENERGY
•ENTHALPY
•THERMODYNAMIC SYSTEM
•HESS’ LAW
•STANDARD ENTHALPIES OF FORMATION
•ENERGY AND THE ENVIRONMENT
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THERMOCHEMISTRY
ENERGY
 Energy is the capacity to do work, or supply
heat.
Energy = Work + Heat
 Kinetic Energy is the energy of motion.
Ek = 1/2 mv2
Ek = kinetic energy
m = mass (kg)
v = velocity (m/s)
(1 Joule = 1 kgm2/s2)
(1 calorie = 4.184 J)
 Potential Energy is stored energy.
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THERMOCHEMISTRY
ENERGY
EXAMPLE 6.1
A regulation baseball weighing 143 grams travels 75 miles per
hour. What is the kinetic energy of this baseball in Joules?
Convert to calories.
Ek = ½ mv2
m = 143g x 1kg = 0.143kg
1000g
v = 75miles x 1hr x 1min x 1609.3m = 33.527 m/s
1 hour 60min 60sec
1 mile
Ek = ½ x 0.143kg x (33.527m/s)2 = 80.3kg.m2/s2 = 80.3J
80.3J x 1 cal = 19.2cal
4.184J
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THERMOCHEMISTRY
ENERGY
EXERCISE 6.1
An electron whose mass is 9.11 x 10-31 kg is accelerated by a
positive charge to a speed of 5.0 x 106 m/s. What is the kinetic
energy of the electron in Joules? In calories?
Ek = ½ mv2
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THE FIRST LAW OF THERMODYNAMICS:
 The law of the conservation of energy: Energy
cannot be created or destroyed.
 The energy of an isolated system must be
constant.
 The energy change in a system equals the work
done on the system + the heat added.
DE = Efinal – Einitial = E2 – E1 = q + w
q = heat, w = work
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 Thermal Energy is the kinetic energy of
molecular motion
 Thermal energy is proportional to the
absolute temperature. Ethermal  T(K)
 Heat is the amount of thermal energy
transferred between two objects at different
temperatures.
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 Enthalpies of Physical Change:
Enthalpy is a state
function, the enthalpy
change from solid to
vapor does not
depend on the path
taken between the
two states.
DHsubl = DHfusion + DHvap
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ILLUSTRATION OF A THERMODYNAMIC SYSTEM
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 In an experiment: Reactants and products are
the system; everything else is the surroundings.
• Energy flow from the system to the surroundings has
a negative sign (loss of energy). (-DE or - DH)
• Energy flow from the surroundings to the system has
a positive sign (gain of energy). (+DE or +DH)
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 Enthalpies of Chemical Change: Often called
heats of reaction (DHreaction).
Endothermic: Heat flows into the system from the
surroundings, heat is absorbed.
 DH has a positive sign.
 Energy added
 q is (+)
Exothermic: Heat flows out of the system into the
surroundings, heat is evolved.
 DH has a negative sign.
 Energy subtracted
 q is (-)
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SIGN CONVENTIONS FOR HEAT, WORK
AND INTERNAL ENERGY
q (heat)
w (work)
DE (change in
internal energy)
+
System gains
System loses
thermal energy
thermal energy
+
Work done on a Work done by the
system
system
+
Energy flows into Energy flows out
the system
of the system
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EXAMPLE 6.2
Barium hydroxide octahydrate reacts with ammonium
nitrate:
Ba(OH)2.8H2O(s) + 2NH4NO3(s)
2NH3(g) + 10H2O(l) + Ba(NO3)2(aq)
When 1mol of barium hydroxide octahydrate reacts with 2
mol of ammonium nitrate, the reaction mixture absorbs
170.8kJ of heat. Is this reaction exothermic or endothermic?
What is the heat of reaction (q)?
Energy is absorbed so this reaction is endothermic.
q is +170.8kJ
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EXERCISE 6.2
Ammonia burns in the presence of a platinum catalyst to
give nitric oxide
4NH3(g) + 5O2(g)
Pt
4NO(g) + 6H2O(l)
4 mol of ammonia is burned and 1170kJ of heat is evolved.
