Chapter 14 Thermochemistry
Now we begin a new area of study called thermochemistry. From the name you get the
idea that we will follow the heat (thermo) in a chemical reaction. Actually it is a bit
more; we will be studying energy flow in chemical reactions. This chapter introduces
you to two measures of energy, internal energy and enthalpy. After we finish this
chapter I will give you a month or so to digest this information, then we return for more
thermodynamics in Chapter 23 to complete your introduction to this important area.
14-1 Energy, Work, and Heat
Our first task is to get some general definitions and concepts down. We will be
studying heat and energy flow so we will divide the entire universe into two parts
Key Definitions:
The system - the chemicals involved in a reaction.
The surroundings - everything else in the universe
We will only recognize two ways that energy can be transferred between the
system and the surroundings,:
Key Definitions
Heat (q) - when energy flows spontaneously from a region of high temperature to
a region of low temperature
Work(w) - When energy exerts a force through a distance W= f x d
There are lots of different kinds of work, mechanical work (force x distance),
electrical work , etc. To keep things simple we will only use mechanical work in
this chapter. Further, we will only use the work of expanding or compressing a
gas
Sketch Figure 14.2 on board, don’t use h use dinitial and dfinal for inital distance 1
and final distance
In this section we will use the symbol Ä extensively.
Key Definition
Ä = final value - initial value
The next thing we have to decide is our convention for sign. Here we will take
the viewpoint of the system. Thus if something comes into the system it has a +
value, and if something leaves the system it will have a negative value
Figure 14.3 on the board (not on powerpoint)
Now let’s put the pieces together
w=fxd
d = Äd
w = f x Äd
2
What will we use for the force? Remembering our definition of pressure?
pressure = F / area
so Force = pressure x area
so w = pressure x area x Äd
but if Äd is +, our cylinder expanded, so it did work on the surroundings, and that
work had to come from our system so what should the sign be?
w= -pressure x area x Ä d
Now what is area x d? If it volume!
So
Key Equation:
w = -P ÄV
and this is the only measure of work we will use in this chapter
Before I turn you loose on some practice problems we have to think a little bit
about units. If work is a type of energy, what should its units be? Joules
What are the units of -PÄV? Atm @ L
So we need a conversion factor 1 L@atm = 101.325 J
You do not have to memorize this factor, it is included with the conversion sheet
you get with your tests
(Where did this conversion come from? I’ll give you some clues
1 n/m2 = 9.87x10-6 atm
1L=1x10-3 m3
1 Joule = 1nt.m)
Practice problem
How much work is associated with a blowing up a 2 L balloon?
w = -PÄV
ÄV = Vfinal - Vinital = 2L - 0L = 2L
P is the pressure of the atmosphere = 1 atm
w = - 1 atm x 2L
= -2 L@atm x 101.325J = -202 J
1 L@atm
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Clicker question?
Have a simple problem set up as above, show 4 -6 different equations to solve
but only 1 equation is correct. Errors to include are forgetting sign, wrong sign
on ÄV, and forgetting conversion to J
Now that we have dealt with work, let’s return to q or heat
We will use the same sign convention
+ heat mean heat is transferred into the system (it got warmer)
- heat means heat left the system (it got colder)
We have special names for chemical reactions that get hotter or colder
Key Concept:
In an exothermic reaction the reaction gets hotter as it releases heat, heat
flows from the system to the to the surroundings, so q is NEGATIVE
(Exo- mean out)
In an endothermic reaction the reaction gets colder so it absorbs heat from the
surroundings, so heat enters the system from the surrounding so q is POSITIVE
(Endo- mean in )
So now we have two ways to transfer energy into or out of our system, q and w.
Now lets look at the total energy in the system U (Note: other texts will refer to
this energy as Internal Energy, E)
If the only ways we have to mover energy into or out of the system are q and w,
and the change in then energy of the system would be ÄU, we can say
Key Equation:
ÄU=q+w
There is something going on in the symbols that I want you to notice. See how q
and w are in small letters and U is a capital letter?
The small letters for q and w are supposed to tell you that these are energy
transfer functions (also called path functions in other texts) and the Capital
letter for U tell you that U is a state function. What is the difference?
