Thermochemistry - HCC Learning Web

Comments

Transcription

Thermochemistry - HCC Learning Web
CHAPTER 5
THERMOCHEMISTRY
Energy
• Energy is the ability to do work or
transfer heat.
– Energy used to cause an object that has mass
to move is called work.
– Energy used to cause the temperature of an
object to rise is called heat.
• This chapter is about thermodynamics,
which is the study of energy
transformations, and thermochemistry,
which applies the field to chemical
reactions, specifically.
Thermochemistry
© 2015 Pearson Education, Inc.
Kinetic Energy
Kinetic energy is energy an object
possesses by virtue of its motion:
Thermochemistry
© 2015 Pearson Education, Inc.
Potential Energy
• Potential energy is
energy an object
possesses by virtue of
its position or chemical
composition.
• The most important
form of potential energy
in molecules is
electrostatic potential
energy, Eel:
Thermochemistry
© 2015 Pearson Education, Inc.
Units of Energy
• The SI unit of energy is the joule (J):
• An older, non-SI unit is still in
widespread use, the calorie (cal):
1 cal = 4.184 J
(Note: this is not the same as the calorie
of foods; the food calorie is 1 kcal!)
Thermochemistry
© 2015 Pearson Education, Inc.
Definitions: System and Surroundings
• The system includes the
molecules we want to
study (here, the hydrogen
and oxygen molecules).
• The surroundings are
everything else (here, the
cylinder and piston).
Thermochemistry
© 2015 Pearson Education, Inc.
Definitions: Work
• Energy used to move
an object over some
distance is work:
• w=Fd
where w is work, F
is the force, and d is
the distance over
which the force is
exerted.
Thermochemistry
© 2015 Pearson Education, Inc.
Energy transformations
– Energy is the capacity for doing work or supplying heat
– Thermochemistry is the study of energy changes that occur
during chemical reactions and changes in state
– Chemical potential energy is the energy stored in chemical
bonds
• Chemical reactions transform substances into different substances
with a different amount of chemical energy
• Energy changes occur as either heat transfer or work, or a
combination of both
– Heat (q), is energy that transfers from one object to another due
to temperature differences
• Heat always flows from a warmer object to a cooler object
Heat
• Energy can also be
transferred as heat.
• Heat flows from
warmer objects to
cooler objects.
Thermochemistry
© 2015 Pearson Education, Inc.
Conversion of Energy
• Energy can be converted from one type to another.
• The cyclist has potential energy as she sits on top of
the hill.
• As she coasts down the hill, her potential energy is
converted to kinetic energy until the bottom, where
the energy is converted to kinetic energy.
Thermochemistry
© 2015 Pearson Education, Inc.
First Law of Thermodynamics
• Energy is neither created nor
destroyed.
• In other words, the total energy of the
universe is a constant; if the system
loses energy, it must be gained by the
surroundings, and vice versa.
Thermochemistry
© 2015 Pearson Education, Inc.
Internal Energy
The internal energy of a system is the sum of all
kinetic and potential energies of all components
of the system; we call it E.
Thermochemistry
© 2015 Pearson Education, Inc.
Internal Energy
By definition, the change in internal energy, E,
is the final energy of the system minus the initial
energy of the system:
E = Efinal − Einitial
Thermochemistry
© 2015 Pearson Education, Inc.
Changes in Internal Energy
• If E > 0, Efinal > Einitial
– Therefore, the system absorbed energy
from the surroundings.
– This energy change is called endergonic.
Thermochemistry
© 2015 Pearson Education, Inc.
Changes in Internal Energy
• If E < 0, Efinal < Einitial
– Therefore, the system released energy to
the surroundings.
– This energy change is called exergonic.
Thermochemistry
© 2015 Pearson Education, Inc.
Changes in Internal Energy
• When energy is
exchanged between
the system and the