Is the reaction endothermic or exothermic? What is the
value of q?
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 The amount of heat exchanged between the
system and the surroundings is given the
symbol q.
q = DE + PDV
At constant volume (DV = 0): qv = DE
At constant pressure: qp = DE + PDV = DH
Enthalpy of Reaction: DH = Hproducts – Hreactants
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PRESSURE-VOLUME WORK
Work = -atmospheric pressure * area of piston * distance piston moves
w = -PDV
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EXAMPLE 6.2A
If a balloon is inflated from 0.100L to 1.85L against an
external pressure of 1.00atm, how much work is done (J)?
w = -PDV
= -1.00atm (1.85L – 0.100L)
= -1.75 L.atm
Conversion 1L.atm = 101.3J
-1.75 L.atm x 101.3J
1L.atm
-177J
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EXERCISE 6.2A
A cylinder equipped with a piston expands against an
external pressure of 1.58 atm. If the initial volume is
0.485L and the final volume is 1.245L, how much work is
done (in J)?
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EXAMPLE 6.3
Aqueous sodium hydrogen carbonate reacts with
hydrochloric acid to produce aqueous sodium chloride,
water and carbon dioxide gas. The reaction absorbs
12.7kJ of heat at constant pressure for each mole of
NaHCO3. Write the thermochemical equation.
NaHCO3(aq) + HCl
NaCl(aq) + H2O(l) + CO2(g)
DH = +12.7kJ
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EXERCISE 6.3
A propellant for rockets is obtained by mixing the liquids
hydrazine (N2H4) and dinitrogen tetroxide. These
compounds react to give gaseous nitrogen and water vapor
and 1049kJ of heat is evolved (at constant pressure when 1
mol reacts). Write the thermochemical equation for this
reaction
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 Reversing a reaction changes the sign of DH for a
reaction.
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DH = –2219 kJ
3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2(g) DH = +2219 kJ
 Multiplying a reaction increases DH by the same
factor.
3 [C3H8(g) + 15 O2(g)  9 CO2(g) + 12 H2O(l)]:
DH = 3(-2219) kJ
DH = -6657 kJ
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EXAMPLE 6.4
When 2 mol H2(g) and 1mol O2 react to give liquid water, 572kJ of
heat evolves:
2H2(g) + O2(g)
2H2O(l) ; DH = -572kJ
Write this equation for 1 mol of liquid water. Give the reverse
equation, in which 1 mol of liquid water dissociates into hydrogen
and oxygen.
2H2(g) + O2(g)
1H2(g) + ½ O2(g)
1H2O(l)
2H2O(l) ; DH = -572kJ
2
1H2O(l) ; DH = -286kJ
Reversing the equation:
1H2(g) + ½ O2(g) ; DH = +286kJ
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EXERCISE 6.4
Write the thermochemical equation for the reaction
described in Exercise 6.3 for the case involving 1 mol N2H4.
Write the reverse reaction.
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Applying Stoichiometry to Heats of Reaction
Grams of A
(reactant or
product)
x
Conversion Factor:
grams of A to mols
of A (using molar
mass)
=
kiloJoules of
Heat
x
Conversion Factor:
mols of A to kJ
(DH)
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EXAMPLE 6.5
How much heat is involved when 9.007x105g of ammonia is
produced according to the following reaction (assuming the
reaction is at constant pressure)?
N2(g) + 3H2(g)
2NH3(g); DH = -91.8kJ
9.07 x 105g x 1 mol NH3 x
17.0g NH3
-91.8kJ = -2.45 x 106 kJ
2 mol NH3
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EXERCISE 6.5
How much heat evolves when 10.0g of hydrazine reacts
according to the reaction described in EXERCISE 6.3?
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 Heat capacity (C) is the amount of heat required to raise the
temperature of an object or substance a given amount.
qcal = Ccal x DT
 Specific Heat: The amount of heat required to raise
the temperature of 1.00 g of substance by 1.00°C at
constant pressure.
q = s x m x Dt
q = heat required (energy)
s = specific heat
m = mass in grams
Dt = Tf - Ti
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SPECIFIC HEATS AND MOLAR CAPACITIES FOR SOME COMMON
SUBSTANCES @ 25oC
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• Molar Heat: The amount of heat required to raise the
temperature of 1.00 mole of substance by 1.00°C.
q = MH x n x Dt
q = heat required (energy)
MH = molar heat
n = moles
Dt = Tf - Ti
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EXAMPLE 6.6
Calculate the heat absorbed by 15.0g of water to raise the
temperature from 20.0oC to 50.0oC (at constant pressure).
