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Homework Problems
Chapter 6 Homework Problems: 8, 12, 24, 40, 46, 52, 58, 64, 68, 70,
72, 74, 78, 82, 94, 96
CHAPTER 6
Thermochemistry
Thermochemistry, Work, and Heat
Thermochemistry is the study of energy flow in chemical
systems.
Work (w) is (force) . (displacement)
Heat, or thermal energy, (q) is the energy that moves from a hot
object to a cold object when placed in contact with each other.
T1 > T3 > T2
heat flows from hot
to cold block
Energy
Energy is the capacity to do work or supply heat. In principal any
kind of energy can be converted into an equivalent amount of work or
heat.
We divide energy into two general types, kinetic (due to motion
in a particular direction) and potential (due to position or composition).
The total amount of all the different kinds of energy for a system
is called the internal energy (E).
In a closed system (no energy in or out) energy can be converted
from one type to another, but the total amount of energy remains
constant. This statement is called conservation of energy.
Units of Energy
All forms of energy can be expressed in the same units. To find
the MKS unit for energy, it is convenient to use the equation for kinetic
energy
EK = mv2 So units are (kg) (m/s)2 = kg.m2 = 1 Joule = 1 J
2
s2
Other common units for energy include:
calorie (cal) Amount of heat needed to raise the temperature of 1 g of
water by 1 C
1 cal = 4.184 J (exact)
1 kcal = 1 food calorie = 1000 cal = 4184. J (exact)
Kilowatt hour (kWh) Amount of energy equal to 1000. J/s (1000 Watt =
1 kWatt) for 1 hour. Note that 1 watt = 1 J/s, a unit of power.
1 kWh = 3.6 x 106 J (exact)
System, Surroundings, and Universe
The system is a part of the universe that we have separated off for
study.
The surroundings are everything not included in the system.
The universe is everything - system + surroundings.
While in principle the surroundings are everything not included
in the system, in practice we can usually focus on that part of the
surroundings in the immediate vicinity of the system.
State Function
A state function is any function whose change in value depends
only on the initial and final state of the system, but which is independent
of the pathway used to go between them. The value for something that is
not a state function depends not only on the initial and final state but also
on the pathway used to travel between them.
Altitude is a state function, distance traveled is not a state function. In
thermodynamics, E is a state function, q and w are not state functions (in
fact, they are not functions at all!)
First Law of Thermodynamics
The first law of thermodynamics gives a relationship between
internal energy, work, and heat.
E = q + w
E = Ef - Ei = change in internal energy
q = heat
w = work
For now, we limit ourselves to mechanical work, where w = - p V.
Note that the first law is based on experimental observation and is
not derived from some other law or principle. The first law relates the
change in a state function (E) to the sum of two things that are not state
functions (q and w).
Mechanical Work
Mechanical work is the work associated with the change in the
volume of a system.
Consider the expansion of a gas
Derivation of the Expression For Mechanical Work
w = F . d (from the definition of work)
But p = F/A, and so F = p . A
So |w| = p . A . d However, from the diagram d = h
So |w| = p . A . h
But A . d = V where V is the change in volume.
So |w| = p . V (absolute value)
But notice that when a gas expands
its volume change is in a direction opposite
to that of the applied pressure. This introduces a negative sign into the expression for
mechanical work. Therefore
w = - p . V
p
d
Sign Convention for Heat and Work
We use the following sign convention for heat and work.
q > 0 heat flows from the surroundings into the system (endothermic)
q < 0 heat flows from the system to the surroundings (exothermic)
w > 0 work is done on the system (system is compressed)
w < 0 work is done by the system (system expands)
Note that qsurr = - qsyst, and wsurr = - wsyst.
Use of the First Law
A gas is confined inside of a cylinder at an initial pressure and
volume pi = 1.00 atm and Vi = 2.000 L. 8000 J of heat is added to the
gas under conditions of constant pressure. The final volume of the gas is
Vf = 6.000 L. What are q, w, and E for the process?