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Key concept:
A State function depends only on the current state of the system. It is easy to
calculate a Ä in a state function because it is simply equal to final state - initial
state
An energy transfer function (or path function) depends on the method (or
path) used to change from one state to another. It is much more difficult to
calculate because you have to integrate that function over all the different steps
you take to go from one state to another.
Examples. Potential energy is a state function. You can calculate the energy of
this piece of chalk located here, about 6 feet above the ground, vs here right on
the ground by simply E = g x Äh
Work (w) is a energy transfer or path function. Say I had the biggest guy in the
room holding the piece of chalk, and you had to arm wrestle him to push the
chalk out of his hand and on to the ground. It would take a lot more work on
your part! And if I chose somebody else to hold the chaulk it would take a
different amount of work.
14-2 Enthalpy
You have just seen that heat (q) is a path function so that means the value you
get depends on the conditions you use in your measurement. This is not good
science because that means different people in different labs will get different
numbers. So what we need is a measure of heat in a reaction that is a state
function.
The function we want is called Enthalpy (H)
Key equation:
Enthalpy (H) = U + PV
For a chemical reaction then
ÄH rxn = Hf - Hi = ÄU + ÄPV
ÄPV = PfVf - PiVi
We will measure H when P is constant, so Pf = Pi = P
ÄPV = PÄV
ÄH = ÄU + PÄV (When pressure is constant)
From the above equation you can see that ÄU and ÄH are usually very close,
unless the volume of the reaction changes
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Continuing our derivation
if ÄU = q + w
then
ÄH = q + w + PÄV
but w = -PÄV
ÄH = q -PÄV + PÄV
ÄH =qp where the p subscript means heat measured at constant P
When we work with chemical reactions will use the equation
ÄH rxn = Hproducts - Hreactants
You can think of this graphically in a figure like figure 14.6
Let’s extend the convention we used earlier
Key Concept:
Exothermic - a reaction that get warmer - H product < H reactant
ÄH rxn -is negative
Endothermic - a reaction that gets colder - H product > H reactant
ÄH rxn -is positive
Can you give me an everyday example of a chemical reaction with + ÄH? - ÄH?
When you use the term ÄH rxn you always have to give a written chemical
reaction as a reference. Why? Back in the first chapter you were introduced to
the terms extensive and intensive properties. Does anybody remember what
that means?
Most of our themodynamic properties like U and H are extensive properties, that
is, they depend on the amount of material in the system. Thus when you give
the ÄH of a reaction, you have to say specifically how many moles of each
reactant and product there were in the reaction
For instance ÄH rxn for 2H2(g) + O2(g) 62H2O(l) is -571.6 kJ
and
ÄH rxn for H2(g) + ½ O2(g) 61H2O(l) is -285.8 kJ
Another part of the symbolism you need to be aware of is the o sign as in
ÄHo
This symbol tells you that all reactant and products are in their standard states
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Key definition:
A standard state of a compound or element is the most stable form (s,l,or g) of
that compound or element that occurs when the pressure is one bar (called the
one bar state).
The standard state of a gas is that gas at a partial pressure of 1 bar
The standard state of an aqueous chemical is a 1M concentration of that
chemical.
Sample Calculation:
How much heat will be evolved if I react 5 g of Al with 5 g of Fe2O3 in the thermite
reaction:
Fe2O3(s) + 2 Al(s) 6Al2O3(s) + 2 Fe(s) ÄH Rxn = -851.5 kJ/mol
Step 1. What is the limiting reactant?
5 g Fe2O3 x 1 mole Fe2O3 x 2 mole Fe = .0626 mol Fe
159.7 g Fe2O3 1 mol Fe2O3
5 g Al x 1 mol Al x 2 mole Fe = .185 mol Fe
26.98 g Al 2 mol Al
So Fe2O3 is limiting reactant
Using the defined stoichiometry for ÄH rxn
1 mole Fe2O3(s) % -851.5 kJ
5 g Fe2O3 x 1 mole Fe2O3 x -851.5 kJ
159.7 g Fe2O3 1 mol Fe2O3
= -26.66 kJ
One other thing to notice. To properly specify these reactions I should be giving
both the temperature and the pressure at which the reaction occurs because
both of these factors will change the ÄH rxn. However the changes are usually
pretty small, and I and not that picky of a detail person, so I usually forget to
state the pressure and temperature in my problems
14-3 ÄHrxn .ÄU rxn
A little while back we had the equation:
ÄH o rxn = ÄU 0 rxn + PÄV
This says that ÄH and ÄU will have close to the same value unless the volume
changes significantly in a reaction.