surroundings, it is
exchanged as either
heat (q) or work (w).
• That is, E = q + w.
Thermochemistry
© 2015 Pearson Education, Inc.
E, q, w, and Their Signs
Thermochemistry
© 2015 Pearson Education, Inc.
Exchange of Heat between System and
Surroundings
• When heat is absorbed by the system from the
surroundings, the process is endothermic.
Thermochemistry
© 2015 Pearson Education, Inc.
Exchange of Heat between System and
Surroundings
• When heat is released by the system into the
surroundings, the process is exothermic.
Thermochemistry
© 2015 Pearson Education, Inc.
State Functions
• Usually we have no way of knowing the internal
energy of a system; finding that value is simply too
complex a problem.
• However, we do know that the internal energy of a
system is independent of the path by which the
system achieved that state.
– In the system below, the water could have reached room
temperature from either direction.
Thermochemistry
© 2015 Pearson Education, Inc.
State Functions
• Therefore, internal energy is a state function.
• It depends only on the present state of the
system, not on the path by which the system
arrived at that state.
• And so, E depends only on Einitial and Efinal.
Thermochemistry
© 2015 Pearson Education, Inc.
State Functions
• However, q and w are
not state functions.
• Whether the battery is
shorted out or is
discharged by running
the fan, its E is the
same.
– But q and w are different
in the two cases.
Thermochemistry
© 2015 Pearson Education, Inc.
Work
Usually in an open
container the only work
done is by a gas
pushing on the
surroundings (or by
the surroundings
pushing on the gas).
Thermochemistry
© 2015 Pearson Education, Inc.
Work
We can measure the work done by the gas if
the reaction is done in a vessel that has been
fitted with a piston:
w = −PV
Thermochemistry
© 2015 Pearson Education, Inc.
Enthalpy
• If a process takes place at constant
pressure (as the majority of processes we
study do) and the only work done is this
pressure–volume work, we can account for
heat flow during the process by measuring
the enthalpy of the system.
• Enthalpy is the internal energy plus the
product of pressure and volume:
H = E + PV
Thermochemistry
© 2015 Pearson Education, Inc.
Enthalpy
• When the system changes at constant
pressure, the change in enthalpy, H, is
H = (E + PV)
• This can be written
H = E + PV
Thermochemistry
© 2015 Pearson Education, Inc.
Enthalpy
• Since E = q + w and w = −PV, we
can substitute these into the enthalpy
expression:
H = E + PV
H = (q + w) − w
H = q
• So, at constant pressure, the change in
enthalpy is the heat gained or lost.
Thermochemistry
© 2015 Pearson Education, Inc.
Endothermic and Exothermic
• A process is
endothermic
when H is
positive.
• A process is
exothermic when
H is negative.
Thermochemistry
© 2015 Pearson Education, Inc.
Enthalpy of Reaction
The change in
enthalpy, H, is the
enthalpy of the
products minus the
enthalpy of the
reactants:
H = Hproducts − Hreactants
Thermochemistry
© 2015 Pearson Education, Inc.
Enthalpy of Reaction
This quantity, H, is called the enthalpy of
reaction, or the heat of reaction.
Thermochemistry
© 2015 Pearson Education, Inc.
The Truth about Enthalpy
1. Enthalpy is an extensive property.
2. H for a reaction in the forward
direction is equal in size, but opposite
in sign, to H for the reverse reaction.
3. H for a reaction depends on the state
of the products and the state of the
reactants.
Thermochemistry
© 2015 Pearson Education, Inc.
THERMOCHEMISTRY
Define the following terms in your
notebook:
1) Thermochemistry
2) Kinetic energy
3) Potential energy
4) Calorimeter
5) Temperature
6) Joule
7) Heat transfer
8) Conduction
9) Convection
10) Radiation
THERMOCHEMISTRY
Study of the transfers of energy as heat that
accompany chemical reactions and physical
changes
 All chemical reactions are accompanied by a
change in energy
 Usually absorb (endothermic) or release
(exothermic) energy as heat

