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chapter
Calculations with Equilibrium Constant
>> Parts of this equation/concept include:
| A. Calculating Equilibrium Constants from Equilibrium Concentrations |
The concentrations in the equilibrium constant expression are the
concentration values at equilibrium. If the concentrations are known,
the equilibrium constant is determined by solving the mass action
expression.
>> Example 1
What is the equilibrium constant for the gaseous reaction
2 N2 + O2 
2 N2O
if, at equilibrium, 0.00335 mole N2, 0.000464 mol
O2 and 0.545 mol N2O are in a 1.00-L container.
Solution:
The mass action expression for this reaction is
Using the given values in the equation,
| KC |
= |
| (0.545)2 |
|
| (0.00335)2(0.000464) |
|
| KC |
= |
| 0.297025 |
|
| (5.20724) x 10–9 |
|
KC = 5.7 x 107
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| B. Calculating Equilibrium Concentrations from Equilibrium Constants |
In the most common type of equilibrium calculations, the problem
gives initial concentration values and equilibrium constant. It
asks for concentrations at equilibrium. The easiest approach is
to create an "ICE" table.
There is a column in the table for each reactant and product that
has a concentration value. (That is, solids, solvents, and pure
liquids are not included.) In the first row, I (for "initial"),
list all the initial values given in the problem under the appropriate
column. If no value is given for the concentration of that substance,
assume that the concentration is zero. (After all, it doesn't make
sense to list everything that is not present.)
In the next row, C (for "change"), first determine reaction
direction. The easiest way to do this is to find a zero in the I
row. Since you can't lose what you don't have, the concentration
of that substance must increase. Therefore put a plus sign in that
column and for all other substances on that side of the reaction.
Put a minus sign for each substance on the opposite side of the
reaction. Using x to represent the unknown amount of substance
that reacts, add to each column the stoichiometric coefficient of
that substance times x.
If no substance has a zero concentration, there are two ways to
determine the direction of the reaction. One way is to calculate
Q and compare that to the equilibrium constant to determine
reaction direction. (See the section on the reaction quotient.)
The other way is to pick a direction and realize that if you choose
incorrectly, the value of x might be negative.
In the last row, E (for "equilibrium"), add the I and C
rows for each column. The values in this row are used in the equilibrium
constant expression.
The equilibrium constant expression is solved for x. There
will be two values obtained for x. One is correct, one is
not. The one that is not correct will predict a negative (and thus
impossible) concentration for some substance. The equilibrium concentrations
can be determined by solving the expression of each E column for
concentration by substituting in the value for x.
>> Example 2
What is the concentration for each substance at equilibrium for
the following gaseous reaction
C2H4 + H2
C2H6 KC
= 0.99
if the initial concentration of ethene is 0.335 M and
that of hydrogen is 0.526 M?
Solution:
The equilibrium constant expression for this reaction is
Since the concentration values in this equation are equilibrium
concentrations, the ICE table is used to determine those values.
Each substance in the reaction is a column. The "initial" line
will have the [C2H4] = 0.335 M and
[H2] = 0.526 M, as described in the problem.
| |
C2H4 |
H2 |
C2H6 |
| Initial |
0.335 |
0.526 |
0 |
| Change |
|
|
|
| Equilibrium |
|
|
|
Since the concentration of ethane, C2H6,
is not mentioned in the problem, its initial concentration is
zero.
Because the ethane concentration is zero, its "change" must be
an increase. Negative concentrations are impossible. Since ethane
is a product, all other products, if there were any, would also
increase in concentration. All reactants will decrease.
Using x as the amount of change, the table becomes
| |
C2H4 |
H2 |
C2H6 |
| Initial |
0.335 |
0.526 |
0 |
| Change |
–x |
–x |
+x |
| Equilibrium |
|
|
|
The equilibrium line is determined by adding the "initial" and
"change" lines for each column.
| |
C2H4 |
H2 |
C2H6 |
| Initial |
0.335 |
0.526 |
0 |
| Change |
–x |
–x |
+x |
| Equilibrium |
0.335 – x |
0.526 – x |
x |
The values from the equilibrium line are substituted into the
mass action expression.