For reactions involving s, l or aq the ÄV are so small we can assume that ÄH.ÄE
It is only when a reaction involves a gas that we need to worry about the
difference.
7
I will ignore the section on determining the ÄV for a reaction, because I don’t
think you need to worry about it at this level.
14-4 ÄH and Hess’s Law
Key Concept:
Hess’s Law: Since H is a state function, if we add two chemical equations
together to create a new chemical equation, we can add the ÄH’s of the
individual reactions together to come up with the ÄH of the new equation.
This principle is known as Hess’s Law and what it means is that you can find the
ÄH of an unknown reaction by summing up the ÄH’s of different known
reactions. To get a concrete example let’s look at the oxidation of nitrogen in
photochemical smog.
Suppose we wanted to obtain the ÄH for the reaction
N2(g) + 2O2(g) 6 2NO2(g)
If you remember this actually is a combination of two reactions, first the oxidation
of nitrogen into nitrogen oxide that occurs in the combustion process in the
engine:
N2 +O2 62NO Ä H1=180kJ
Followed by the spontaneous oxidation of nitrogen oxide into nitrogen dioxide
2NO + O2 6 2NO2
ÄH2=-112 kJ
If we add the two reactions together we get
N2 +O2 + 2NO + O262NO + 2NO2
Remove the spectator NO’s and we have
N2 +2O2 62NO2
The ÄH for the overall reaction is just the sum of the ÄH’s for the individual
reactions:
ÄH = ÄH1 + ÄH2
=180 +(-112) = 68kJ. (N2 +2O2 62NO)
You can use this principle in very elaborate schemes to get new ÄH values
There are also some algebraic tricks you can use as well.
In this case we had two reactions that both when forward and we added their
ÄH’s together, what would happen if we had to reverse a reaction? Say we
needed the reaction
2NO 6 O2 + N2
We already had
So
N2 +O2 62NO Ä H1=180kJ
2NO 6 O2 + N2 = -180 kJ
8
Key concept:
If you reverse a reaction you change the sign of the reaction’s ÄH
Now what if, to get the stoichiometry straight, you have to change a coefficient
like
4NO 6 2O2 + 2N2
Key Concept:
If you change the coefficient of a reaction, you multiply the ÄH by that coefficient.
In the above case, if 2NO 6 O2 + N2 = -180 kJ
and you want
4NO 6 2O2 + 2N2
Since you multiply the stoichiometric coefficients of the first reaction by 2 to get
the second, you can multiply the ÄH of the first reaction by 2 to get the ÄH of the
new reaction.
ÄH = 2(-180) = -360kJ
Let’s try another example. You know that carbon is an element used in every
living thing, Pure elemental carbon has two forms, graphite the lead in your
pencil or a fine, non-oily lubricant, and diamond, the brilliant=hard gem!. If the
alchemist tried long and hard to change lead to gold, and couldn’t do it, we know
now that is because they are different elements, and you can’t change one
element into another without some drastic nuclear reaction. However, if they had
tried to turn graphite into diamond, they might have had a chance! Let’s see if
you modern day alchemists can turn graphite into diamonds
What is the ÄH for the conversion of graphite into diamond. Well you aren’t
going to find a ÄH for this reaction in the back of the book, so you have to figure
it out for yourself. The whole idea of Hess’s law is that since the pathway
doesn’t matter, you can use any pathway you want, even if they seem quite
strange. In this case, you may not find the ÄH for some reactions involving
diamond and graphite. What you can find are the ÄH’s for the combustion of
these substances :
Cgraphite +O2 6CO2 ÄH=-394 kj
Cdiamond +O2 6CO2 ÄH=-396 kj
Now to get the reaction we want you have to have the first reaction going
forward, and the second going in reverse, that is:
Cgraphite +O2 6CO2
CO2 6Cdiamond +O2
For a net of
Cgraphite +O2 + CO2 6Cdiamond +O2 + CO2
Remove spectator molecules”
Cgraphite6Cdiamond
Now the idea is that the ÄH for the reaction is the sum of the ÄH’s of the
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individual reactions. But remember, when we combined reaction we used the
diamond reaction in reverse, therefor we subtract its ÄH ie
ÄH = -394 -(-396)
For a net of +2kJ. (Plus meaning we have to add heat to the system, So all we
have to do is add a little heat and some pressure)
Calculations like this typically require you to put several reactions together to get
the desired overall reaction
Putting together the right combination is sometimes tricky, and involves a lot of
trial and error. Always write down the final reaction you want, and slowly work
backwards from there in a systematic manner, and you will get to a final answer.