A gas is confined inside of a cylinder at an initial pressure and
volume pi = 1.00 atm and Vi = 2.000 L. 8000 J of heat is added to the
gas under conditions of constant pressure. The final volume of the gas is
Vf = 6.000 L. What are q, w, and E for the process?
From the first law, E = q + w.
Heat is added to the system, so q = + 8000. J
For constant pressure w = - p V
= - (1.00 atm) (6.000 L – 2.000 L)
= - 4.00 L.atm 101.3 J = - 405. J *
1 L.atm
Finally, E = q + w = 8000. J + (- 405. J) = 7595. J
*Note: To find the conversion between L.atm and J we do the following
# J = 1 L.atm 1 m3
1000 L
1.013 x 105 N/m2 = 101.3 N.m = 101.3 J
1 atm
Heat Capacity
Consider adding a known amount of heat to a substance. One
would expect that the temperature of the substance would increase.
Heat capacity (C) is a measure of how much heat is required to
cause the temperature of a substance to change by a given amount.
Heat capacity is defined as
C = q/T
where q = amount of heat added
T = Tf - Ti = change in temperature
Notes on Heat Capacity
1) Since C = q/T, the units for C are (energy)/(temperature).
The derived MKS units for C are J/K, though it is often given in J/C.
2) Note that the numerical value for C when expressed in J/K or
J/C is identical. That is because we are using the change in temperature.
Since the size of a degree Kelvin and a degree Celsius is the same, the
numerical value for T is the same whether expressed in K or C.
Example: Ti = 15.0 C (288.2 K) and Tf = 21.5 C (294.7 K)
T = Tf – Ti = 21.5 C – 15.0 C = 6.5 C
= 294.7 K - 288.2 K = 6.5 K
3) C is not a state function, since q is not a state function.
However, if we restrict ourselves to processes occurring under conditions
of constant pressure then C is a state function. This is sometimes
indicated by using the symbol Cp.
Specific Heat and Molar Heat Capacity
There are two quantities that are closely related to heat capacity.
Specific heat capacity, (specific heat, Cs) - The heat capacity per
gram of substance.
Molar heat capacity, (Cm) - The heat capacity per mole of
substance.
The relationships between Cs, Cm, and heat capacity are as
follows:
Cs = C/m
where m= mass of substance
Cm = C/n
where n = moles of substance
So Cs has units of J/g.C and Cm has units of J/mol.C.
Note that C is an extensive property, but Cs and Cm are intensive
properties.
Finding C, Cs, and Cm
We may use the above relationships to find heat capacity and
related quantities from experimental data.
Example: 1000. J of heat is added to a 20.0 g sample of water
(H2O, M = 18.01 g/mol). The initial and final temperatures of the water
are Ti = 19.3  and Tf = 31.3 C. What are C, Cs, and Cm?
Example: 1000. J of heat is added to a 20.0 g sample of water
(H2O, M = 18.01 g/mol). The initial and final temperatures of the water
are Ti = 19.3 C and Tf = 31.3 C. What are C, Cs, and Cm?
C = q/T
T = Tf - Ti = 31.3 C - 19.3 C
= 12.0 C
So C = 1000. J = 83.3 J/C
12.0 C
Cs = C/m = (83.3 J/C) = 4.17 J/g.C
20.0 g
Cm = C/n
n = 20.0 g 1 mol = 1.11 mol
18.01 g
Cm = (83.3 J/C) = 75.0 J/mol.C
1.11 mol
Specific Heat For Common Substances
The above table is at T = 25. C. Note that water has an unusually large value for heat capacity, which acts to moderate temperatures for
cities surrounded by large bodies of water.
Constant Volume Processes
Consider some general process taking place at constant volume.
From the first law
E = q + w
But for mechanical work, w = - pV
so at constant volume, V = 0, so w = 0
Therefore, for a process carried out at constant volume
E = qV (V = constant)
What does this mean? For a process carried out at constant
volume, q is a state function, and so no information is needed concerning
path. This makes it far easier to calculate and keep track of heat flow for
these kinds of processes.
