Download Molar Heat of Combustion

Chemistry 3202
Thermo Notes 2016 (Text Ch 16 and 17, page 624)
Thermodynamics is the study of energy changes with matter when it melts or
boils (physical change) or burns (chemical change).
-Thermo is greek for hot, dynamics means movement.
-Energy - the ability to do work, “en” is greek word for in, & ergon is for work.
So energy is the “work in” something.This energy can be changed from one form to
another, for example electrical energy can be changed to light, but it cannot be lost
and was created in the big bang. (1St Law of Thermodynamics)
TWO (2) Types of energy:
1) Kinetic - Greek for movement is kinema (like cinema for moving
pictures ) where temperature is used to describe the kinetic energy of
particles. The kinetic energy of an object is due to the motion of the
object, which depends on the mass and velocity (speed) of the object:
K.E. = ½mv2
2) Potential - stored energy in the bonds of compounds, or the
intermolecular forces between molecules in phase changes . It is the
energy of an object associated with its chemical composition. For
example, water on top of a waterfall has a high potential energy. As this
water cascades down the mountain, its potential energy decreases, but
the kinetic energy increases as it falls with an increasing speed.
Energy Units
A calorie (cal) is defined as the quantity of heat required raising the
temperature of one gram of pure water by one degree Celsius. The SI
unit of energy is Joule (J), where 1 J = 1 kg.m2/s2, and 1 cal = 4.184 J.
The kilojoule , is where 1000 J = 1KJ] (also; calories is the old system still used for
food calories but is food is actually a Kilocalorie)
Physical Change
Example using a Physical Change of state: H2O(l) + energy (40.7 KJ/mol)  H2O(g)
The energy, in this case, is absorbed and thus endothermic. For the process:
H2O(l)  H2O(s) + energy (6.02 KJ/mol) (I call this freezing)
In this process, freezing, heat is lost and the phase change is exothermic. It
involves IMF’s and is a potential energy change.
Chemical Change- where the chemicals change. Unlike in a physical chage, where
the chemicals stay the same, as in melting and boiling.
In a chemical change, one substance(s)(reactants) with a specific amount of energy, is
changed into another substance(s)(products) (with a different amount of energy). Ex:
H2(g) + 1/2 O2(g)  H2O(g)
-Energy changes involve changes in bonding; thus the energy involved is potential
energy.
-Actually, reactant bonds break and rearrange to make new product bonds.
-For all reactions, energy is absorbed for bonds to break reactant bonds and is
released when new product bonds form.
Conversion from one form of energy to another obeys the first Law of
Thermodynamics, no energy is lost or made when we change forms of energy, say
the chemical change of the burning of coal in a power station into:
 Thermal (heat) energy from the burning
 Electrical energy from the spinning turbines coils of wire (in a magnetic field)
sent to your home.
 Mechanical (kinetic + potential) energy of the trbines
 Light energy for you to read this by
 Chemical energy again , if you are reading this by a battery charged laptop!
Example of a chemical change:
1. Propane burns:
C3H8(g)
+
5O2(g) 
H2O(g) + Energy
(Exothermic Reaction)
3CO2
+
4
2. Photosynthesis (MEMORIZE THIS REACTION) . It is often asked.
6CO2(g) + 6 H2O(g) + 3177.6 kJ

C6H12O6(l) + 6O2(g)
Photosynthesis is an endothermic reaction requiring 3177 kJ of energy for the
formation of one mole of glucose. The energy is stored in the glucose as chemical
potential energy.The heat of the reaction can also be written OUTSIDE of the
equation and then a sign (here positive for Endo thermic) is used:
6CO2(g) + 6 H2O(g) 
C6H12O6(l) + 6O2(g)
ΔH = +3177 kJ
The opposite of photosynthesis, cellular respiration (a form of burning), has an
energy release.
C6H12O6(l)
+ 6O2(g)

6CO2(g) +
6 H2O(g)
+
3177 kJ
OR
C6H12O6(l)
+
6O2(g) 
6CO2(g) + 6 H2O(g)
ΔH = -3177 kJ
Thus, in an endothermic reaction, the gaining of heat energy can be written in two
different ways:
(1) In the equation, on the reactant side ; or (2) Outside the equation as ΔH = +
In an exothermic reaction, the releasing of heat energy can be written:
(1) In the equation, on the product side or (2) Outside the equation as ΔH = Historical Background to the development of the Joule
Energy is the ability to do work. Physics students calculate work as: W=F x d (m x
a x d). This is easy for Physics students to measure, heck , all they need is a
meter stick and balance. In chemistry we have a hard time catching heat, and
measuring it is even harder. Heat is not tangible, you can’t grab it.