The specific heat of water is 4.18 J/g.oC
Dt = 50.0oC – 20.0oC = 30.0oC
q = s x m x Dt
q = 4.18J x 15.0g x 30.0oC = 1.88 x 103 J
g.oC
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EXERCISE 6.6
Iron metal has a specific heat of 0.449 J/(g.oC). How
much heat is transferred to a 5.00 g piece of iron, initially
at 20.0oC when placed in a pot of boiling water? (Assume
that the temperature of the water is 100.0oC and the
water remains at this temperature, which is the final
temperature of the iron).
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 Calorimetry is the science of measuring heat
changes (q) for chemical reactions. There are two
types of calorimeters:
• Bomb Calorimetry: A bomb calorimeter measures the
heat change at constant volume such that q = DE.
• Constant Pressure Calorimetry: A constant pressure
calorimeter measures the heat change at constant
pressure such that q = DH.
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Constant Pressure
Bomb
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EXAMPLE 6.7
Suppose 0.562g of graphite is placed in a calorimeter with an
excess of oxygen at 25.00oC and 1atm pressure. Excess O2
ensures that all carbon burns to form CO2. The graphite is
ignited, and it burns according to the following equation
C(graphite) + O2(g)
CO2(g)
On reaction, the calorimeter temperature rises from 25.00oC to
25.89oC. The heat capacity of the calorimeter and it’s contents
was determined in a separate experiment to be 20.7 kJ/oC. What
is the heat of reaction at 25.00oC and 1 atm pressure? Express in
a thermochemical equation.
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EXAMPLE 6.7 continued…
qrxn = -CcalDt = -20.7kJ/oC x (25.89oC – 25.00oC)
= -20.7kJ/oC x 0.89oC
= -18.4kJ
0.562g x 1 mol = 0.0468 mol
12 g C
-18.4kJ
0.0468mol
= -3.9 x 102 kJ/mol
C(graphite) + O2(g)
CO2(g) ; DH = -3.9 x 102 kJ
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EXERCISE 6.7
Suppose 33mL of 0.120M HCl is added to 42mL of a solution
containing excess sodium hydroxide in a coffee cup
calorimeter. The solution temperature, originally at 25oC, rises
to 31.8oC. Give the enthalpy change DH for the reaction:
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
Assume that the heat capacity and the density of the final
solution in the cup are those of water.
s = 4.184kJ/g.oC
d = 1.000g/mL
Express the answer as a thermochemical equation.
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THERMOCHEMISTRY
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.(not a physical change,
chemical change)
3 H2(g) + N2(g)  2 NH3(g) DH° = –92.2 kJ
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THERMOCHEMISTRY
HESS’S LAW
 Reactants and products in individual steps can be
added and subtracted to determine the overall
equation.
(1) 2 H2(g) + N2(g) 
N2H4(g)
DH°1 = ?
(2) N2H4(g) + H2(g)  2 NH3(g)
DH°2 = –187.6 kJ
(3) 3 H2(g) + N2(g)  2 NH3(g)
DH°3 = –92.2 kJ
DH°1 + DH°2 = DH°reaction
Then DH°1 = DH°reaction - DH°2
DH°1 = DH°3 – DH°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ
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THERMOCHEMISTRY
HESS’S LAW
EXAMPLE 6.8
What is the enthalpy of reaction, DH, for the formation of
tungsten carbide (WC) from the elements?
