Intro Physical Chemistry
Homework – Heat, Work and Internal Energy
Questions to Consider:
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What are heat and work?
What is the internal energy of a molecular system?
The temperature scale is an arbitrary construct. Therefore, we need a constant that ties
energy to temperature. What is that constant?
What is the baseline reference point for enthalpy? Why was it chosen to be what it is?
What is the difference between a “standard” state and a “reference” state?
How does Hess’s Law relate to the fact that enthalpy is a state function?
How does heat capacity relate to molecular structure? The heat capacities we’ve
calculated are for ideal gases. What happens to the heat capacity when non-ideal effects
of inter-particle forces are included?
Why do we use enthalpy instead of the internal energy?
What is the “Enthalpy of Formation” and how are the values determined?
1.) What is heat? Simply put, when energy is transferred to or from a system or substance,
heat is the part of the energy that goes into kinetic motion. Thus, heat is NOT
temperature (and vice versa) but is intimately related to temperature. This problem
examines this point.
One of the more important, albeit simple, expressions from kinetic theory is that which
connects kinetic energy to temperature. That relationship is given here as:
𝐸=
3
𝑘 𝑇
2 𝑏
where kb is Boltzmann’s’ constant of 1.3801 x 10-23 J/K
This expression is for kinetic energy of translational motion. We will examine other
motions in a moment. For 1 mole of particles, we may use:
𝐸=
3
𝑅𝑇
2
where R is the universal gas constant of 8.3145 J/mol-K
a.) Verify that kb when multiplied by Avogadro’s number yields the gas constant, R.
b.) Consider a simple atomic system. If we wish to raise the temperature of one mole of
molecules by 1 K. How much is the kinetic energy increased? The heat capacity of
an atomic system such as Helium has a value of 12.47 J/mol-K. How does this
compare to your result and what does it tell you?
Kinetic energy takes on other forms as well. For example, if a nonlinear polyatomic
molecule is allowed to rotate, then by kinetic theory its rotational kinetic energy is
3
2
increased by 𝑘𝑏 𝑇 . Similarly, we know that the molecule can vibrate as well and
kinetic theory tells us that the kinetic energy of vibration increases by k bT for each
way in which the molecule vibrates. (polyatomic molecules vibrate in many ways)
c.) Consider a linear molecule that can only translate (move linearly through space) and
rotate. If we wish to increase the temperature by 1 K again for 1 mole of these
molecules, what will be the increase in kinetic energy total?
d.) Thinking conversely now. Consider that we wish to put INTO a molecule that only
rotates and translates that amount of energy you calculated in c.) . Doing so will
increase the temperature by 1 K. Now consider an atomic system again that has no
rotation, but only translation. If we put that amount of energy you calculated into that
system, what would be the temperature rise?
What you have just examined is the relationship between kinetic energy, motion and
temperature. They are related in the way you examined. Energy is partitioned among
different motions for an ideal system. (Ideal meaning no inter-particle attractions).The
more ways there are to place energy (such as translation and rotation) then the
kinetic energy increase for each motion is less!
Let us now consider a full diatomic ideal molecule. Such a molecule can translate in 3
directions (x, y and z) and has 1 way in which it can vibrate. Being diatomic, it can only
rotate 2 ways as seen below. A rotation around the axis, X, is a spin and requires very
little energy and thus has negligible contribution.
Therefore, we can write by the Kinetic Theory model, the total kinetic energy as:
𝐸𝑇𝑜𝑡𝑎𝑙 = 𝐸𝑇𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛 + 𝐸𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛 + 𝐸𝑉𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛 =
3
2
7
𝑘𝑏 𝑇 + 𝑘𝑏 𝑇 + 𝑘𝑏 𝑇 = 𝑘𝑏 𝑇
2
2
2
e.) If we wish to increase the temperature of 1 mole of such molecules by 1 K, how
much will the kinetic energy increase? How much energy, then, will we need to put
in?