James Joules experiment solves all that. He catches heat and converts the
POTENTIAL energy of a raised brick into the KINETIC energy of a moving brick.
This mechanically generated value, gives 4.184 J /g C (the mechanical equivalent
of heat). (Note: In 1850, Joule published a refined measurement of 772 ft·lb or Btu
(4.1 J/cal) or 772 pounds raised 1 foot generates 1 Fahrenheit temperature
change in 1 pound of water. We use Metric today or 4.184 J/g C
The First Calorimeter (as seen in the film absolutezero.)
Temperature Meaasures Kinetic Energy
Temperature (T): a measure of the average kinetic energy of the particles. Or a
thermometer is a molecular speedometer.
Systems, Surrounding, and Universe
The First Law of Thermodynamics Energy is not created or destroyed, it is
transformed, and conserved, so this is also known as the Law of Conservation of
Energy. Or, more simply, heat is transferred from a hotter object to () a colder
object, again, a one way street. So if we have 2 blocks of Al (same mass) one at
80 oC touching the other one at 20 oC , they will equal out to 50 oC as the hot block
losses it’s energy to the cold one, NOT the other way around. HOT TO COLD, a
one way street.
The Second Law of Thermodynamics A rock will fall if you lift it up and then let
go. Hot frying pans cool down when taken off the stove. Air in a high-pressure tire
shoots out from even a small hole in its side to the lower pressure atmosphere. Ice
cubes melt in a warm room. No matter how often you organize your “stuff”, your
room, it eventually just ends up cluttered again!
A simple way of stating the second law: Energy spontaneously disperses from
being concentrated to becoming spread out. Entropy means being being
scattered or disordered.
Systems and Surroundings
System: The part being studied, the chemical. If energy leaves the system it is
EXOthermic (it is easier if you always take the perspective of the system (the
chemicals)). Surroundings: The parts of the universe, the surroundings, with which
the system is in contact.
Types of Systems: NEED to look at 2 things, Energy and Matter (mass)
1. Open Systems:
Both energy and matter can be transferred between a system and its
surroundings. Example: An open beaker of hot water loses heat to the
surroundings (exo)
System = water molecules
.
and the
Surroundings = beaker + air near it
2. Closed Systems:
Energy can be exchanged between the system and surroundings; matter cannot.
Example: A stoppered erlenmeyer flask of hot water loses heat to the
surroundings, however, water vapor cannot escape.
System = water molecules and Surroundings = beaker + stopper + air near it
3. Isolated Systems:
Neither energy nor matter is exchanged between the system and its
surroundings.
Thermally isolated, well now come to think of it, TOTALLY isolated.
Example: A vacuum flask or vacuum thermos of hot water approximates an
isolated system.
SURROUNDINGS
System = liquid contents
and
Surroundings = thermos + air near it
Note: A vacuum contains no matter, it cannot transfer heat by conduction or
convection.
Molar Heat of Combustion
Burning one mole of a substance is called the molar heat of combustion.
An Example: The molar heat of combustion of methane gas is 890.4 kJ.
CH4(g)
+ 2O2(g)
 CO2(g)
+
H2O(g)
ΔH = -890.4 kJ
PHASE CHANGES (Molar Heats)
A change in the state of matter without any change in the chemical compostion.
Usually that is melting or boiling, however PHYSICAL changes may be
sublimation, which occurs when you go from solid to liquid, quickly skipping
through the liquid phase as with :
I2 (s)  I2 (g) or dry ice, in smoke machines….. OR CO2(s) dry ice  CO2 (g)
PHYSICAL changes may also be simply dissolving NaCl (s)  NaCl (aq)
which further dissociates to Na+(aq) + Cl - (aq)
In the lab we disslove NaOH all the time and you remember it gave of a LOT of
heat.
Ex:
NaOH(s)  Na+(aq) + OH - (aq) + ENERGY
However, the major calculations, for physical changes involve Phase changes (…
plus the Kinetic changes ) . When there is a phase change, there is no change in
the kinetic energy of the particles, or no temperature changes. A change in
potential energy is taking place due to changes in the attractions between particles
(IMF, intermoloceular forces, for molecular substances).