Enthalpy
Most processes in the laboratory are carried out at constant
pressure instead of constant volume. It would be nice to have a state
function whose change in value was equal to q for constant pressure
processes.
We define enthalpy, H, as follows
H = E + pV
Since E, p, and V are state functions, it follows that enthalpy is
also a state function.
Constant Pressure Processes
Consider some general process taking place at constant pressure.
From the definition of enthalpy
H = E + pV
H = E + (pV) = E + (pfVf - piVi)
Now, if pressure is held constant, then pf = pi = p, and so
H = E + p(Vf - Vi) = E + pV
Now, from the first law
E = q + w
If we only have mechanical work
E = q - p V
If we now substitute into the expression for enthalpy, we get
H = E + pV = (q - p V) + p V
or (finally!)
H = qp (p = constant)
What does this mean? For a process carried out at constant
pressure, q is a state function, and so no information is needed
concerning path. This makes it far easier to calculate and keep track of
heat flow for these kinds of processes.
To summarize
E = qV (constant volume processes)
H = qp (constant pressure processes)
Since q is something that can be measured experimentally, we now have
a way to relate this information to changes in state functions (internal
energy or enthalpy).
Calorimetry
Calorimetry is the experimental method used to measure the heat
produced or taken up in a chemical reaction. The measured value for q
can be related to Erxn (for constant volume) or Hrxn (for constant
pressure).
Bomb Calorimetry
In bomb calorimetry a measured mass of a chemical substance
reacts with excess oxygen to form combustion products. If the heat
capacity of the calorimeter apparatus is known (and this can be
determined experimentally) then
q = - C T
C = heat capacity of calorimeter
T = Tf - Ti = change in temperature
The negative sign in the above equation occurs because we are
measuring the value of q for the surroundings, and qsyst = - qsurr.
Since the combustion process occurs under conditions of constant
volume, q = E, the change in internal energy.
Example: 1.412 g of carbon (M = 12.01 g/mol) are burned in a
bomb calorimeter (C = 10325. J/C). The temperature of the calorimeter
changes by 4.48 C. What are E (change in internal energy) and Em
(change in internal energy per mole of carbon) for the process.
Example: 1.412 g of carbon (M = 12.01 g/mol) are burned in a
bomb calorimeter (C = 10325. J/C). The temperature of the calorimeter
changes by 4.48 C. What are E (change in internal energy) and Em
(change in internal energy per mole of carbon) for the process.
q = - C T = - (10325. J/C) (4.48 C) = - 46300. J
For a process carried out at constant volume q = E, so
E = - 46300. J
Finally, the change in internal energy per mole of carbon is
Em = E/n
n = 1.412 g 1 mol = 0.1176 mol
12.01 g
Em = - 46300. J = - 393000. J/mol = - 393. kJ/mol
0.1176 mol
Constant Pressure Calorimetry
Just as bomb calorimetry is carried out under conditions of
constant volume, we can also do calorimetry under conditions of
constant pressure.
In a coffee cup calorimeter
a chemical reaction is carried out
under conditions of constant
pressure.
For a reaction in solution
qsoln = msoln . Cs,soln . T
H = qsyst = - qsoln
Standard Conditions and Standard State
It is convenient to report thermodynamic data for a particular set
of conditions. These conditions are called the thermodynamic standard
conditions, and are usually (but not always) taken to be p = 1.00 atm*, T
= 25.0 C = 298.2 K.
The standard state for an element is the most stable form of the
element for standard conditions.
carbon C(s)
nitrogen N2(g)
bromine Br2()
oxygen O2(g)
iron Fe(s)
argon Ar(g)
mercury Hg ()
chlorine Cl2(g)
sulfur S(s)
__________________________________________________________
* Technically, standard pressure is now taken to be 1.00 bar = 0.987 atm. This makes only a
minor difference in values for thermodynamic quantities.