TYPES OF ENERGY

Kinetic Energy – the energy an object has
because of its motion


Thermal Energy – The kinetic energy of
molecules, energy in the form of heat
Potential Energy – the amount of energy
stored in an object

Chemical Energy – energy stored within
chemical bonds
Calorimetry
• Since we cannot know
the exact enthalpy of the
reactants and products,
we measure H through
calorimetry, the
measurement of
heat flow.
• The instrument used to
measure heat flow is
called a calorimeter.
Thermochemistry
© 2015 Pearson Education, Inc.
TEMPERATURE
The measure of the
average kinetic energy
of the particles in a
sample of matter
 The greater the kinetic
energy, the higher the
temp, and the hotter it
feels
 Measured in °Celsius
or Kelvins

Heat Capacity and Specific Heat
The amount of energy required to raise the
temperature of a substance by 1 K (1 C) is its
heat capacity, usually given for one mole of the
substance.
Thermochemistry
© 2015 Pearson Education, Inc.
HEAT
-the energy transferred
between samples of matter
because of a difference in their
temperatures
-energy moves spontaneously
from higher to lower temp; when
the temps are equal, energy is
no longer transferred
Heat Capacity and Specific Heat
We define specific
heat capacity (or
simply specific heat)
as the amount of
energy required to
raise the temperature
of 1 g of a substance
by 1 K (or 1 C).
Thermochemistry
© 2015 Pearson Education, Inc.
Heat Capacity and Specific Heat
Specific heat, then, is
Thermochemistry
© 2015 Pearson Education, Inc.
Constant Pressure Calorimetry
• By carrying out a reaction in
aqueous solution in a simple
calorimeter, the heat change for
the system can be found by
measuring the heat change for
the water in the calorimeter.
• The specific heat for water is
well known (4.184 J/g∙K).
• We can calculate H for the
reaction with this equation:
q = m  Cs  T
Thermochemistry
© 2015 Pearson Education, Inc.
Bomb Calorimetry
• Reactions can be carried
out in a sealed “bomb”
such as this one.
• The heat absorbed (or
released) by the water is
a very good approximation
of the enthalpy change for
the reaction.
• qrxn = – Ccal × ∆T
Thermochemistry
© 2015 Pearson Education, Inc.
Bomb Calorimetry
• Because the volume
in the bomb
calorimeter is
constant, what is
measured is really the
change in internal
energy, E, not H.
• For most reactions,
the difference is very
small.
Thermochemistry
© 2015 Pearson Education, Inc.
At the boiling point energy is
being transferred to the
system, but until all the water
is vaporized the temperature
will not increase.
The transfer of
heat may not
affect the
temperature.
This is
particularly true
during phase
changes. In this
example the
temperature of ice
at the melting
point will not
increase until all
the ice has
melted. The same
is true at the
boiling point.
PRACTICE
Label these as CONDUCTION, CONVECTION OR RADIATION :
1. Light waves traveling through empty space.
2. You sit in on a hot leather seat in a car during
the middle of the summer & your legs burn.
3.
How air is heated in our atmosphere.
4. The transfer of heat between substances
directly in contact with one another.
5. Raising warm air currents.
6. You get sunburned.
HEAT TRANSFER

Conduction - the transfer of heat energy within
an object, or between objects that are directly
touching each other, due to collisions between the
particles in the objects
HEAT TRANSFER
Convection - the transfer of heat from one place
to another caused by movement of molecules
 Radiation - a process by which energetic
electromagnetic waves move from one place to
another

LAW OF CONSERVATION OF ENERGY
Energy cannot be created nor destroyed
 It only changes forms