Heating Curve for Water
GENERAL STEPS:
Leg A. Heating of Solid: As the temperature increases, kinetic energy increases
and potential energy remains constant.
The energy goes to Increased vibration of particles about their fixed positions in
solid.
Leg B. Fusion (Melting) of Solid: Here the temperature remains constant and
there is no change in the kinetic energy, however the potential energy increases.
The added energy weakens the attractions between the particles, so they move
more freely; now the substance changes into a liquid
Leg C. Heating of Liquid: The temperature again rises causing an increase in
kinetic energy but the potential energy remains constant.
The particles moving more freely, but the attractions are still strong enough to hold
particles in the liquid phase.
Leg D. Vaporization (Boiling) of Liquid: Again, like #B, this is a phase change so
the temperatue remains constant. This means the kinetic energy is constant and
the potential energy is increasing.
Again, the energy is used to overcome the attractions between particles as the
particles break free and the substance changes into the gas phase.
Leg E. Heating of Gas: The temperature begins to rise above its boiling point
causing the kinetic energy to increase (potential energy remains constant.) Adding
heat speeds up the gas particles.
Heating Curve Notes
Below is a diagram showing a typical heating/cooling curve for water. It reveals a
wealth of information about the structure and changes occurring in water as it is
heated or cooled through all three phases of matter at different temperatures. At
the top of the diagram are pictures representing the typical particle arrangement as
substances change through their states.
1. Identify by letter (A-E) in which section the following are found:
a. ____B___ Freezing (if cooling)
b. ___E____ Particles farthest apart
c. ___E____ Boiling
d. ____A___ Particle motion is most restricted
e. ____B___ Heat of fusion
f. ___B & D_ All areas where energy change is potential only
g. ____D___ Heat of vaporization
h.___A____ Least kinetic energy
i ____E___ most potential energy
j. ____B,D___ All areas where phase changes occur
k. _A,C,E______ All areas in which the heat is making the particles move faster
l. ____B,D___ All areas in which the heat is breaking the attractions or bonds
between particles
Calculating Kinetic Energy:
Kinetic energy involves movement, thus there will be a temp change as a
thermometer is a molecular speedometer. To get energy we need 3 things, speed
, type and number of molecules.
q = mcT
(or with mass and c put together and to be C in the equation
q = CT
)
q = mct is for when WATER is given in the calorimeter, HOWEVER, when we look at the
specific heat of the whole calorimeter, the water AND the calorimeter can itself, we look at
heat capacity (C ).
Heat Capacity (C):
The quantity of heat required to change the temperature of an object by 1 degree
Celsius, or q = C∆t (really the same as q = mc∆t , just that mc are combined or
already multipied for you)
specific heat (c) = q m-1 T-1 the heat needed to increase 1 g of material by 1 oC, as in
water = 4.184J g-1 oC-1
heat capacity (C) = q T-1 is the amount of heat needed to increase an object by 1 oC
(notice mass is incorporated into C) so we have just J oC-1 increase an object by 1 oC
(notice mass is absent or incorporated into C)
Example: A 1.32 g sample of sucrose, C12H22O11, is burned in a calorimeter with a heat
capacity of 9.43 x 103 J oC-1. If the temperature of the calorimeter changed from 25.00 oC to
27.31 oC, calculate the MOLAR heat of the reaction in units of kJ mole-1 of sucrose.
Answer:
C = 9.43 x 103 J oC-1 , (THIS IS NOT specific heat, it is HEAT CAPACITY) , T = 2.31 oC
q = C x T
q= 21.8 kJ
moles sucrose = mass / molar mass
= 1.32 g C12H22O11 /
342 g . mole-1
= 3.8596 x 10-3 mole C12H22O11
H = - 21.8 kJ / 3.8596 x 10-3 mole C12H22O11
H = - 5643.8 kJ mole-1
or
H = - 5.64 x 103 kJ mole-1
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Molar Heat: The total energy of a system is called the enthalpy (H) of the system, also
referred to as Molar Enthalpy
FOUND BY: q = n H (the enthalpy change involving one mole of a substance, the
units = kJ/mol. )
Hrx = q / # of moles
or
Hrx = mcT / # of moles
Example: Calculate the heat required to melt 50.0 g of ice at 0.0°C.