Enthalpy of Reaction (Hrxn)
Because of the importance of enthalpy, we define enthalpy
changes for specific kinds of processes.
Hrxn (enthalpy of reaction) - The enthalpy change when one
mole of a specific reaction is carried out at p = 1.00 atm (); equals qp for
the process.
Examples:
CaCO3(s)  CaO(s) + CO2(g)
Hrxn = + 178.3 kJ/mol
CaO(s) + CO2(g)  CaCO3(s)
Hrxn = - 178.3 kJ/mol
2 CaCO3(s)  2 CaO(s) + 2 CO2(g)
Hrxn = + 356.6 kJ/mol
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O() Hrxn = - 55.8 kJ/mol
Notice that changing the direction of the reaction changes the
sign for Hrxn , and multiplying a reaction by a constant multiplies the
value for Hrxn .
We can see this as follows:
CaCO3(s)  CaO(s) + CO2(g)
Hrxn = + 178.3 kJ/mol
CaO(s) + CO2(g)  CaCO3(s)
Hrxn =
?
__________________________________________________________
CaCO3(s)  CaCO3(s)
Htotal = 0.0 kJ/mol
Carrying out the first process followed by the second process
means that our final state is the same as our initial state. There is no
change in the system, and so Htotal = 0.0 kJ/mol. Therefore, the change
in enthalpy for the second process must be the same as the change in
enthalpy for the first process, except for the sign.
Relationship Between Hrxn and Erxn
For a process carried out at constant pressure
H = E + pV
If
V > 0 then H > E
V = 0 then H = E
V < 0 then H < E
Now consider a chemical reaction. Since Vg >> Vs, V it follows
that
V = Vf - Vi  Vg,f - Vg,i
If the gases are ideal, then
p V = pVg,f - pVg,i = ng,fRT - ng,iRT = ngRT
where ng = ng,f - ng,i
So Hrxn = Erxn + pV  Erxn + ngRT
Use of Thermodynamic Relationships
When ignited magnesium burns with a bright white flame. The
balance chemical equation for the process is
2 Mg(s) + O2(g)  2 MgO(s)
Consider carrying out the
above process at standard
thermodynamic conditions. What
are q, w, and E for the process?
Also, do you expect Hrxn to be
greater than, about the same, or
less than Erxn?
Hrxn = - 1203.2 kJ/mol
When ignited magnesium burns with a bright white flame. The
balance chemical equation for the process is
2 Mg(s) + O2(g)  2 MgO(s)
Hrxn = - 1203.2 kJ/mol
Consider carrying out the above process at standard thermodynamic
conditions. What are q, w, and E for the process?
Process is carried out at constant pressure, so
q = H = - 1203.2 kJ/mol
H = E + ngRT , so
E = H - ngRT
ng = 0 - 1 = -1
(- 1 mol/mol)
E = - 1203.2 kJ/mol - (-1)(8.314 x 10-3 kJ/mol.K)(298. K)
= - 1200.7 kJ/mol
Finally, E = q + w, so
w = E - q = (-1200.7 kJ/mol) - (- 1203.2 kJ/mol) = + 2.5 kJ/mol
Enthalpy of Formation (Hf)
The formation reaction for a substance is defined as the reaction
that produces one mole of a single product out of elements in their
standard state. Because of the way we have defined the formation
reaction, we may have to use fractional stoichiometric coefficients for
some or all of the reactants. The enthalpy change for this reaction is
defined as the enthalpy of formation for the substance.
H2(g) + 1/2 O2(g)  H2O()
Hf(H2O()) = - 285.8 kJ/mol
H2(g) + 1/2 O2(g)  H2O(g)
Hf(H2O(g)) = - 241.8 kJ/mol
C(s) + O2(g)  CO2(g)
Hf(CO2(g)) = - 393.5 kJ/mol
Pb(s) + C(s) + 3/2 O2(g)  PbCO3(s) Hf(PbCO3(s)) = - 699.1 kJ/mol
3/
2
O2(g)  O3(g)
Hf(O3(g)) = + 143.0 kJ/mol
Example: Write the formation reaction for acetone (CH3COCH3()).
