Journal of Chemical Education
Software
Print
•
Software
•
Online
•
Books
General Chemistry Collection
Equilibrium Calculator for Windows
Windows-compatible computers
Suggested Exercises
Exercises
User's Guide
Guide.pdf
Suggested Exercises for Equilibrium Calculator for Windows
Note to Equilibrium Calculator 2.0 Users
The figures in these exercise are from the earlier verson of the program.
Suggested Exercises
Equilibrium Calculator can be used to quickly provide accurate answers to the
exercises below.
1. It is observed in nature that when a chemical reaction occurs in a system that
contains a fixed number of atoms, the final (equilibrium) mixture of reactants
and products always has the same composition regardless of whether the initial
ingredients were all reactants, all products, or some mixture of reactants and
products. The purpose of the first few exercises is to show that the chemist’s
mathematical model of chemical equilibrium also has this property.
−−−> 2 NH
Consider the reaction: N2 + 3 H2 <−−−
3
for which the equilibrium constant at room
temperature is Kc = 0.016. If in one vessel
the initial concentration of N2 is 4 M, the
initial concentration of H2 is 12 M and no
NH3 is present initially, when the chemicals are allowed to react, increasing
amounts of ammonia are formed until all
the chemicals reach their equilibrium concentrations. (Note that at every point in the
reaction, the concentration of nitrogen atoms is 8 M and the concentration of hydrogen atoms is 24 M). These equilibrium
concentrations are readily calculated using
Equilibrium Calculator as is illustrated in
Figure 2.
−−−> 2 NH reFigure 2. Results for N2 + 3 H2 <−−−
3
action with no product initially present.
In a second vessel, the initial concentration of NH3 is 8 M and no N2 or H2 are
present initially. (Note that as in the previous example, the concentration of
nitrogen atoms is 8 M and the concentration of hydrogen atoms is 24 M.) When
the chemicals are allowed to react, increasing amounts of hydrogen and nitrogen
will be formed until all the chemicals reach their equilibrium concentrations.
These concentrations are readily calculated using Equilibrium Calculator as is
illustrated in Figure 3.
Since the equilibrium concentrations are exactly the same in both examples, the
mathematical model is in agreement with the experimental observation. The
position of a chemical equilibrium does not depend on whether the reaction
User Reproducible
JCE Software • General Chemistry Collection
2
Suggested Exercises for Equilibrium Calculator for Windows
starts with all reactants or all products as
long as the total amounts of the elements
are the same in both cases.
To test your understanding of this principle, choose another chemical reaction with
a known equilibrium constant and use
Equilibrium Calculator to demonstrate
that the position of the chemical equilibrium is always the same regardless of the
starting position.
2. A second important principle is that
while the position of the equilibrium is
always the same, many different, but related, pairs of chemical equations and equilibrium constants can be used to describe
the same equilibrium.
−−−> 2 NH reFigure 3. Results for N2 + 3 H2 <−−−
3
action with no reactants initially present.
Consider the reaction: N2 + 3 H2 −−−>
<−−− 2 NH3 for which the equilibrium constant
at room temperature is Kc = 0.016. The equilibrium expression is:
[NH3 ] 2
= Kc
[N2 ] [H2 ] 3
The mathematical expression for this equilibrium when the initial concentration
of N2 is 4 M, the initial concentration of H2 is 12 M and initially no NH3 is present
can be written:
(2x)2
(4−x) (12−3x)3
= 0.016
where x = the number of reaction units the reaction proceeds to the right to come
to equilibrium. The equilibrium concentrations for this example have already
been given in Figure 2. Now, suppose the same chemical reaction is written as:
1/2 N2 + 3/2 H2 −−−>
<−−− NH3. The equilibrium expression is:
[NH3]
[N2] ⁄2 [H2]
1
3⁄
2
= Kc′
With the same initial concentrations as above, the mathematical expression for
this equilibrium can be written:
x
= Kc′
3x 3
x 1⁄2
(4 − ) (12 − ) ⁄2
2
2
where x = the number of reaction units the reaction proceeds to the right to come
to equilibrium.