Answer
Notice the sign of the heat value. Melting (fusion) ice
melting is an endothermic process - energy is required to break the intermolecular
forces that hold water molecules together a ice.
Types of Problems
1) Physical Changes
Ex: Hot metal is dropped into 1000.0 g ( always be sure to check the units of mass as
they must match the units of specific heat, KJ with KJ, and J with J ) of water. If the
temperature of the water rose from 22.1 oC to 35.9 oC, how much heat energy did the
water gain?
q = mcT
13.8 oC
where m = 1000.0 g and
c = 4.18 J g-1 oC-1
T = (35.9 oC – 22.1 oC) =
q = (1000.0 g)( 4.18 J g-1 oC-1)( 13.8 oC)
q = 57.7 kJ or 5.77 x 104 J
q = 57684 J
NOTE - this is an exothermic change. Please do Practice Problems: Page 636: #’s 7 – 10
(in the book)
2) Chemical Changes
Example: A student placed 50.0 mL of 1.00 M HCl at 25.5 oC in a coffee cup calorimeter.
To this, 50.0 mL of 1.00 M NaOH also at 25.5 oC was added. The mixture temperature rose
to a maximum of 32.4 oC. Calculate the heat of the reaction (heat of neutralization).
mAcid = 50.0 mL x 1.0 g mL-1 = 50.0 g we also have 50 grams base OR 100 grams acid
and base which when reacted is water.
c = 4.18 J g-1 oC-1
T = (32.4 oC – 25.5 oC) = 6.9 oC
q = mcT
q = (100.0 g water (acid and base combo))( 4.18 J g-1 oC-1)( 6.9 oC) and so q = 2.88 kJ
H per mole of HCl …
n=Cxv
or
n = ( 1.00 M )( 0.0500 L)
n = 0.0500 mole HCl
H = - 2.88 kJ/ 0.0500 mole HCl
= - 57.7 kJ mole-1
There are 3 types of changes in Thermodynamics you need to know:
We are concerned with 3 types of changes in:
a) Physical changes - Tens of KJ/mol (LITTLE HEAT)
b) Chemical Changes - Hundreds of KJ/mol ( A LOT OF HEAT)
c) Nuclear changes - Millions and even Billions KJ/mol (HOOCHIE MAMA,
AWESOME HEAT)
physical
H2(l)  H2(g)
chemical
2 H2(g) + O2(g)  2 H2O(l)
nuclear
Hnucleus + Hnucleus  Henucleus
10 to 100 kJ/mole (Tens)
100 to 10,000 kJ/mole
(Thousands)
1010 to 1012 kJ/mole
(Billions)
CHEMICAL CHANGE CALCULATIONS
Knowing the heat change associated with a chemical change involving one mole of
a chemical allows you to calculate the heat change associated with any measured
amount of that chemical.
Example : Calculate the heat change when 10.4 g of CaCl2 is formed from its
elements . The molar heat for this FORMATION reaction can be obtained from
molar heats of formation tables, but is given here as -795.4 KJ/mol. THIS is a
CHEMICAL change
Ca(s)
+ Cl2 (g)

CaCl2 (s)
Example : A 5.00 g sample of benzene was burned at SATP conditions resulting
in the release of 209.2 kJ of heat. Calculate the molar heat of combustion for
benzene.
The molar heat of combustion of benzene is -3267 kJ/mol.
Heat of Reaction and Simple Calorimetry
Heat changes associated with aqueous chemical reactions can be determined by
simple calorimetry. This information can then be used to calculate heat of reaction.
HCl
+
NaOH

HOH
+
NaCl
Example: A 75.00 mL sample of 0.500 M hydrochloric acid at 25.00°C was mixed
with 75.00 mL of 0.500 M NaOH in a simple calorimeter. The highest temperature
recorded after mixing was 27.19°C. Calculate the molar heat of reaction for
hydrochloric acid.
