Intro Physical Chemistry
Homework – Gibbs Free Energy - Equilibrium
Questions to Consider






The development of the reaction quotient and the subsequent equilibrium
constant involves the new quantity called the “chemical potential” What is this
quantity?
What assumptions are made in the development of the equilibrium constant?
What is the “standard chemical potential”. What does it define?
Equilibrium constants are unitless. Why?
What happens when use an equilibrium constant in terms of another unit, say
torr?
In the phase transitions above, do these obey all the equilibrium conditions
discussed? What is G for any phase transition at equilibrium? Is this a
thermodynamic criteria?
Equilibrium is clearly a major concern in chemistry. Once again, it’s the Gibbs
Free Energy that governs this process. It must be understood, however, what
approximations have been used! The equilibrium constant was developed using the
expressions for Gibbs Free Energy but we assumed substances were ideal gases!
This, clearly is a problem for non-ideal gases and for solution systems. We will address
this at a later time, however, we need to see how the thermodynamics govern the
equilibrium in a system.
We found that the equilibrium constant can be calculated through the standard
Gibbs Free Energy change for a reaction. Note that this means that all substances are
in their standard state, which is the substance in it’s state at 1 bar pressure. Note that
temperature is not a part of this standard, but is a reference state that we can use.
We have, then:
𝐾𝑒𝑞 = 𝑒
−∆𝐺 𝑜⁄
𝑅𝑇
where T = 298.15 K.
Note that one cannot change the temperature to get a new Keq! When we do this, we
are assuming that Go is the same at all temperatures which is not the case! The T goes
with the Go , thus if we had Go values at 273.15 K, then we would use that
temperature for T!
From the standard Gibbs Free Energy change, we can calculate the Free Energy
change under non-standard conditions. We use the relationship..
∆𝐺 = ∆𝐺 𝑜 + 𝑅𝑇𝑙𝑛(𝑄)
Recall that when G = 0, then Q becomes Keq!
When we apply this, the rules still apply; if the Free Energy change is less than
zero, the reaction will proceed to the right and conversely for a positive Free Energy.
This give us an opportunity! If the Free Energy change is zero, we are at equilibrium!
This is the criteria we used to get the equilibrium constant.
Note the inclusion of the word “change”. A chemical system will adjust so as
to minimize the Gibbs Free Energy. When the Free Energy is at its lowest value, then
the change will be zero, not the Free Energy value.
Temperature Dependence:
Dealing with the temperature dependence on equilibrium is not directly
straightforward. We approach this using a relationship called the “Gibbs-Helmholtz”
equation. The approach is beyond what we will discuss in this course, however, the
result is easily presented as:
𝐾2
−∆𝐻𝑟𝑥𝑛 1
1
𝑙𝑛 ( ) =
[ − ]
𝐾1
𝑅
𝑇2 𝑇1
There are still assumptions applied here. We are assuming the Hrxn is temperature
independent. We know this is not true, however, this assumption will produce acceptable
results over a wide range of temperatures. For exacting work, more extensive
relationships can be produced. We’ll leave it at that.
Note: If the reaction is endothermic, then Hrxn is a positive quantity. From the relation
above, we see that the right hand side of the equation will be a positive number if T 2 >
T1. If this is true, then we must have the left hand side be positive as well which requires
K2 > K1. This is an equation example of Le Chatelier’s Principle! If energy is required
(endothermic), then if the temperature is increased the equilibrium will shift to the right.
The converse is true as well. If the reaction is exothermic, then the equilibrium will shift
to the left!
The following problems have you examine and practice such an applications.
1.) Consider the reaction:
CO2(g) + H2(g)  CO(g) + H2O(g)
a.) Calculate the standard Gibbs Free energy at 298.15 K.
b.) Calculate the Gibbs Free Energy change for this reaction at the following nonequilibrium pressures:
PCO2 = 1.5 bar PH2 = 0.15 bar PCO = 0.0015 bar and PH2O = 0.0040 bar
In which direction will the reaction shift under these conditions?
c.) At 960 K, the Kp for this reaction is 0.534 and at 1260 K, Kp = 1.571. From this
calculate the enthalpy of the reaction.
d.) Calculate the reaction enthalpy from standard tables and compare with d.) . Why are
the results slightly different?
Let’s examine the answer to the why question another way.
e.) Using your result for H in d.) and calculating the entropy change for this reaction
from standard tables, calculate the Gibbs Free Energy and Kp for this reaction at
1260 K using G = H - TS. Compare the Kp with literature value given in c.). State
any assumptions made. What is wrong? Why did you get an incorrect answer?
2.) Consider the reaction: N2O4  2 NO2.