User Reproducible
JCE Software • General Chemistry Collection
3
Suggested Exercises for Equilibrium Calculator for Windows
Obviously, since the left-hand side of this
expression is different from the one given
above, Kc ≠ Kc′. Because all the stoichiometric coefficients in the first chemical equation have been divided by 2 to give the
second chemical equation, the second
mathematical expression is equal to the
first one raised to the 1/2 power. Thus, Kc′
= (Kc)1/2 = 0.1264911. If the mathematical
model is correct, the chemical equation: 1/2
N2 + 3/2 H2 −−−>
<−−− NH3 with Kc = 0.1264911
and the same initial concentrations as were
used in Figure 2 should give the same equilibrium concentrations when entered into
Equilibrium Calculator. This is shown to be
true in Figure 4.
Figure 4. Dividing the equation for the ammonia equilibrium by 2 gives identical equilibrium
concentrations when Kc is adjusted accordingly.
Since the equilibrium concentrations are exactly the same in both examples, the
mathematical model is in agreement with the experimental observation. The
position of a chemical equilibrium is correctly predicted regardless of how the
chemical reaction is written so long as the numerical value of the equilibrium
constant is adjusted to be consistent with the different ways of writing the
stoichiometric coefficients in the balanced chemical reaction.
To test your understanding of this principle, choose another chemical reaction
with a known equilibrium constant and use Equilibrium Calculator to demonstrate that the position of the chemical equilibrium can be calculated correctly
independent of the form in which the chemical reaction is written.
3. A related principle is that if the direction in which a chemical reaction is
written is reversed, the value of the new equilibrium constant is the reciprocal
of the old one.
Consider the reaction: N2 + 3 H2 −−−>
<−−− 2 NH3 for which the equilibrium constant
at room temperature is Kc = 0.016. The equilibrium expression is:
[NH3]2
[N2] [H2]3
= Kc
The mathematical expression for this equilibrium when the initial concentration
of N2 is 4 M, the initial concentration of H2 is 12 M and initially no NH3 is present
can be written:
(2x)2
(4 − x) (12 − 3x)3
= 0.016
where x = the number of reaction units the reaction proceeds to the right to come
to equilibrium.
User Reproducible
JCE Software • General Chemistry Collection
4
Suggested Exercises for Equilibrium Calculator for Windows
The equilibrium concentrations for this example have already been given in
Figure 2. Now suppose the same chemical reaction is written as: 2 NH3 −−−>
<−−− N2
+ 3 H2. The equilibrium expression is:
[N2] [H2]3
[NH3]2
= Kc′′
With the same initial concentrations as above, the mathematical expression for
this equilibrium can be written:
(4 − x) (12 − 3x)3
(2x)2
= Kc′′
where x = the number of reaction units the reaction proceeds to the left to come
to equilibrium.
Obviously, since the left-hand side of this
expression is different from the one given
above, Kc ≠ Kc″. Because the second
mathematical expression is equal to the
reciprocal of the first one, Kc ″ = (Kc)-1 =
62.5. If the mathematical model is correct,
the chemical equation: 2 NH3 −−−>
<−−− N2 + 3 H2
with Kc = 62.5 and the same initial concentrations as were used in Figure 2 should
give the same equilibrium concentrations
when entered into Equilibrium Calculator.
This is shown to be true in Figure 5.
Since the equilibrium concentrations are Figure 5. Reversing the equilibrium equation
exactly the same in both examples, the still produces the same equilibrium concentramathematical model is in agreement with tions when the equilibrium constant is adjusted accordingly.
the experimental observation. The position
of a chemical equilibrium is correctly predicted regardless of whether the chemical reaction is written forward or backwards so long as the numerical value of the equilibrium constant is adjusted to
be consistent with the direction of chemical reaction.
To test your understanding of this principle, choose another chemical reaction
with a known equilibrium constant and use Equilibrium Calculator to demonstrate that the position of the chemical equilibrium can be calculated correctly
independent of the direction in which the chemical reaction is written.