Example:
(CH3COCH3()).
Write
the
formation
reaction
3 C(s) + 3 H2(g) + ½ O2(g)  CH3COCH3()
for
acetone
Note that based on the above definition it follows that the
enthalpy of formation of an element in its standard state is 0.0 kJ/mol.
We may see this as follows:
The formation reaction for N2(g), by definition (formation of one
mole of a single product out of elements in their standard state), is
N2(g)  N2(g)
But nothing happens in the above process, and so it follows that
Hrxn = Hf(N2(g) ) = 0.0 kJ/mol
This will be true for any element in its standard (thermodynamically
most stable) state.
Enthalpy of Combustion (Hc)
The combustion reaction for a substance is defined as the reaction
of one mole of a single substance with O2(g) to form combustion
products. Because of the way in which we have defined the combustion
reaction we may have to use fractional coefficients for some of the
reactants and products. The enthalpy change for this reaction is defined
as the enthalpy of combustion for the substance.
Combustion products for common elements are
C  CO2(g)
H  H2O()
N  N2(g)
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O()
Hc(CH4(g)) = - 890.3 kJ/mol
C6H6() + 15/2 O2(g)  6 CO2(g) + 3 H2O()
Hc(C6H6()) = - 3267.4 kJ/mol
Writing Formation and Combustion Reactions
We may use the definition of formation reaction and combustion
reaction to write own the balanced chemical equations corresponding to
these reactions.
Example: Hexane (C6H14()) is a hydrocarbon often used as a
solvent in organic reactions. Write the formation reaction and the
combustion reaction for hexane.
CH3CH2CH2CH2CH2CH3
Example: Hexane (C6H14()) is a hydrocarbon often used as a
solvent in organic reactions. Write the formation reaction and the
combustion reaction for hexane.
Formation - One mole of a single product out of elements in their
standard state.
 C6H14()
6 C(s) + 7 H2(g)  C6H14()
(balanced)
Combustion - One mole of a single reactant, plus oxygen, to form
combustion products.
C6H14() + ? O2(g) 
C6H14() + ? O2(g)  6 CO2(g) + 7 H2O()
C6H14() + 19/2 O2(g)  6 CO2(g) + 7 H2O()
(balanced)
Enthalpy Change For Phase Transitions
There are three phase transitions that can occur by adding heat to
a substance.
s   fusion (melting)
Hfus
  g vaporization
Hvap
s  g sublimation
Hsub
The enthalpy change when one mole of a substance undergoes the
transition at the normal transition temperature is defined as the enthalpy
change for the phase transition. It represents the amount of heat required
to convert one mole of the substance from the initial phase to the final
phase.
H2O()  H2O(g)
Hvap(H2O) = 40.7 kJ/mol at T = 100.0 C
Unlike other processes, the temperature for a phase transition is
usually taken to be the temperature for which the two phases exist at
equilibrium when p = 1.0 atm.
Hess’ Law
We previously noted that the change in the value for a state
function depends only on initial and final state and is independent of the
path used to travel between the two states. We may put this in a more
formal manner in terms of Hess’ law.
Hess’ law – The change in value for any state function will be the
same for any process or combination of processes that have the same
initial and final state.
We are particularly interested in applying Hess’ law to chemical
reactions.
Hess’ Law For a Chemical Reaction
Let us use Hess’ law to find the value for (Hrxn) for the
following chemical reaction:
CaCO3(s)  CaO(s) + CO2(g)
Hrxn
We may obtain this same reaction as follows:
Ca(s) + ½ O2(g)  CaO(s)
Hf(CaO(s))
C(s) + O2(g)  CO2(g)
Hf(CO2(g))
CaCO3(s)  Ca(s) + C(s) + 3/2 O2(g)
- Hf(CaCO3(s))
CaCO3(s)  CaO(s) + CO2(g)
Hrxn
So by Hess’ law,
Hrxn = Hf(CaO(s)) + Hf(CO2(g)) - Hf(CaCO3(s))
We can get Hrxn using only data on enthalpies of formation!