W(s) + C(graphite)
2W(s) + 3O2(g)
2WO3(s) ;
C(graphite) + O2(g)
2WC(s) + 5O2(g)
CO2(g) ;
WC(s)
DH = -1685.8 kJ
DH = -393.5 kJ
2WO3(s) + 2CO2(g) ; DH = -2391.8kJ
LAST EQUATION NEEDS TO BE REVERSED
2CO2(g) + 2WO3(s)
2WC(s) + 5O2(g) ; DH = 2391.8kJ
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THERMOCHEMISTRY
HESS’S LAW
EXAMPLE 6.8 cont…
W(s) + C(graphite)
2W(s) + 3O2(g)
2
2WO3(s) ;
C(graphite) + O2(g)
2CO2(g) + 2WO3(s)
2
CO2(g) ;
WC(s)
DH = -1685.8 kJ
2
DH = -393.5 kJ
2WC(s) + 5O2(g) ; DH = 2391.8kJ
2
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EXAMPLE 6.8 cont…
W(s) + C(graphite)
W(s) +
3/2 O2(g)
C(graphite) + O2(g)
CO2(g) + WO3(s)
W(s) + C(graphite)
WC(s)
DH = -842.9 kJ
WO3(s) ;
CO2(g) ;
WC(s) +
WC(s)
DH = -393.5 kJ
5/2 O2(g) ;
DH = 1195.9kJ
DH = -40.5kJ
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EXERCISE 6.8
Manganese metal can be obtained by the reaction of
manganese dioxide and aluminum
4Al(s) + 3MnO2(s)  2Al2O3(s) + 3Mn(s)
What is the DH for this reaction? Use the following data:
2Al(s) + 3/2O2(g)  Al2O3(s) ;
DH = −1676 kJ
Mn(s) + O2(g)  MnO2(s) ;
DH = −520 kJ
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THERMOCHEMISTRY
STANDARD ENTHALPIES OF FORMATION
 Standard Heats of Formation (DH°f): The
enthalpy change for the formation of 1 mole
of substance in its standard state from its
constituent elements in their standard states.
 The standard heat of formation for any
element in its standard state is defined as
being ZERO.
DH°f = 0 for an element in its standard state
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THERMOCHEMISTRY
STANDARD ENTHALPIES OF FORMATION
Some Heats of Formation, DHf° (kJ/mol)
CO(g)
-111
C2H2(g)
227
Ag+(aq)
106
CO2(g)
-394
C2H4(g)
52
Na+(aq)
-240
H2O(l)
-286
C2H6(g)
-85
NO3-(aq)
-207
NH3(g)
-46
CH3OH(g)
-201
Cl-(aq)
-167
N2H4(g)
95.4
C2H5OH(g)
-235
AgCl(s)
-127
HCl(g)
-92
C6H6(l)
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Na2CO3(s)
-1131
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THERMOCHEMISTRY
STANDARD ENTHALPIES OF FORMATION
 Thermodynamic Standard State: Most
stable form of a substance at 1 atm pressure
and 25°C; 1 M concentration for all
substances in solution.
 These are indicated by a superscript ° to the
symbol of the quantity reported.
 Standard enthalpy change is indicated by the
symbol DH°.
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THERMOCHEMISTRY
STANDARD ENTHALPIES OF FORMATION
 Calculating DH° for a reaction:
DH° = DH°f (Products) – DH°f (Reactants)
 For a balanced equation, each heat of formation
must be multiplied by the stoichiometric
coefficient.
aA + bB  cC + dD
DH° = [cDH°f (C) + dDH°f (D)] – [aDH°f (A) + bDH°f (B)]
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THERMOCHEMISTRY
STANDARD ENTHALPIES OF FORMATION
EXAMPLE 6.9
Use the values of DHof to calculate the heat of vaporization
DHovap of carbon disulfide at 25oC. The vaporization process
is:
CS2(l)
CS2(g)
DHof = 89.7kJ/mol
DHof = 116.9kJ/mol
DHovap = Sn DHof(products) – Sm DHof (reactants)
DHof[CS2(g)] - DHof [CS2(l)]
116.9kJ - 89.7kJ = 27.2kJ
47
THERMOCHEMISTRY
STANDARD ENTHALPIES OF FORMATION
EXERCISE 6.9
Calculate the heat of vaporization, DHovap of water using
standard enthalpies of formation
48
THERMOCHEMISTRY
STANDARD ENTHALPIES OF FORMATION
EXAMPLE 6.10
Large quantities of ammonia are used to prepared nitric
acid. The first consists of the catalytic oxidation of ammonia
to nitric oxide
4NH3(g) + 5O2(g)
45.9kJ
0kJ
4NO(g) + 6H2O(g)
90.3 6kJ
-241.8kJ
What is the standard enthalpy change for this reaction?