If you jump in right in the middle and start guessing, you will never get there.
Final Example
Calculate ÄH for the reaction:
2N2(g) + 5 O2(g) 6 2N2O5(g)
Given the following reactions:
H2(g) + 1/2 O2(g) 6 H2O (l)
N2O5(g) + H2O(l) 6 2HNO3(l)
1/2 N2(g) + 3/2 O2(g) + 1/2 H2(g) 6 HNO3(l)
ÄH = -285.8
ÄH = -76.6
ÄH= -174.1
To get N2O5 on the left side of the left side of the equation
2 HNO3(l) 6 N2O5(g) + H2O(l)
ÄH = +76.6
But I need 2 N2O5 so:
4 HNO3(l) 62 N2O5(g) + 2 H2O(l)
ÄH = +76.6x2
To cancel out the 4 HNO3 from the right
=153.2
4/2 N2(g) + (4x3)/2 O2(g) + 4/2 H2(g) 64 HNO3(l)
ÄH=-174.1x4
My net reaction at this point is:
= -696.4
2N2 + 6O2 + 2H2 62 N2O5(g) + 2H2O(l)
So I need to get rid of the 2 waters on the left + some O2 and H2 on the right
2H2O (l) 62 H2(g) + 2/2 O2(g)
ÄH = +285.8x2
=571.6
So my net ÄH = 76.6(2) -174.1(4) + 285.8(2) = 28.0 kJ/mol
14-5 ÄH of formation
Yet another way to come up with a ÄH rxn is to use the standard molar enthalpy of
formation.
Key Definition:
A standard molar enthalpy of formation,
, is defined as the heat gained
or lost when a compound is formed in its one bar standard state from the most
stable forms of its constituent elements in their one bar standard states.
10
In simpler terms, it is the heat gained or lost when your make a compound from
its elements.
Chemists have gone to great pains to either determine or calculate the ÄH of of a
great many compound and tabulate their results Appendix D or Table 14.3
Why? Because we can use the ÄH formation and Hess’s law to calculate the ÄH for
may reaction you can think of!
Let’s try this on our previous NO2 example. We were trying to get the ÄH for the
reaction
N2 + 2O2 6 2NO2
And we did it by combining the reactions
N2 +O2 62NO And
2NO + O2 6 2NO2
If everything works as advertised we should be able to get it directly from the
heat of formation. Let’s see – Well doesn’t this work out nicely, we don’t need to
use the two equations, Our two reactants are elements, so all we need to do is to
look of the ÄHfo of NO2! And it is 34kJ/mole
But hold it, our previous answer was 68kJ. What is wrong?
(In this reaction we generate 2 moles so need to double the energy!)
That was too easy, so let’s try another example. In that problem we said that the
ÄH of the reaction
2NO + O2 6 2NO2 is H=-112 kJ
Can we confirm this? This problem is a bit trickier because we have some
compounds on both sides of the equation, not just elements.
The book goes into a long explanation where we use Hess’s law to combine the
following attack.