Before we examine the results, there is a simpler way to describe this process. I’ve
arbitrarily had you use 1 K increases. If I change the question to “How much will the
kinetic energy change per mole per degree Kelvin?”, then we arrive at THE definition of
a heat capacity. We then write, instead of ETotal:
𝐶𝑣𝑇𝑜𝑡𝑎𝑙 = 𝐶𝑣𝑇𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛 + 𝐶𝑣𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛 + 𝐶𝑣𝑉𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛 =
3
2
7
𝑅+ 𝑅+𝑅 = 𝑅
2
2
2
7/2 R is the number you calculated in e.). Verify for yourself. Also, Cv means that this the
heat capacity for a system held at constant volume.
Experimentally, the value for a diatomic molecule such as O2 is 20.81 J/mol-K
f.) Calculate the difference in multiples of R between your calculated value and the
experimental value.
Note the magnitude of the difference you calculated. It is almost exactly R different!
Have we made an error? Well, no. But what is the source of the error we observe?
For reasons we will see later, the difference is due to vibrational motion. Kinetic theory
does indeed tell us that we can store kbT amount of energy per molecule or RT amount
of energy per mole in vibrational motion. However, it does not tell us whether or not
we can access that storage! Imagine that you have space in the attic of your house to
store stuff but no ladder to reach it. The storage is available, but not accessible!
Vibration is like this. We will find that only when we have lots of energy to store will that
motion become accessible.
However, not all vibrations of a molecule are equal! Heavy molecules are able to bend
more easily than they are able to stretch and lighter molecules and the strengths of
bonds and steric interactions all play a role.
The model we’ve been using is the kinetic model called the “Dulong-Petit” model which
holds that:
𝐶𝑣𝑇𝑜𝑡𝑎𝑙 =
3
𝑅
2
+ 2 𝑅 + (3𝑁 − 6)𝑅
3
𝐶𝑣𝑇𝑜𝑡𝑎𝑙 =
3
𝑅
2
+ 2 𝑅 + (3𝑁 − 5)𝑅
2
for a nonlinear polyatomic molecule
for a linear polyatomic molecule.
The (3N – 6) and (3N – 5) terms are the number of unique ways a molecule can vibrate
where N is the number of atoms in the molecule.
g.) Calculate the constant volume heat capacity for CO, N2O and CH4 using the DulongPetit law. The literature values are:
Cv(CO) = 20.83 J/mol-K, Cv(N2O) = 30.14 J/mol-K,
Cv(CH4) = 27.00 J/mol-K
For each of the molecules, calculate the heat capacity contribution from vibration
only in multiples of R. Verify that the difference between the experimental and
calculated values is equal to or less than this number.
Many years after the development of the Dulong-Petit law, better expressions were
developed for the description of heat capacity. First by Einstein and then demonstrated
using quantum statistics as we will see later, an improved model is for a diatomic
molecule:
5
𝐶𝑣 = 𝑅 +
2
ℎ𝜈𝑜
⁄𝑘 𝑇
𝑏
ℎ𝜈𝑜 2
2 (𝑘 𝑇 ) 𝑅
ℎ𝜈𝑜
𝑏
⁄𝑘 𝑇
𝑏 − 1)
(𝑒
𝑒
where h = Planck’s constant 6.62608 x 10-34 J-s and o is the frequency at which the
molecule vibrates.
h.) Shown here is the heat capacity of the Einstein model vs. temperature for NO2 gas.
Verify that as the temperature gets sufficiently high, the Einstein model converges to
the Dulong-Petit model. Why is this happening?
i.) For N2, o = 6.99 x 1013 s-1 and for Cl2, o = 1.67 x 1013 s-1 Experimental values are:
Cv(N2)exp = 20.81 J/mol-K and Cv(Cl2)exp = 25.60 J/mol-K. Use the expression above
to verify that it produces much more accurate answers compared to experiment.
Take T = 298.15 K. The remaining differences are due to non-ideal effects.
Work
2.) The heat capacity will be extremely important going forward. The values we examined in
problem 1.) were those at constant volume. We tend to carry out calculations at constant
pressure. How does this change the heat capacity?
Any system held at constant pressure when energy is introduced MUST increase the
volume. Thus, this requires energy. So as above, when we saw the energy going into a
system partitioned among the difference motions, we must contribute some of that
energy in order to do the work. The end result is that we must put in MORE energy for
each degree rise in temperature as some is used to do work. Although it can be proven
rigorously to be true for an ideal gas and a very good approximation for a non-deal gas,
we find that
Cp = Cv + R
or the heat capacity of a substance at constant pressure is R larger than the system held
at constant volume.
