Example Problems: (part III) - Avon Chemistry

So now our equation looks like this:
This makes the algebra a LOT EASIER…
Let’s finish, and solve for “x”…
3
H
N
NH
3
3x x
2.75
K 5.78
10
2 2 5
2 2
3
Then:
Then:
Then:
Then:
3
2
2.75
4
3x x
5
27x
5.7810
7.5625
5.7810 7.5625 27x
5 4
5.7810 5
7.5625
27
1.6189 10 x
5 4
x
4
4 5
1.6189 10 x 0.0634
Now let’s find the concentrations of each at equilibrium:
H
x ; N
x ;
Okay, let’s check for the rule of 5%:
2
3 3 0.0634 0.1902
2
0.0634
NH3 2.75
3
H
N
NH
x x
3 3
3 0.1902 0.0634
K 5.768
10
2.75 2.75
2 2 5
2 2 2
3
5 5
calculated true
5.76810 5.7810
% Error 100 100 0.208 %
5
true
5.7810
That my friend is well within the 5% rule !!! And, this method saved us a great deal of time trying to
solve a very interesting algebra problem… However, this will NOT work for every equilibrium problem, but
if you NOTICE a very small K value (usually at least to the negative three power ( 10 -3 )) you may use the 5%
rule… If the K value is not very small, you may need to use the quadratic equation (see examples, part II from
today).