1. Break all reactants into elements
2. Build all products up from elements
3. Add it all together
When all these pieces are added together you get the final result that:
If we want the ÄH of a reaction
ÄHrxn =ÄH f products - ÄH f reactants
If we include stoichiometry we have
Key equation:
11
Example Calculation:
Determine ÄH rxn for 2NO + O2 6 2NO2
ÄHf product = ÄHf NO2 = 33.2 kJ/mol (Find these in Appendix 4)
2 mole of product so 2x33.2 kJ
ÄHf reactant = ÄHf NO = 91.3 kJ/mol
2 moles of reactant so 2x91.3
ÄHrxn =(2x33.2)- (2x91.3) = -116.2 kJ/mole
A little off from previous example because of table that I took my values from
Another example
Determine the ÄH for the combustion of octane (gasoline) and methanol, an
additive now being used in gas.
Octane = C8H18 so the combustion equation would be:
2 C8H18(l) + 25 O2 (g)6 16CO2 (g)+ 18H2O(l)
ÄHf for various compounds
C8H18(l)
-269 kJ/mol
O2(g)
?
CO2(g)
-393.5
H2O(g)
-241.8
ÄHf = [16(-393.5) + 18(-241.8)] - [2(-269) +25(0)]
= -6296 -4352.4 +538 -0
= -10110 kJ/ 2mol
= -5.06 MJ/1 mol
The ÄH combustion for Methanol is -22.7 kJ/g so it releases about half as much
energy!
14-6 ÄH of bonds
This semester we have used the concept of covalent bonds many times, so how
much have your retained? Why does a covalent bond hold a molecule together?
(Electron orbital between nuclei holds the nuclei together as a unit. Energy of
isolated nuclei is higher that energy of molecule)
Thinking in reverse, it takes energy to take a molecule and turn it into isolated
atoms
H2O(g) 62H(g) + O(g)
+925 kJ
12
Knowing that H2O has 2 OH single bonds you can then say that
it takes 925/2 or 462.5 kJ to destroy a single OH bond.
Thinking on the formation side if you start with one mole of O atoms and 2
mole of H atoms 925 kJ of energy will be released as two moles of OH bond
form to hold the molecule together
Key definition:
The molar bond energy is the energy needed to break one mole of a given bond
This gives us yet another way to come up with the energy of a reaction, look at
the total energy gained or lost when all the bonds of the reactants are broken
and all the bonds of the products are formed
This method is a bit less accurate because the OH bond in a water molecule has
a slightly different energy then the OH bond in methanol, and that is slightly
different than the energy of an OH bond in ethanol, and when we make a table of
bond energies like table 14.5 we give an average value based on many different
compounds.
(Looking at the table notice the different in E of single vs double vs triple bonds)
While using bond energies to calculate the ÄH of a reaction is a bit less accurate
than the other methods we have used so far, it is make for a pretty good
estimate.
So how does this method work?
Figure 14.15 and equations
First you total up the energy needed to break all the bonds in the reactants, then
you subtract out the energy that is given off as all the bonds in the products are
formed.
Notice the big change in this equation form our usual way of doing business.
Here we are subtracting products from reactants. This occurs because we
have defined our molar bond energy as the energy needed to break covalent
bonds. I think this is the only time in this course we will do reactant - products so
be careful!!
Key Equation
ÄH oRxn = sum of all bonds broken in reactants - sum of all bonds fomred in
products
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Example calculation
Let’s demonstrate this with a reaction that we know releases lots of energy, the
combination of oxygen and hydrogen to form water.
2H2 + O2 6 2 H2O (H-H and O=O goes to H-O-H)
2 (H-H) + 1(O=O) -4(O-H)
= 2(435) + 1(502) -4(464)
=870 + 502 - 1856 = -484 kJ for 2 moles of H2O or -242 kJ/1 mole water
Clicker question
Set up similar question with one correct set up and 3 bad set ups - like product reactants not counting total number of bonds, not finding double or triple bonds
14-7 Heat capacity
In the next section I will show you how to determine the ÄH for a reaction
experimentally. But before you can do that I must introduce a new quantity
called heat capacity
Key definition:
The heat capacity of a substance is the heat energy required to raise the
temperature of that substance by one degree Celsius (or 1K, they are the same)
Mathematically, if our change in temperature is something other than exactly 10C
we can use the equation
cp = qp/ÄT
where cp is the heat capacity measured at a constant pressure
Remember that qp = ÄH we can write
qp = ÄH = cp@ÄT
This is your clue as to how we will measure ÄH in the next section
cp is always a positive number.