At the temperatures given, the following partial pressures of the gases were measured at
equilibrium and are provided in the table. Use this data to answer the following:
Temperature P(N2O4)
350 K
0.362
400 K
0.0666
P(NO2)
1.28
1.87
a.) Calculate the value of Kp at each temperature
b.) Calculate the enthalpy of the reaction.
c.) Calculate the entropy of the reaction at each temperature. State any assumptions.
The “Extent of Reaction” approach
The approach that follows is called the “Extent of Reaction” approach. To apply
this approach, we assign “x” to be an arbitrary variable that determines how far the
reaction proceeds. Notice that if x = 0, nothing has reacted but when x = 1, the reaction
has gone to completion. Thus, x is a measure of the extent of the reaction, giving us the
method name. In terms of material, x is related to mole fractions. For example, consider:
I.C.E. result
N2O5(g)
 NO2(g)
1–x
x
+
NO3(g)
x
The total amount of material at any point in the reaction is then:
1–x+x+x=1+x
From this, the mole fractions are:
1−𝑥
)
1+𝑥
𝑥
)
1+𝑥
𝑋𝑁2 𝑂5 = (
𝑎𝑛𝑑 𝑋𝑁𝑂2 = (
Finally, from Dalton’s Law, the partial pressures of each are: 𝑃𝑖 = 𝑋𝑃𝑇𝑜𝑡
Applying..
1−𝑥
) 𝑃𝑡𝑜𝑡
1+𝑥
𝑃𝑁2 𝑂5 = 𝑋𝑁2 𝑂5 𝑃𝑇𝑜𝑡 = (
𝑥
) 𝑃𝑡𝑜𝑡
1+𝑥
𝑎𝑛𝑑 𝑃𝑁𝑂2 = 𝑋𝑁𝑂2 𝑃𝑇𝑜𝑡 = (
Finally, we write the equilibrium constant as:
𝑃𝑁𝑂
𝑃𝑁𝑂
( 𝑃 2) ( 𝑃 2)
𝑜
𝑜
𝐾𝑃 =
=
𝑃𝑁2 𝑂5
𝑃𝑜
𝑥
𝑥
(
) 𝑃𝑡𝑜𝑡
(
)𝑃
1
+
𝑥
1
+
𝑥 𝑡𝑜𝑡 )
(
)
(
𝑃
𝑃
𝑜
𝑜
1−𝑥
(1 + 𝑥 ) 𝑃𝑡𝑜𝑡
𝑃𝑜
𝑥 2 𝑃𝑡𝑜𝑡 2
)
(1 + 𝑥)2 𝑃𝑜2
𝑥 2 𝑃𝑡𝑜𝑡
=
=
1−𝑥
1 − 𝑥 2 𝑃𝑜
(1 + 𝑥 ) 𝑃𝑡𝑜𝑡
𝑃𝑜
(
Thus we see that Kp is now directly related to PTot and therefore, so is x. We are
free now to solve for x given Kp and a total pressure.
Apply this method to the following:
3.) Consider the reaction: N2 + 3 H2
 2 NH3 .
For this reaction 1 bar of N2 and 3 bars of H2 are introduced into a 10.0 L vessel at 455
K. From ICE diagrams, we see that at equilibrium, we have:
N2
+
(1 – x)
3 H2

2 NH3 .
3(1 – x)
2x
Let us maintain this system at a constant pressure of PT = 4 bar.
a.) From the above ICE diagram, produce an expression in terms of the extent of
reaction, x, that gives the total amount of material in the system at any point.
b.) From the above ICE diagram and your answer in a.), produce expressions for the
mole fractions of each species in terms of the extent of reaction, x.
c.) From your result in b.), produce expressions for the partial pressures of each species
using Dalton’s Law.
d.) From your results above, produce an expression for the equilibrium constant, Kp in
terms of x, PTot and the standard pressure, Po
The Gibbs Free Energy for this system is plotted below vs “x” the “extent of reaction” defined
above.
Gibbs Free Energy as a Function of Reaction Progress
9
Gibbs Free Energy (kJ/mol)
7.571
6.143
4.714
3.286
1.857
0.429
1
0
0.1
0.2
0.3
0.4
0.5
0.6
Extent of Reaction
0.7
0.8
0.9
1
e.) Using the graphic above, calculate the partial pressures of all species at equilibrium.
f.) Calculate the equilibrium constant from your results at 455 K.
g.) From standard tables, calculate the standard Gibbs Free Energy for this reaction and
use the result to calculate Kp at 298.15 K. Gf(NH3) = -16.45 kJ/mol.
h.) From standard tables, calculate the reaction enthalpy for this reaction at 298.15 K.