DHovap = Sn DHof(products) – SmDHof (reactants)
= [4(90.3) + 6(-241.8)]kJ - [4(-45.9) + 5(0)]kJ
= -906kJ
49
THERMOCHEMISTRY
STANDARD ENTHALPIES OF FORMATION
EXERCISE 6.10
Calculate the enthalpy change for the following reaction:
3NO2(g) + H2O(l)
2HNO3(aq) + NO(g)
50
THERMOCHEMISTRY
STANDARD ENTHALPIES OF FORMATION
EXERCISE 6.11
Calculate the standard enthalpy change for the reaction of
an aqueous solution of barium hydroxide with an aqueous
solution of ammonium nitrate at 25oC.
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THERMOCHEMISTRY
FUELS
FUELS is any substance that is burned or react to
provide heat and other forms of energy.
FOODS AS FUELS
FOSSIL FUELS
COAL GASIFICATION AND LIQUEFICATION
ROCKET FUELS
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THERMOCHEMISTRY
FUELS
Sources of energy consumed in the United States.
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Example 2: Work
How much work is done (in kilojoules) and in
which direction, as a result of the following
reaction?
54
Example 3: Work
The explosion of 2.00 mol of solid TNT with a
volume of approximately 0.274 L produces
gases with a volume of 489 L at room
temperature. How much PV (in kilojoules)
work is done during the explosion? Assume P
= 1 atm, T = 25°C.
2 C7H5N3O6(s)  12 CO(g) + 5 H2(g) + 3 N2(g) + 2 C(s)
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Example 5:
Is an endothermic reaction a favorable process
thermodynamically speaking?
1) Yes
2) No
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Example 6: Hess’s Law
 The industrial degreasing solvent methylene
chloride (CH2Cl2, dichloromethane) is prepared
from methane by reaction with chlorine:
CH4(g) + 2 Cl2(g) CH2Cl2(g) + 2 HCl(g)
Use the following data to calculate DH° (in kilojoules)
for the above reaction:
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
DH° = –98.3 kJ
CH3Cl(g) + Cl2(g)  CH2Cl2(g) + HCl(g)
DH° = –104 kJ
57
Example 7: Standard heat of formation
Calculate DH° (in kilojoules) for the reaction of
ammonia with O2 to yield nitric oxide (NO)
and H2O(g), a step in the Ostwald process for
the commercial production of nitric acid.
58
Example 8: Standard heat of formation
Calculate DH° (in kilojoules) for the
photosynthesis of glucose and O2 from CO2
and liquid water, a reaction carried out by all
green plants.
59
Example 9:
Which of the following would indicate an
endothermic reaction? Why?
1. -DH
2. + DH
60
Example 10: Specific Heat
What is the specific heat of lead if it takes 96 J
to raise the temperature of a 75 g block by
10.0°C?
61
Example 11: Specific Heat
How much energy (in J) does it take to increase
the temperature of 12.8 g of Gold from 56C
to 85C?
62
Example 12: Molar Heat
 How much energy (in J) does it take to increase
the temperature of 1.45 x104 moles of water
from 69C to 94C?
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 How much heat (in kilojoules) is evolved or absorbed in
each of the following reactions?
a) Burning of 15.5 g of propane:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
DH = –2219 kJ/mole
b) Reaction of 4.88 g of barium hydroxide octahydrate with
ammonium chloride:
Ba(OH)2·8 H2O(s) + 2 NH4Cl(s)  BaCl2(aq) + 2 NH3(aq) + 10 H2O(l)
DH = +80.3 kJ/mole
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Heat of Phase Transitions from DHf
Calculate the heat of vaporization, DHvap of water,
using standard enthalpies of formation
H2O(g)
H2O(l)
DHf
-241.8 kJ/mol
-285.8 kJ/mol
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Bromination vs. Chlorination
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