HEAT CAPACITY
-the quantity of energy transferred as heat depends on the
nature of the material changing temp, the mass changing temp,
and the size of the temp change
The equation to solve for heat capacity is:
Q = m cpT
Q stands for heat capacity in the unit of
Joules or J, while m is mass in grams, cp
is for specific heat usually in J/g ºC, and
T is for temperature change in ºC
SPECIFIC HEAT
-is the amount of energy required to raise the temperature of
one gram of substance by one Celsius degree or one Kelvin
To solve for specific heat: rearrange
the previous equation for cp
cp
Specific heat is
measured in J/g K
or J/g ºC (and is
under constant
pressure)
=
q
m T
Mass
in
grams
Energy in
Joules
Change in
temp in ºC or K
ASSIGNMENT #2: SHOW ALL MATH
Q = M C T
1. A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and
its temperature changes from 25°C to 175°C. Calculate the
specific heat capacity of iron.
2. How many joules of heat are needed to raise the temperature of
10.0 g of aluminum from 22°C to 55°C, if the specific heat of
aluminum is 0.90 J/g°C?
3. 3. Calculate the specific heat capacity of a piece of wood if
1500.0 g of the wood absorbs 67,560 joules of heat, and its
temperature changes from 32°C to 57°C.
4. 4. 100.0 g of 4.0°C water is heated until its temperature is
37°C. If the specific heat of water is 4.18 J/g°C, calculate the
amount of heat energy needed to cause this rise in
temperature.
5. 5. 25.0 g of mercury is heated from 25°C to 155°C, and absorbs
455 joules of heat in the process. Calculate the specific heat
capacity of mercury.
–review heat capacity calculations:
SHOW YOUR MATH WORK
1. What is the specific heat capacity of silver metal if 0.055 Kg of the
metal absorbs 47.3J of heat and the temperature rises 15.0°C?
2. What mass of water will change its temperature by 3 0C when 525 J
of heat is added to it? The specific heat of water is 4.18 J/g 0C
3. A 13 g piece of copper is heated and fashioned into a bracelet. The
amount of energy transferred by heat to the copper is 275 J. If the
specific heat of copper is 0.390 J/g 0C, what Is the final temperature if
the initial temperature was 25 0C?
Endothermic and Exothermic Reactions
Potential Energy
System
Surroundings
(Products)
2 mol NO
∆(𝑃𝐸)
Heat Energy Absorbed
Endothermic
1 mol O2 + 1 mol N2
(Reactants)
Potential Energy
System
Surroundings
(Reactants)
1 mol CH4 + 2 mol O2
∆(𝑃𝐸)
2 mol H2O + 1 mol CO2
(Products)
Heat Energy Released
Exothermic
HOW DO YOU KNOW IF THE PROCESS IS
ENDOTHERMIC OR EXOTHERMIC?
Endothermic = energy is taken up by the reaction in the form
of heat (feels cold to touch)
Exothermic = energy is released by the reaction in the form of
heat (feels hot to touch)

Classify each of the following as an exothermic or
endothermic process on a sheet of paper to turn in.
 1) Melting ice cubes –
 2) Burning a candle –
 3) Evaporation of Water –
 4) Baking Bread –
 5) Splitting a gas molecule apart –
 6) Formation of snow in clouds –
LOGIN IN TO EDPUZZLE.COM USING YOUR
SCHOOL GOOGLE ACCOUNT TO:


Watch, listen and answer the questions in the
video called “Endothermic and Exothermic
reactions”
If new to my class then you may need to use the
password dewpuse
IDENTIFY IF THESE ARE EXOTHERMIC OR
ENDOTHERMIC PROCESS:
1) Candle feels warm
2) Heat is a product of the reaction.
3) A + B ---> C + heat.
4) heat absorbed (feels cool)
5) heat is a reactant in the reaction.
6) heat + C ---> A + B
ENDOTHERMIC OR EXOTHERMIC?
Exothermic or endothermic:
Candle feels warm?
Exothermic
Exothermic or endothermic:
Exothermic
Heat is a product of the
reaction.
Exothermic or endothermic: A Exothermic
+ B ---> C + heat.
Exothermic or endothermic:
Endothermic
heat absorbed (feels cool)
Exothermic or endothermic:
heat is a reactant in the
reaction.
Exothermic or endothermic:
heat + C ---> A + B
Endothermic
Endothermic
STANDARD HEAT OF FORMATION
ΔH FORMATION
-the change in enthalpy that accompanies the formation of 1
mole of the compound in their standard state from its
elements in their standard states.
We will look up the heat of formation on a table that will be provided
for you on the test.
** free elements in their standard states is assigned
ΔH formation = 0.0 kJ for example O2
These reactions follow the law of conservation of energy.
HEAT OF THE REACTION:
ΔHREACTION = HPRODUCTS - HREACTANTS
ΔH is the energy absorbed or released as heat during a
chemical reaction at constant pressure
H is the symbol for a quantity called enthalpy.
Enthalpy has traditionally been defined as the heat
content of a system at constant pressure.
ΔH is read as “change in enthalpy” An enthalpy
change is the amount of energy absorbed or lost by a
system as heat during a process at constant pressure.
If ΔH is a (-) value then the
process is exothermic & if it is
understood to be (+) then it is
endothermic
Show your work & solve for the heat of reaction:
SiH4 (g)
+ 2 O2 (g)  SiO2 (s)
TABLE VALUES:
CO
Hof kJ/mol
+ 2 H2 O
(l)
Hof kJ/mol
(g)
-110.5
Pb
CO2 (g)
-393.5
PbO
Fe2O3 (s)
-824.2
SiH4 (g)
34.3
FeS2 (s)
-177.5
SiO2 (s)
-910.9
H2O
-285.8
SO2 (g)
-296.8
(l)
O2 (g)
0
(s)
(s)
0
-217.3
REVIEW OF HEAT OF FORMATION
CALCULATION:
Calculate ∆H for the reaction of sulfur dioxide with oxygen.
2SO2 (g) + O2 (g)  2SO3 (g)
Given values from table:
∆Hf for SO2 (g) = -296.8 kJ
∆Hf for SO3 (g) = -395.7 kJ
USE THE HEAT OF FORMATION TABLE &
ΔHREACTION = HPRODUCTS - HREACTANTS
TO SOLVE ENTHALPY IN THE FOLLOWING:
(FIRST COPY EACH)
a) NaOH(s) + HCl(g) ----> NaCl(s) + H2O(g)
-133.8 kJ
b)2 CO(g) + O2(g) ---> 2 CO2(g)
-566.0 kJ
c) CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(l)
-890.2 kJ
d) 2 H2S(g) + 3 O2(g) ---> 2 H2O(l) + 2 SO2(g)
-1123.6 kJ
e) 2 NO(g) + O2(g) ---> 2 NO2(g)
-122.6 kJ
HEAT OF FORMATION TABLE
Compound
DHf (kJ/mol)
Compound
Hf (kJ/mol)
CH4(g)
-74.8
HCl(g)
-92.3
CO2(g)
-393.5
H2O(g)
-241.8
NaCl(s)
-411.0
SO2(g)
-296.1
H2O(l)
-285.8
NH4Cl(s)
-315.4
H2S(g)
-20.1
NO(g)
+90.4
H2SO4(l)
-811.3
NO2(g)
+33.9
MgSO4(s)
-1278.2
SnCl4(l)
-545.2
MnO(s)
-384.9
SnO(s)
-286.2
MnO2(s)
-519.7
SnO2(s)
-580.7
NaCl(s)
-411.0
SO2(g)
-296.1
NaF(s)
-569.0
SO3(g)
-395.2
NaOH(s)
-426.7
ZnO(s)
-348.0
NH3(g)
CO (g)
-46.2
-110.5
ZnS(s)
-202.9
USE THE HEAT OF FORMATION TABLE & SOLVE:
ΔHREACTION = HPRODUCTS - HREACTANTS
NaOH(s) -426.7
NaCl(s) -411.0
HCl(g) -92.3
H2O(g) -241.8
a) NaOH(s) + HCl(g) ----> NaCl(s) + H2O(g)
This equation literally means that the change in enthalpy for a reaction is
equal to the sum of bonds formed in the products minus the sum of the
bonds broken in the reactants.
What if we don’t have a heat of formation table but are given heat of
formation for the parts of the reaction?????????????????????
WHAT IF WE DON’T HAVE A HEAT OF FORMATION TABLE BUT
ARE GIVEN HEAT OF FORMATION FOR THE PARTS OF THE
REACTION?????????????????????
Calculate ∆H for the reaction
4NH3 (g) + 5 O2 (g)  4 NO (g) + 6H2O(g)
N2 (g) + O2 (g) 2 NO (g)
N2 (g) + 3 H2 (g) 2 NH3 (g)
2 H2 (g) + O2 (g)  2 H2O (g)
∆H = ?
∆H = -180.5 kJ
∆H = -91.8 kJ
∆H = -483.6 kJ
THERMOCHEMICAL EQUATION
2H2(g) + O2(g)  2H2O(g) + 483.6 kJ
This is an example of a thermochemical equation, an
equation that includes the quantity of energy released or
absorbed as heat during the reaction as written.