Hf(NH3) = -46.11 kJ/mol.
i.) Using your results in g.) and h.), calculate Kp at 455 K and compare with your
estimated result that you obtained from the graphic provided. Since the temperature
is elevated and involve gases, use the extended equation that gives the proper
temperature dependence on the enthalpy given here as:
∆𝐶𝑝 ∙ 𝑇1 − ∆𝐻𝑇1
1
1
∆𝐶𝑝
𝑇2
𝐾𝑝 (𝑇2 ) = 𝐾𝑝 (𝑇1 ) ∙ 𝑒𝑥𝑝 [(
)∙( − )+
∙ 𝑙𝑛 ( ) ]
𝑅
𝑇2 𝑇1
𝑅
𝑇1
Cp(N2) = 29.125 J/mol-K, Cp(H2) = 28.824 J/mol-K, Cp(NH3) = 35.06 J/mol-K
Non-ideal Systems
The equilibrium constant and reaction coefficient were produced from fundamental
equations defining the Gibbs Free Energy. In doing so, we made the assumption that the
materials involved were ideal gases. We found earlier that gases are non-ideal and often
behave in a non-ideal fashion.
We found that if we tried to develop an equilibrium system using a non-ideal gas, the
result is too cumbersome to use effectively. We LIKE the form of the equilibrium
constant as it provides for us a convenient way to determine equilibria and reaction
directions. So what to do?
To keep the form if Q and Keq, the solution is to introduce an “activity”, a. The activity is
related to the experimental measurements through an “activity coefficient”, . The
relationship is:
a = P for gases or a = C for solution systems where C is the concentration.
Important: The P or C is the experimental measure; the activity is the
thermodynamic measure and  connects the two together.
All thermodynamic tables relate to the activity. So the process is as follows:



Use the thermodynamic tables to calculate the equilibrium constant.
Use the equilibrium constant to determine the activities
Use the activity coefficient to determine the pressure or concentration
The problem, of course, is what is ? The determination of this value will be addressed after we
introduce solution systems. For gases, the procedure is straightforward although beyond the
scope of this course.
This problem introduces the use of activities in order to determine an experimental pressure.
4.) Consider the decomposition reaction: CaCO3(s)  CaO(s) + CO2(g)
a.) Using the thermodynamic tables, calculate Go and then Kp for this reaction.
b.) Calculate H for this reaction and use it and your results in a.) in order to calculate
Kp at 1200 K.
c.) Write the equilibrium constant expression for this reaction in terms of activities. Then
replace the activity with the activity coefficient and pressure of CO2 in your
expression.
Without proof, let us use the following derived expression for the activity coefficient of
CO2. If proof is desired, I recommend the full course in Physical Chemistry..it is a
wonderful course.
𝛾=
1
1−𝐵∙𝑃
where B is a constant having a value of 0.05 bar-1
If we try to use this, we see we have a problem. We can’t calculate the equilibrium
pressure of CO2 until we know “”. But we don’t know “” until we know the pressure of
CO2!
The approach is to carry out the following steps in order::
 Calculate the pressure of CO2 assuming ideality, that is,  = 1.
 Use this pressure to calculate a  value.
 Use the new  to calculate a new pressure of CO2.
 Calculate a new 
 Repeat the above sequence until the numbers all converge to a single value.
d.) Assuming  = 1, calculate a value for PCO2. Use this to calculate  from the
expression above. Recalculate PCO2 and repeat until the pressure converges to a
single value.
Now, having examined our first non-ideal system, our next task will be to examine
solution systems.
Consider: We REALLY want to keep the form of the equilibrium constant! If we were to
introduce non-ideal equations of state into our derivation of an equilibrium constant, the result
becomes unmanageable. Fortunately, this is not needed. We only need to connect the form of
the equilibrium constant that we want to keep to the laboratory measurements that we make,
thus connecting the thermodynamics to the lab saving our precious equilibria criteria. To do this,
we introduce the “activity” This quantity replaces the pressures and concentrations we formerly
used in the equilibrium constant and is thus connected directly to the thermodynamics. We
connect this quantity to the concentrations and pressures through the “activity coefficient”.
Unfortunately, having a simple model that will allow us to do this is not there. We have
reasonable methods for gases and ionic solutions but they only have limited use. We are going
to have to go into the lab and make measurements to connect the two. Once we do, we will then
have all the tools needed in order to continue our work at predicting the outcome of a reaction
system.
Selected Solutions:
1.) b.) 2.5 kJ/mol
c.) 36.18 kJ/mol
2.) b.) 57.06 kJ/mol
3.) e.) P(N2) = 0.571 bar at a PT = 4 bar
4.) b.) 178.3 kJ/mol
d.) During the recursion, the first two pressures calculated are: 4.83 and 3.65 bar. Repeat until
converged.