Solve the equation for x.
| 0.99 |
= |
| x |
|
| (0.17621 – 0.861x + x2) |
|
0.1744479 – 0.85239x + 0.99x2
= x
0.1744479 – 1.85239x + 0.99x2
= 0
Using the quadratic equation to solve for x,
| x |
= |
| 1.85239 ± [(1.85239)2
– 4(0.99)(0.1744479)]1/2 |
|
| 2(0.99) |
|
| x |
= |
| 1.85239 ± [3.43135 –
0.69081]1/2 |
|
| 1.98 |
|
| x |
= |
| 1.85239 ± [2.73325]1/2 |
|
| 1.98 |
|
x = 0.1006 or 1.7705
Determine which value for x is correct by calculating
equilibrium concentrations
[C2H4] = 0.335 – x = 0.335
– 0.1006 = 0.234 M
or
[C2H4] = 0.335 – x = 0.335
– 1.7705 = –1.435 M
Since a negative concentration is impossible, the x =
1.77 value must be incorrect.
So
[C2H4] = 0.234 M
[H2] = 0.526 – x = 0.526 – 0.101
= 0.425 M
[C2H6] = 0.101 M
>> Example 3
What is the equilibrium concentration of silver ion in 1.00 L
of solution with 0.010 mol AgCl and 0.010 mol Cl–
in a solution with the equilibrium reaction of
AgCl(s)
Ag+(aq) + Cl–(aq)
Solution:
The equilibrium constant expression for the reaction is
K = [Ag+][Cl–] = 1.8 x 10–10
Recall that as a solid, AgCl is not included in the expression.
Therefore it need not be included in the ICE table either. Using
the ICE table to determine equilibrium concentrations,
| |
Ag+ |
Cl– |
| Initial |
0 |
0.010 mol/1.00 L = 0.010 M |
| Change |
+x |
+x |
| Equilibrium |
x |
0.010 + x |
Since silver ion is not discussed, its initial concentration
is zero. When the reaction approaches equilibrium, it must increase
in concentration.
The initial concentration of chloride ion is given. Recall that
M = mol/L and that calculation is shown in the "initial"
box. Because it is on the same side of the reaction as silver
ion, it also increases in concentration.
Using the equilibrium concentrations in the equilibrium constant
expression
1.8 x 10–10 = [x][0.010 + x]
Solve for x
1.8 x 10–10 = 0.010x + x2
0 = x2 + 0.010x – 1.8 X 10–10
Using the quadratic equation
| x |
= |
| –0.010± [(0.010)2
– 4(1)(–1.8 x 10–10)]1/2 |
|
| 2 |
|
x = 0 if you round
x = 1.8 x 10–8 if you don't
An easier alternative!
The value of K as 1.8 x 10–10 implies
that the amount of product at equilibrium will be quite small.
If you add a small number to a large number, you get the large
number back. (This doesn't work for multiplication!)
In this reaction, because of the equilibrium constant, the value
for x must be small. Therefore if x is added or
subtracted to a number, you get that number back. Using this assumption
in the equilibrium constant expression,
1.8 x 10–10 = [x][0.010 + x]
becomes
1.8 x 10–10 = [x][0.010]
which is a lot easier to solve
1.8 x 10–8 = x
The same answer!
As a rule, if the equilibrium constant is small (<10–4
is a good cutoff), this assumption is valid. Nevertheless, you
should always check that the assumption is valid by comparing
the answer (x) to the value it was added or subtracted
from. In this example, the assumption was
0.010 + x = 0.010
Using the value for x from the calculation
0.010 + 1.8 x 10–8 = 0.010
when rounding to the appropriate number of decimal places according
the the significant figure rules, the statement is true. Therefore
you can skip the quadratic and keep 1.8 x 10–8
as the final answer for x.