Heat capacity is another extensive property, so it depends on how much material
you have in your system. Because of this we want to modify the number just a
little bit more to come up with an intensive parameter that doesn’t vary with
amount. Just as we divided qp by ÄT when the temperature change is something
other than 1, we will divide our cp by the amount of material used in the
experiment to come up with a new parameter
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Key definitions:
The heat capacity per mole of material is called the molar heat capacity, Cp
The heat capacity per gram of a substance is called the specific heat capacity csp
Key equations:
qp = cp@ÄT; Cp = cp/n; csp = cp/m
qp = n@Cp@ÄT
(n = # of moles of material)
qp = m@csp@ÄT
(m= # of grams of material)
Sample calculations
If 4.219 J of heat is needed to raise the temperature of 36.0 g of water from
10.00 0C to 38.05 oC, what are the specific and molar heat capacities of water.
Specific heat capacity
qp = m@csp@ÄT
4.219 kJ = 36.0 g @ csp @(38.05-10.00)
4.219 kJ = 36.0 g @ csp @28.05 oC
4.219 kJ/(36.0 g @28.05 oC) = 4.178x10-3 kJ/oC@g = 4.178J/oC@g
Molar heat capacity
qp = n@Cp@ÄT
(n = # of moles of material)
36 g H2O x 1 mol H2O = 2 mol H2O
18.016g H2O
4.219 kJ = 1.998 mol @ Cp @(38.05-10.00)
4.219 kJ = 1.998 mol @ Cp @ 28.05 K
4.219 kJ/(1.998 mol @28.05K) = .0753 kJ/mol@K = 75.3 J/mol@K
Another type of problem you will see when dealing with heat capacity is to
determine what happens when you mix a hot material and a cold material
together. For instance you place a 50 g aluminum block @ 110 oC in a bucket
containing 950. g of water @ 10oC
To handle this kind of system you start by assuming that this is all done is a
perfect insulator, so no heat is gained or lost to the surroundings. This is called
an adiabatic system.
If no heat flows to the surroundings then all the heat in the hot material is used to
warm the cold material and
qh(lost) + ql(gained) = 0
where qh is the heat that will be lost from the high temperature material
and ql is the heat that will be gained by the low temperature material
I will define TF as the final temperature oaf the system after it comes to
equilibrium. This will be somewhere between the temperatures of the
high and low temperature materials
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If qh = n@Cp@ÄT and ÄT = Tf-TH
qh = nh × Cph ×(TF-TH)
Note that since TF< TH ÄT in this equation is - consistent with the
warm material losing heat
ql = nl ×Cpl × ÄT and ÄT = TF - Tl
ql = nl × Cpl × (TF - Tl)
In this case ÄT is + consistent with heat flowing into the cold
system
Putting this together
qh(lost) + ql(gained) = 0
qh + ql =0
nh × Cph ×(TF-TH) + nl × Cpl × (TF - Tl) =0
Filling in the numbers
Cp for Al = Cph = 24.4 J/K@mol
nh = moles of Al = 50 g Al x 1 mol Al = 1.853 mol Al
26.98g Al
Cp for water = Cpl = 75.3 J/K@mol
nl = moles water = 950 g x 1 mol
= 52.72 mol H2O
18.02 g
TH = 110
TL = 10
nh × Cph ×(TF-TH) + nl × Cpl × (TF - Tl) =0
[1.853×24.4×(X-110)] + [52.72×75.3×(X-10)]=0
[45.2×(X-110)] + [3969(X-10)] = 0
45.2X - 4972 + 3969X -38690 = 0
45.2X +3969X = 4972 + 38690
4014.2 X =43662
X = 43662/4014.2 = 10.88oC
14-8 Calorimetry
ÄH and ÄU for chemical reactions are usually measured in devices called
calorimeters .