In any thermochemical equation the coefficients are treated as
moles. The quantity of energy released as heat depends on the
amounts of reactants and products. Producing twice as much
water would require twice as many moles of reactants and would
release 2 x 483.6 kJ of energy. Producing ½ as much would
require ½ as many moles of reactants and would release ½ as
much energy. Fractional coefficients are sometimes used.
EXOTHERMIC REACTIONS
• reactants have a higher energy than the products
• ∆H is negative (system loses energy)
• energy is included in the products and is measured in
kJ/mol
2H2(g) + O2(g)  2H2O(g) + 483.6 kJ/mol
• for this equation ∆H = -483.6 kJ/mol
EXOTHERMIC REACTION PATHWAY
The energy of the
products is less than
that of the reactants;
therefore energy in
the form of heat is
released and ∆H is
negative.
ENDOTHERMIC REACTIONS
• products have a higher energy than reactants
• ∆H is always positive (the system gains energy)
• energy is included in the reactants
2H2O(g) + 483.6 kJ  2H2(g) + O2(g)
-for this reactions ∆H = 483.6 kJ/mol
The physical states of reactants and products must always
be included in thermochemical equations because they
influence the overall amount of energy exchanged. EX:
the energy needed for the decomposition of water would be
greater than 483.6 kJ if we started with ice because extra
energy would be needed to melt the ice and to change the
liquid into a vapor.
ENDOTHERMIC REACTION PATHWAY
The reactants have
less energy than the
products meaning the
reactants have
absorbed energy and
∆H is positive.
Hess’s Law
• H is well known for many reactions,
and it is inconvenient to measure H
for every reaction in which we are
interested.
• However, we can estimate H using
published H values and the
properties of enthalpy.
Thermochemistry
© 2015 Pearson Education, Inc.
Hess’s Law
• Hess’s law: If a reaction is
carried out in a series of
steps, H for the overall
reaction will be equal to the
sum of the enthalpy changes
for the individual steps.
• Because H is a state
function, the total enthalpy
change depends only on the
initial state (reactants) and the
final state (products) of the
reaction.
Thermochemistry
© 2015 Pearson Education, Inc.
Enthalpies of Formation
An enthalpy of formation, Hf, is
defined as the enthalpy change for the
reaction in which a compound is made
from its constituent elements in their
elemental forms.
Thermochemistry
© 2015 Pearson Education, Inc.
Standard Enthalpies of Formation
Standard enthalpies of formation, ∆Hf°, are
measured under standard conditions (25 ºC
and 1.00 atm pressure).
Thermochemistry
© 2015 Pearson Education, Inc.
Calculation of H
C3H8(g) + 5 O2(g)  3 CO2(g) + 4
H2O(l)
• Imagine this as occurring
in three steps:
1) Decomposition of propane to
the elements:
C3H8(g)  3 C(graphite) + 4 H2(g)
Thermochemistry
© 2015 Pearson Education, Inc.
Calculation of H
C3H8(g) + 5 O2(g)  3 CO2(g) + 4
H2O(l)
• Imagine this as occurring
in three steps:
2) Formation of CO2:
3 C(graphite) + 3 O2(g) 3 CO2(g)
Thermochemistry
© 2015 Pearson Education, Inc.
Calculation of H
C3H8(g) + 5 O2(g)  3 CO2(g) + 4
H2O(l)
• Imagine this as occurring
in three steps:
3) Formation of H2O:
4 H2(g) + 2 O2(g)  4 H2O(l)
Thermochemistry
© 2015 Pearson Education, Inc.
Calculation of H
C3H8(g) + 5 O2(g)  3 CO2(g) + 4
H2O(l)
• So, all steps look like this:
C3H8(g)  3 C(graphite) + 4 H2(g)
3 C(graphite) + 3 O2(g) 3 CO2(g)
4 H2(g) + 2 O2(g)  4 H2O(l)
Thermochemistry
© 2015 Pearson Education, Inc.
Calculation of H
C3H8(g) + 5 O2(g)  3 CO2(g) + 4
H2O(l)
• The sum of these
equations is the overall
equation!
C3H8(g)  3 C(graphite) + 4 H2(g)
3 C(graphite) + 3 O2(g) 3 CO2(g)
4 H2(g) + 2 O2(g)  4 H2O(l)
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
© 2015 Pearson Education, Inc.
Thermochemistry
HESS’S LAW CALCULATIONS
Solve ΔH for 2 S(s) + 3 O2 (g)  2SO3 (g)
When you are given:
S (s) + O2 (g)  SO2 (g) ΔH = -297 kJ
2SO3 (g)  2SO2 (g) + O2 (g) ΔH = 198kJ
CALCULATING ΔH FROM TWO REACTIONS
WHEN NO TABLE IS GIVEN:
Take the known reactions and arrange (flip if
needed), to make the unknown reaction
 Multiply a reaction by as many times needed to
create the unknown reaction
 Known Reactions:

N2O4 (g)  2 NO2 (g)
 2 NO (g) + O2 (g)  2NO2 (g)


H = 58 kJ
H = -112 kJ
Unknown Reaction
Calculate the ΔH for the process
 2 NO (g) + O2 (g)  N2O4 (g)
H = ?

CALCULATING ΔH FOR TWO REACTIONS

Take reaction (2) first, to make the reactants side of
the unknown reaction:


2 NO (g) + O2 (g)  2NO2 (g)
Add to it, a flipped reaction (1), to make the products
side, (notice that the sign changed on H).
2 NO2 (g)  N2O4 (g)
 Result:





H = -112 kJ
H = - 58 kJ
2 NO (g) + O2 (g) + 2 NO2 (g)  N2O4 (g) + 2NO2 (g)
H = -112 kJ + (-58 kJ)
No equation needs to be multiplied because the
unknown reaction is already produced (see below)
Take out “like” values on both sides of the equation
2NO2 (g) – removed
 2 NO (g) + O2 (g)  N2O4 (g) H = -170 kJ = -112 + -58

WHAT IS HF?



Heat of formation: the amount of heat needed to
produce 1 mole of a given compound.
Use the Hf to caluculate the Hreactants or the Hproducts.
Example:

Determine the standard enthalpy change for the following
reaction:






CH4(g) + 4 Cl2(g)  CCl4(g) + 4 HCl(g)
Standard Hf for the following:
 CH4 = -74.9 kJ/mol; Cl2 = 0 kJ/mol; CCl4 = -95.7 kJ/mol; HCl
= -92.3 kJ/mol
ΔH = Hproducts – Hreactants
= (-95.7 + (4 * -92.3)) – (-74.9 + (4 * 0)) = -390 kJ/mol
This is an exothermic reaction
More Practice Problems on Netschool Moodle
REVIEW for TEST:
Use Q = m cpT
Complete in your COMPOSITION NOTEBOOK. Copy down the questions in your
notebook, and show all work in your answers.
1. Explain the difference between heat capacity and specific heat. Explain your answer
by deciding if heat capacity and specific heat of 50g of a substance will be the same
as, or greater than, the specific heat of 10 g of the same substance.
2. Using calories, calculate how much heat 32.0 g of water absorbs when it is heated
from 25.0 ⁰C to 80.0 ⁰C. How many Joules is this? hints: water’s specific heat is 1 cal
/ (g * ⁰C), 1 cal = 4.184 J
3. When 435 J of heat is added to 3.4 g of olive oil at 21 ⁰C, the temperature increases
to 85 ⁰C. What is the specific heat of olive oil?
4. How much heat is required to raise the temperature of 250.0 g of mercury 52 ⁰C?
hint: specific heat capacity for mercury is 0.14 J / (g * ⁰C)
5. How many kilojoules of heat are absorbed when 1.00 L of water is heated from 18 ⁰C
to 85 ⁰C? hint: 1000 J = 1 kJ, water’s specific heat is 4.184 J / (g * ⁰C)

Similar documents