Calculate the constant pressure heat capacity for N2 and Cl2 using your calculated
Einstein expression results in problem 1 and the expression above and compare with
literature values. Calculate the % error.
Cp(N2)exp = 29.125 J/mol-K
and
Cp(Cl2)exp = 33.91 J/mol-K.
3.) Work is the energy expended applying motion against a resisting force. It is NOT a
property or “state function” of a system! The extremely important (and troublesome)
effect of this is that the amount of work done depends on HOW the motion is applied.
Consider the gas forming reaction: Zn(s) + 2 HCl(aq)  H2(g) + ZnCl2(aq) carried
out at a constant temperature of 298.15 K.
a.) Let us carry out this reaction in an open vessel subject to a constant atmospheric
pressure, 1 atm. If 1.5 grams of Zn are reacted in an excess of concentrated HCl,
calculate the work done by the expansion of the gas produced.
b.) Now, let us carry out this reaction in a piston system in which the pressure outside
the piston is balanced with the pressure of the gas as it is formed in the reaction.
That is, the expansion is done reversibly. Take the initial volume of the cylinder prior
to expansion to be 100 mL.
Consider: Work and heat are NOT properties of a system. They are, therefore, NOT
state functions. Because of this, HOW we carry out a process becomes extremely
important. We need to know EVERY micro step of every reaction in order to calculate
these quantities. This is troublesome as how can we possibly know this?! How can we
circumvent this problem? We will examine this in a moment. First, let’s examine the first
Law of Thermodynamics.
First Law of Thermodynamics – The Internal Energy, U
Internal energy is the TOTAL amount of kinetic energy within a system AND the potential
energy of interaction. The internal energy IS a property of the system!! It is,
therefore, a “state” function.
The most important consequence of a state function is that
it does not matter HOW a process is carried out, only
knowledge of the initial and final state of the system!
Adiabatic Processes:
An adiabatic process is one that does not exchange heat with the surroundings. (The
system is insulated so that for the process q = 0). Such processes are important for
understanding cooling and heating processes involved in many areas including
liquefaction and weather. As the system is isolated, any change will necessarily produce
a change in P,V and T simultaneously.
Without proof, it can be shown that for an adiabatic process the relationship between
temperatures, pressures and volumes are as follows:
𝑇
𝑇1
𝑉
𝑉2
𝑅⁄
𝐶𝑣
( 2) = ( 1)
𝑇
𝑇1
𝑃
𝑃1
𝑅⁄
𝐶𝑣
( 2) = ( 2)
𝑎𝑛𝑑
𝑃
𝑃1
𝑉 𝛾
𝑉2
( 2) = ( 1)
𝑤ℎ𝑒𝑟𝑒 𝛾 =
𝐶𝑝
⁄𝐶
𝑣
In lecture, we condensed the calculation of q, w, and U for various process of
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Constant pressure
Constant temperature
Constant Volume
Adiabatic
4.) Calculate the final T, then q, w and U for the adiabatic expansion of 1 mole of H2,
initially at 300 K from 1.2 L to 4.0 L. Cp for H2 = 28.824 J/mol-K.
5.) Let us carry out the following cycle. Cycles are processes used in engines in order to
turn heat into work. This problem examines this.
Consider 1 mole of an ideal monatomic gas carried through the following process.
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Isothermal expansion at 300 K from 1 L to 3 L.
Constant volume cooling to 144.22 K
Adiabatic compression to the starting point of 1 L and 300 K
Notice that the gas starts and ends in the same place. Calculate the work, heat and
internal energy change for each step and for the cycle. Does this engine do work? What
is the heat cost to do the work? Does this demonstrate that heat and work are not state
functions but that internal energy is?
6.) Calorimetry is the process by which reaction heats are measured and values tabulated
for substances. A bomb calorimeter is kept at constant volume and, as such, the work is
zero for the process. The heat evolved, then, IS the internal energy change for the
process!
a.) A calorimeter is set up as shown below. 0.769 grams of benzoic acid is placed in the
bomb and the system is charged with oxygen gas. When ignited, the calorimeter is
found to undergo a temperature rise of 2.145 oC. Calculate the heat capacity of the
calorimeter given that Ucomb = -3228 kJ/mol.
b.) The calorimeter, now calibrated, is used to combust 1.221 grams of phenol, C6H5OH.