More good news: Generally, even if the assumption turns
out not to be valid, this quick calculation is a good way to estimate
the answer. (The smaller the value of K, the better the estimate.)
You can use it to check your calculation with the quadratic.
>> Example 4
What are the equilibrium concentrations of all products and reactants
for the following aqueous reaction:
HSO4–(aq) + H2O(l)
H3O+(aq) + SO42–(aq) K
= 0.012
where the initial concentrations are [HSO4–]
= 0.50 M, [H3O+] = 0.020 M,
[SO42–] = 0.060 M?
Solution:
The mass action expression for the reaction is
Water, as the solvent, is not included.
Setting up the ICE table,
| |
HSO4– |
H3O+ |
SO42– |
| Initial |
0.50 |
0.020 |
0.060 |
| Change |
|
|
|
| Equilibrium |
|
|
|
Since there is no column with a zero, Q is used to determine
the reaction direction.
Note that the only difference between Q and K is
that Q uses the initial line of the table and K
uses the equilibrium line.
Since Q < K, the concentration of products must
increase and the reactants decrease. Therefore the table is
| |
HSO4– |
H3O+ |
SO42– |
| Initial |
0.50 |
0.020 |
0.060 |
| Change |
–x |
+x |
+x |
| Equilibrium |
0.50 – x |
0.020 + x |
0.060 + x |
Using the values in the equilibrium constant expression
| K |
= |
0.012 |
= |
| (0.020 + x)(0.060 + x) |
|
| 0.50 – x |
|
= |
| 0.0012 + 0.080x + x2 |
|
| 0.5 – x |
|
Assuming x is small probably won't work, K = 0.012.
In addition, all the x's would cancel, so unfortunately,
the simplifying assumption won't work.
0.0060 – 0.012x = 0.0012 + 0.080x + x2
0 = x2 + 0.092x – 0.0048
| x |
= |
| –0.092± [8.464 x 10–3
– 4(1)(–0.0048)]1/2 |
|
| 2 |
|
x = 0.0372 or –0.129
Calculating equilibrium concentrations,
[HSO4–] = 0.50 – 0.0372 = 0.46
M
or
[HSO4–] = 0.50 – (–0.129)
= 0.63 M
and
[H3O+] = 0.020 + 0.0372 = 0.057 M
or
[H3O+] = 0.020 + (–0.129) = –0.109
M
Since this value is impossible, [H3O+]
= 0.057 M and [HSO4–] = 0.046
M are correct and
[SO42–] = 0.060 + 0.0372 = 0.097
M
>> Example 5
What is the concentration of H+ if the initial concentration
of H2CO3 is 0.14 M in the following
reaction?
H2CO3
2 H+ + CO32– K
= 2.4 x 10–17
Solution:
The mass action expression for this reaction is
The ICE table is
| |
H2CO3 |
H+ |
CO32– |
| Initial |
0.14 |
0 |
0 |
| Change |
–x |
+2x |
+x |
| Equilibrium |
0.14 – x |
2x |
x |
Note that the change for H+ is 2x. This is
determined from the stoichiometry, where there are twice as many
H+ ions as carbonate ions. In the change line, use
the stoichiometric coefficient times x.
Substituting the equilibrium concentrations into the equilibrium
constant expression
Since the equilibrium constant is very small, it is reasonable
to assume that x is small. Therefore 0.14 – x
= 0.14. This is fortunate because there is not an easy way to
solve a cubic.
3.4 x 10–18 = 4x3
8.4 x 10–19 = x3
9.4 x 10–7 = x
Checking the assumption, 0.14 – 9.4 x 10–7
= 0.14. Yes, x is small!
The concentration of H+ is 2x. Therefore [H+]
= 2x = 2(9.4 x 10–7) = 1.9 x 10–6
M.
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