The calorimeter used to determine ÄH can be very simple to slightly complex.
The calorimeter shown in Figure 14.17 is a high precision device that would be
used in an advanced Physical Chemistry class. On the other hand, the device
you will use in the lab consists of a couple of styrofoam cups and a digital
thermometer.
In using these calorimeters we assume that the dewar in the fancy calorimeter,
or the sytrofoam in our simple calorimeter we assume adiabatic conditions, that
is, no heat exchange with the surroundings
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This kind of calorimeter is used for reactions that take place in solution, and the
solvent for these solutions is usually water.
Thus, since we have to heat exchange outside of the calorimeter, and we do this
experiment on the lab bench where we assume the pressure is constant over the
course of the experiment
qwater + qreaction occuring in water =0
and qrxn p = ÄH = -qwater p
Example Problem:
Have you ever made up a NaOH solution in the lab from pellets? It gets very
hot. Let’s try to measure this in our constant pressure calorimeter.
Let’s fill it with 100 grams of water and put in 1 gram of NaOH. If the
temperature goes up from 24.1 to 26.8, what is the molar enthalpy for the
dissolution of NaOH?
qp water = m × csp H2O x ÄT
= 100g x 4.18 (J/Cg) x (26.8-24.1)
= 100g x 4.18 (J/Cg) x 2.7 oC
= 1,129 J
qp H2O + qp rxn =0
1,129 J + qp rxn =0
This is heat gained by the water, so heat lost or released from the chemical
reaction is
qp rxn = -1,129J
This was the energy released by one gram of NaOH. How many moles of NaOH
does this represent?
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14-9 Bomb Calorimeters
Figure 14.18 need to get from CD!
A more complicated type of calorimeter is called the bomb calorimeter
This type of calorimeter is use for reactions that involve gases, and it is used to
measure ÄU instead of ÄH
The most important part of its construction is that it is made out of ½ inch thick
steel. This keeps the volume of the reaction constant. Why is that important?
Remember that U = q + w
if the only work we do is -PÄV, and ÄV = 0 then
qV = Urxn
where qv is heat measured at constant volume.
Where does the term bomb come from? The typical experiment you do in one of
these calorimeter is involves putting some solid in the bottom, filling it with O2,
then using a spark to ignite the solid. If you miscalculate the amount of the
materials you can literally turn the device into a bomb!
With a bomb calorimeter you usually do a separate experiment to determine the
heat capacity of the bomb calorimeter as a whole, and designate this value as
cv,cal
Since this device is a bit expensive, we won’t use it in the gen chem lab, but you
may see it in the Physical Chem lab. Since you won’t use it, let’s skip the math
until you take P chem and actually need it.
14-10 Molecular basis of Heat Capacity
In section 14-5 we discussed where the heat of a reaction comes from; energy
has to be put into the reactants to break the bonds, but then energy is released
from the system as new bonds are formed. So that is the molecular basis of the
heat of a reaction.
But what is the molecular basis for the heat capacity, IE the fact that some
materials will only have a small change in temperature the heat is added, but
other materials my have a much different temperature change for the same
amount of heat?
Let’s start at the low end. He(g), Ne(g) Ar(g) and Hg(g) all have the lowest molar
heat capacities (Cp) ~ 20.8 kJ
What happens on the molecular level when these atoms absorb heat? Well,
going back to the last chapter, at T8, KE8, so the individual atom simply start to
travel faster (translational motion). Your text goes on and shows how the Cp of
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monoatomic ideal gases can be derived from R our gas constant
In poly atomic gases you can absorb energy in other ways than simply going
faster. You can make molecules rotate and you can make atoms at each end of
a bond vibrate. In general the larger, more complicated gas molecules will have
higher Cp because of these added pathways to absorb energy.
In solids and liquids thing become even more complicated because now each
atom or molecule is in intimate contact with a neighbor, and interacts with it in
some way. These interactions provide yet another mechanism for the material to
absorb heat. In fact, take a look at liquid water. It has the largest heat capacity
of any material in table 14.6 largely because of at the intermolecular interactions
that occur in the liquid state (which we will study in chapter 16!).