The temperature rise is measured to be 4.185 K. Calculate the internal energy of
combustion for phenol.
Enthalpy
From our work above, we see that work is troublesome. We need to know the
entire process of forces and motions in order to calculate this quantity. We see also, that
the internal energy change for a process is relatively easy if we keep a system at
constant volume as no work is done and all we need is a thermometer. However, we
rarely carry out reactions at constant volume! We carry them out usually in an open
vessel at constant pressure. Thus, we defined the new function, Enthalpy, H. As shown
in lecture, this quantity is the energy value of a process carried out at constant pressure.
And since it is a state function, we only need to know the starting and ending
points in a process!
Also, since the process does not matter, we can CHOOSE any convenient process
we like in order to calculate the enthalpy change, H. No matter what process we
choose, we will get the correct answer!
We would like to know the enthalpy change for any reaction we choose. So, we could
perform calorimetry on every possible reaction and tabulate it, but this would be
ridiculous. Instead, let us choose a convenient reference point. Our choice is arbitrary as
since enthalpy is a state function, it does not matter. Our current selection of reference
values is that of elements in their standard states. We arbitrarily assign these states
an enthalpy value of zero and then calculate the energy required to produce 1 mole of
molecules from elements in their standard states and define this and tabulate those
values as “Heat of Formations” for a substance.
The next few problems examine using and establishing values to place in the tables
along with examination of the arbitrariness of the selection.
7.) Consider the reaction:
CH4(g) + Br2(g)  CH3Br(g) + HBr(g)
a.) Calculate the enthalpy required in order to turn the reactants into elements in their
standard states. (Take note of the proper sign!)
b.) Calculate the enthalpy required in order to produce the products from elements in
their standard states. Take Hf(CH3Br(g)) = -35.40 kJ/mol
c.) Write the overall cycle of the process and, using your two values in a.) and b.),
calculate the enthalpy for the entire cycle.
Since enthalpy is a state function..the enthalpy change for the process you just
calculated is exactly the same as the reaction carried out because the initial and
final states are the same! You should see that by defining a convenient reference
point, we can condense our work into convenient calculations.
What if we chose a different reference state? Let’s examine this next.
Let us define a new reference state as “elements as monatomic gas atoms” Thus we will
define our formation heats now as being “The enthalpy required to form 1 mole of
molecules from elements as monatomic gases.”
In order to do this, we need these values. Let’s create a table.
Using the following energy values, calculate the new formation value for each of the
reactants and produce in the previous reaction.
½ Br2(l)  Br(g)
207.10 kJ/mol
½ H2(g) 
H(g)
217.97 kJ/mol
C(graphite)

C(g)
716.68 kJ/mol
With this, I’ll do the first. We wish to have the enthalpy required to form 1 mole of CH4(g)
from atoms as monatomic gases, or
C(g) + 4 H(g)

CH4(g)
From the values provided and the heats of formation, we have:
Hf = (1 x -74.81 kJ) - (1 x 716.68 kJ + 4 x 217.97 kJ) – = -1663.37 kJ/mol
This value is entered into the table below. Using the same method, the remaining values
are provided
Substance
CH4
Br2(g)
CH3Br(g)
HBr(g)
New Hf
-1663.37 kJ/mol
-383.29 kJ/mol
-1613.09 kJ/mol
-461.47 kJ/mol
d.) Now, using the new values and using the products minus reactants approach,
calculate the enthalpy of the reaction. Compare with the previous values.
We’ve seen that the reference values for substances are elements in their standard
states. But, how do we determine these? We do so using calorimetry as done above.
We begin with executing the following reactions in a bomb calorimeter.
C(graphite) + O2(g)  CO2(g)
and
H2(g) + ½ O2(g)  H2O(l)
As can be seen, these are both combustion AND formation values! So measurement
directly gives the following:
Hf(CO2(g)) = -393.51 kJ/mol
and
Hf(H2O(l)) = -285.83 kJ/mol.