CHEMICAL CHANGE CALCULATIONS WITHOUT CALORIMETERS
( 3 METHODS)
1) FLIP
2) Math (formula) (ΔH°rxn = Σ nΔH°f, products - Σn ΔH°f, reactants)
3) Bond Energies
1) Hess's Law: (FLIP THE EQUATION METHOD)
Mr. Hess states, if two or more equations can be added together to produce an
overall equation, the sum of the enthalpy terms equals the TOTAL enthalpy change
Ex:
Given these thermochemical equations:
Use Hess’s law of heat summation calculate the molar enthalpy for the incomplete
combustion of graphite:
2) Hess's Law: (THE MATH METHOD)
There is a second way to use Hess' Law :
ΔH°rxn = Σ nΔH°f, products - Σn ΔH°f,
reactants
Problem #1: Calculate the standard enthalpy of combustion for the following
reaction: C2H5OH (l) + (7/2) O2 (g) ---> 2 CO2 (g) + 3 H2O (l)
The ΔH°f values I will now give as, FOR 1 Mole:
-393.5 for CO2,
water,
-278 for ethanol
AND
zero for ALL elements
-286 for
ΔH°combustion = [ 2 (-393.5) + 3 (-286) ] minus [ (-278) + (7/2) (0) ]
Doing the math gives us ΔH°comb = -1367 kJ/mol of ethyl alcohol (C2H5OH).
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3) Bond Energies. The process of measuring the changes in the enthalpy of a
system is called calorimetry. The change in enthalpy, H, is related to what is
happening in the reaction.
Ex:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
We can burn methane and the heat it gives off can be absorbed and measured by a cup
of water (calorimeter)
Bonds being broken (THIS ABSORBS ENERGY):
:
CH4(g) + 2 O2(g)  Products
4 C-H bonds are being broken
require energy
AND
2 O-O bonds are being broken these
Bonds being formed (RELEASES ENERGY):
Reactants

CO2(g) + 2 H2O(l)
2 C=O bonds being formed
releasing energy
AND
4 H-O bonds being formed
As the type and number of bonds being broken and formed change, so will the change in
enthalpy.
H = Hbond energy reactants - H bond energy products
Calculating Bond Energies:
Bond breaking is endothermic, because energy is absorbed to separate bonded
particles. The potential energy of the separated particles is greater than that of the
bonded particles.
The amount of energy needed to separate one mole of a bonded species is called
the bond energy. Keeping the sign convention in mind, all bond energies are
positive!
Example: Using average bond energies, calculate the enthalpy change for the
combustion of ethanol.
Enthalpy Diagrams
Potential energy diagrams, also known as enthalpy diagrams, show the relative
potential energies of the species involved in a chemical (or physical) change. Molar
enthalpy is the potential energy change associated with a physical or chemical
change involving one mole of a substance. You may heard this as the heat of
reaction. Consider the following EXOTHERMIC diagram.
The down arrow shows the reaction is exothermic. The potential energy of
products are less than the reactants. The P.E. of the system decreases and is
released to the surroundings as heat.
Now consider this ENDOTHERMIC diagram.
The enthalpy diagram for an endothermic reaction always shows the arrow pointing
upwards. The potential energy of the system increases as energy is absorbed
from the surroundings.
Fueling our Bodies STSE - Caloric Intake
Before going further we must clarify a potentially confusing unit related to the
Calories quoted on food labels. A calorie (cal) is the amount of energy required to
increase the temperature of one gram of water by degree Celsius:
1 cal = 4.184 J
yet this is not the same "calorie" as the food Calorie (Cal). One food Calorie is
equal to one kilocalorie(note the uppercase "C" in the unit for food Calorie
1 food Calorie = 1 Cal = 1000 cal = 1 kcal = 4.184 kJ
Sample Problem
A 2.50g sample of bar is placed in a bomb calorimeter with a heat capacity of
13.17 kJ/oC. The temperature of the calorimeter contents increases by 3.86 oC
upon complete combustion of the sample.
Determine:
(a) the food value, and
(b) number of Calories in a 23.0g serving (one granola bar).
Solution
(a) First, determine the heat of the calorimeter surroundings:
qsurr = CT = (13.17 kJ/oC ) (3.86oC) = 50.8 kJ
Second, we could state the heat evolved by the system (the granola bar) as:
qsys = - qsurr = - 50.8 kJ
Next.
FV = Food kJ = 50.8 kJ/2.50 g = 20.3 kJ/g
To calculate the total Calories in a 23.0g serving:
Food kJ = FV x m = ( 20.3 kJ/g ) x ( 23.0g ) = 467 kJ.
In case you are "counting calories", since 1 Cal = 4.184 kJ, the same food energy
in the old unit of Calories is:
Cal = (467 kJ ) x ( 1 Cal / 4.184 kJ) = 112 Cal
The 23g granola bar contains 467 kJ, or 112 Calories.
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