4. Calculating the pH of a dilute solution of a weak acid is another example of
where Equilibrium Calculator is more convenient than using the exact solution,
which involves solving a quadratic equation. For the general acid dissociation
+
–
HA(aq) −−−>
<−−− H (aq) + A (aq), with a dissociation constant, Ka, the equilibrium
expression is:
User Reproducible
JCE Software • General Chemistry Collection
5
Suggested Exercises for Equilibrium Calculator for Windows
[H+] [A−]
= Ka
[HA]
Which can also be written:
x2
= Ka
[HA] o− x
where x = the amount of H+(aq) and A–(aq) formed and [HA]o – x is the amount
of HA(aq) remaining at equilibrium.
When the initial acid concentration, [HA]o, is high and the concentration of
dissociated ions, x, is low, [HA]o – x ≈ [HA]o so the equilibrium expression is
often approximated as:
1
x2
= Ka and thus x = [H+] ≈ (Ka ⋅ [HA]o) ⁄2
[HA]o
Since this approximate expression is much easier to solve for [H+], the question
arises as to within what range of acid concentrations and dissociation constants
is this approximation valid. If we take Ka = 0.0010 and [HA]o = 0.10 M, we have
a typical example of where the choice of the best method of solution is not clear.
Using [H+] ≈ (Ka ⋅ [HA] o) ⁄2 gives [H+] =
0.010 M but is this sufficiently accurate for
the problem at hand? The exact solution
using Equilibrium Calculator is illustrated in Figure 6.
1
Thus, the exact solution is [H+] = 0.0095 M
and the approximate solution is [H+] =
0.010 M giving an error of 5%. Many texts
suggest using the 5% rule in these situations, i.e., [H+] will be calculated with an
accuracy of 5% by the approximate formula whenever [HA]o / Ka > 100. In the
previous example, the ratio was exactly Figure 6. Equilibrium Calculator gives exact
100 and the error 5% which illustrates the results for acid dissociation calculations.
correctness of this rule. In Figure 7 below,
this 5% error point occurs at a concentration of 0.10 M which corresponds to –log [HA]o = 1. At this point on the graph,
the true and approximate pH points are almost indistinguishable, but, as you
can see, when the acid concentration decreases—to the right on the graph—the
two lines rapidly diverge until at a concentration of 10–6 M, the error is 1.5 pH
units or, in units of concentration, the error is more than 30 fold.
The percent dissociation of a weak acid is defined as % Dissociation = [H+] /
[HA]o * 100 and thus can be calculated accurately using the [H+] values from
Equilibrium Calculator. For example, at the 5% error point (where [HA]o = 0.10
M in this example) the percent dissociation = 9.5%.
User Reproducible
JCE Software • General Chemistry Collection
6
Suggested Exercises for Equilibrium Calculator for Windows
To graphically illustrate this example, you are to prepare a graph of [H+] versus
[HA]o for a weak acid with a dissociation constant chosen to be between 10–4 and
10–2 over a range of [HA]o concentrations from 10–6 M to 10 M. Because of the
wide range of concentrations necessary to illustrate the difference in the two
calculation methods, it is much more informative if the axes of the graph are
defined and plotted logarithmically. Thus, pH and not [H+] should be plotted on
the Y-axis and correspondingly –log [HA]o and not [HA]o should be plotted on
the X-axis. The graph should have two curves, one where the values of [H+] are
1
calculated using [H+] ≈ (Ka ⋅ [HA]o ) ⁄2 and one where the values of [H+] are
calculated using Equilibrium Calculator. A third line showing the percent
dissociation versus concentration can also be
plotted on the same
graph if a second Y-axis
is defined for the % Dissociation. Plotting the
three lines on the same
graph shows that when
the percent dissociation
is less than 10%, the error in [H+] will be less
than 5% and the error in
pH will be 0.02 units or
less which is hard to detect in the laboratory. A
Figure 7. Approximating equilibrium calculations can result in signifitypical such graph for Ka
cant error.
= 0.001 is given in Figure
7.
User Reproducible
JCE Software • General Chemistry Collection
7