General Approach To Solving Chemical Equilibria Problems
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1
2
3General Approach to Solving Chemical Equilibria Problems
Ù
The general form of a chemical reaction is:
aA + bB
cC +dD
where A and B are reactants in the forward direction and C and D are products in the forward direction. The lower case letters are the
stoichiometric coefficients for the balanced equation. The general form of the equilibrium constant equation is then
c
d
a
b
K
= [C]
[D]
/[A]
[B]
eq
Problems involving chemical equilibria can be placed into a matrix format with two kinds of concentrations identified: the initial or
non-equilibrium concentration C
and the equilibrium concentration [X]. (Note the bracket versus the capital C. It is the equilibrium
x
concentrations, [X], that are used to calculate K
.
eq
A
B
C
D
initial
C
C
C
C
A
B
C
D
change
C
-[A]
C
-[A]
[C]- C
[D] - C
A
A
C
D
equilibrium
[A]
[B]
[C]
[D]
There are a variety of equilibrium problems that can be solved using the construction above. In some cases the problems are solved
directly, with one unknown per equation. Others problems require significant algebraic manipulation. All of them are easy if you can
identify the type of information used and place it in the matrix.
Examples:
Model Equilibrium System As the chemical system for all the problems on this work sheet, we will use
Ù
3
2
2NH
3H
+ N
K
= 3.8 = [H
]
[N
]/[NH
]
3
2
2
eq
2
2
3
Problem 1. Calculating K
from equilibrium concentrations.
eq
In these problems it is necessary to identify all the bottom row equilibrium concentrations (the ones in [X]) so that K
is determined
eq
directly. For the ammonia equilibrium above, equilibrium concentrations are [N
] = .10 M, [H
] = .50 M, [NH
] =.057 M. What is
2
2
3
K
? This is the easiest of problems because you are told what all the bottom row equilibria are.
eq
2NH
3H
N
3
2
2
initial
change
equilibrium
.057
.50
.10
1
3
2
K
= [.10]
[.50]
/[.057]
= 3.8
eq
Note that for constant temperaure, this value for K never changes. You will see it used in every example in the handout.
Problem 2. Using stoichiometry to complete the array.
A more complex type of problem provides some of the initial and some of the equilibria concentrations and you are asked to solve for
the rest of the concentrations in the array. These problems are accomplished using stoichiometry and simple substitutions for
unknowns.
Given K
= 3.8, [N
] = 0.3 M, [H
] = 0.2 M and C
= 0.04 M. What is the initial concentration of C
?
eq
2
2
NH3
N2
2NH
3H
N
3
2
2
initial
.04
?
change
final
.2
.3
3
2
First find [NH
] at equilibrium from the K
expression: 3.8 = ((.2)
(.3)/[NH
]
and [NH
]= .025 M.
3
eq
3
3
Then use the stoichimetric relationships to back substitute throughout the matrix. Follow the steps shon.
2NH
3H
N
3
2
2
initial
.185 step 3
.295 step 3
.035
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