Now, bomb calorimetry gives us internal energy values and we wish to have enthalpy
values. How is this done? To do this, go to the definition for enthalpy as:
H = U + PV or in terms of change, H = U + (PV)
Enthalpy and internal energy differences then are the difference in the change in
pressure multiplied by the volume or work! Since solids and liquids do not undergo a
significant change in volume and we are at a constant pressure, we can apply (PV)
only to the gases. From the ideal gas law, PV = nRT, so we can write that:
(PV) = (nRT)  nRT. Where n is the change in the number of moles of gases. IN
general, we use T = 298.15 K for this conversion.
Example: for H2(g) + ½ O2(g)  H2O(l)
n = nproducts – nreactants = 0 – 3/2 = -1.5 moles
8.) The enthalpy of combustion of phenol, C6H5OH is -3054 kJ/mol. Calculate the internal
energy of combustion using the expression above as follows:
a.) Write a balance reaction for the combustion of 1 mole of phenol and from your
balanced reaction, determine the change in the number of moles of gases, n.
b.) Calculate the internal energy of combustion from your result in a.)
Let’s put it all together now…
9.) A bomb calorimeter having a heat capacity of 12.045 kJ/K is used to combust 0.813
grams of ethyl acetate liquid (CH3COOC2H5).The calorimeter is found to undergo a
change in temperature of 1.713 K.
a.) Calculate the molar internal energy of combustion of ethyl acetate.
b.) Write a balance reaction for the combustion of ethyl acetate, determine a value of n
and use that to calculate the molar enthalpy of combustion of ethyl acetate.
c.) Using the formation values for H2O(l) and CO2(g) calculate the enthalpy of formation
for ethyl acetate. Compare to literature.
At this point, we have a convenient state function and a method by which to measure
values. Since it is a state function, we can apply it to any process we like.
10.)
Calculating enthalpy:
a.) Use the heat of formation values to calculate the heat of vaporization of H2O(l)
Hf (kJ/mol)
H2O(l)
-285.83

H2O(g)
-241.82
The literature value is 40.7 kJ/mol. Why is your result different? (We will examine this in a
following problem)
b.) Given the following reactions, calculate the enthalpy of the reaction:
N2O(g) + NO2(g)  3 NO(g)
N2(g) + O2(g)  2 NO(g)
H = +180.7 kJ
2 NO2(g)  2 NO(g) + O2(g)
H = 113.1 kJ
2 N2(g) + O2(g)  2 N2O(g)
H = 163.2 kJ
c.) Consider the following reactions:
C2H4(g) + H2(g)  C2H6(g)
H = -132 kJ/mol
C6H6(g) + 3 H2(g)  C6H12(g)
H = -246 kJ/mol
The first reaction is the hydrogenation of a pure double bond whereas the second
reaction is the hydrogenation of an aromatic system. From this information, calculate
the enthalpy of the hydrogenation of benzene if it contained three isolated double
bonds. Calculate the resonance energy of benzene.
Temperature variation
11.)
The problems above were all carried out using temperatures at which the
tables are listed, usually 298.15 K. However, we often wish to know the enthalpy of a
reaction at other temperatures. Problem 10 a.) is an example. The value calculated in
this problem was the heat of vaporization of water at 298.15 K. When water boils, it is at
373.15 K. Let’s calculate the enthalpy at this temperature.
Given Kirchhoff’s formula and the heat capacities for water liquid and vapor given,
calculate the enthalpy for the vaporization of water at its boiling point. Compare with
your answer in 10 a.) and with literature.
Kirchoff’s Formula: ∆𝐻(𝑇2 ) = ∆𝐻(𝑇1 ) + ∆𝐶𝑝 (𝑇2 − 𝑇1 )
Cp_water(liq) = 75.29 J/mol-K
Partial Solutions:
Cp_water(gas) = 33.58 J/mol*K
1.
c
g
20.79 J/mol-K
CO2 – Cvib = 4R, Exp Difference = 3 R
2
For CO Calculated Cv = 29.11 J/mol-K
3
wrev = -107.4 J
5
Step 1, w = -2.74 kJ Step 3: w = 1.943 kJ
9
a.) Ucomb = -2263 kJ/mol