General Method For Finding Hrxn
Based on the same procedure used in the previous example the
following general relationship can be derived
Hrxn = [  Hf(products) ] - [  Hf(reactants) ]
Notice what this means. If we have a table for formation
enthalpies we can find the value for Hrxn for any chemical reaction. In
fact, the same general procedure can be used to find the values for the
change in any state function.
Also note that when we use the superscript  this indicates not
only standard conditions but also standard concentrations for reactants
and products.
gases p = 1.0 atm
solutes [M] = 1.0 mol/L
solids, liquids, solvents must be present in the system
Example: Find Hrxn for the conversion of acetylene into dichloroethane
C2H2(g) + 2 HCl(g)  CH2ClCH2Cl()
Example: Find Hrxn for the conversion of acetylene into dichloroethane
C2H2(g) + 2 HCl(g)  CH2ClCH2Cl()
Hrxn = [Hf(CH2ClCH2Cl()) ] - [Hf(C2H2(g)) + 2 Hf(HCl(g))]
= [ (- 165.2 kJ/mol) ] - [ (226.7 kJ/mol) + 2 ( - 92.3 kJ/mol) ]
= - 207.3 kJ/mol
Thermodynamic data are given in Appendix II-B.
Other Uses of Hess’ Law
Hess’ law can be used to find Hrxn for any process that can be
written as a combination of other chemical reactions whose values for
Hrxn are known.
Example (6.71): Consider the following formation reaction
5 C(s) + 6 H2(g)  C5H12()
Hrxn = ?
Using the following information find Hrxn for this reaction
(1) C5H12() + 8 O2(g)  5 CO2(g) + 6 H2O(g) Hrxn = - 3505.8 kJ/mol
(2) C(s) + O2(g)  CO2(g)
Hrxn = - 393.5 kJ/mol
(3) 2 H2(g) + O2(g)  2 H2O(g)
Hrxn = - 483.5 kJ/mol
Note all of the above are combustion reactions, which are particularly
easy to carry out experimentally.
Example (6.71): Consider the following formation reaction
5 C(s) + 6 H2(g)  C5H12()
Hrxn = ?
Using the following information find Hrxn for this reaction
(1) C5H12() + 8 O2(g)  5 CO2(g) + 6 H2O(g) Hrxn = - 3505.8 kJ/mol
(2) C(s) + O2(g)  CO2(g)
Hrxn = - 393.5 kJ/mol
(3) 2 H2(g) + O2(g)  2 H2O(g)
Hrxn = - 483.5 kJ/mol
reverse 1st reaction
5 CO2(g) + 6 H2O(g)  C5H12() + 8 O2(g) Hrxn = + 3505.8 kJ/mol
5 times the 2nd reaction
5 C(s) + 5 O2(g)  5 CO2(g)
Hrxn = - 1967.5 kJ/mol
3 times 3rd reaction
6 H2(g) + 3 O2(g)  6 H2O(g)
5 C(s) + 6 H2(g)  C5H12()
Hrxn = - 1450.5 kJ/mol
Hrxn = + 87.8 kJ/mol
End of Chapter 6
“In this house we obey the laws of thermodynamics!”
- Homer Simpson
“...the Dutch physicist Heike Kamerlingh Onnes gave H the name
enthalpy, from the Greek  (in) and  (heat), or from the single
Greek word  (enthalpos), to warm within.” K. J. Laidler, The
World of Physical Chemistry
“[Thermodynamics] is the only physical theory of universal
content which, within the framework of the applicability of its basic
concepts, I am convinced will never be overthrown.” Albert Einstein