Example Problems: (part III) - Avon Chemistry

AP CHEMISTRY
TOPIC 6: EQUILIBRIUM, PART E EXAMPLES ( PART III ) Day 69:
Solving Equilibrium Problems ( 5% Rule )
Solving Equilibrium Problems: (Procedure for solving equilibrium problems)
1. Write the balanced equation for the reaction.
2. Write the equilibrium expression.
3. List the initial concentrations (for reactants AND products)
4. Calculate the Q, and determine the direction of the shift to equilibrium.
5. Define the change needed to reach equilibrium (I.C.E.), and define the equilibrium concentrations by applying the
change to the initial concentrations.
6. Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknowns.
7. Check your calculated equilibrium concentrations by making sure they give the correct value of K.
Example: Synthesis of hydrogen gas and nitrogen gas from ammonia, NH 3 (all gases), 2.75 mol NH 3 is mixed in a
1.00 L flask. The equilibrium constant for this reaction (at this temperature) is 5.78 x 10 -5 .
Step 1: write the balanced equation:
Step 2: write the equilibrium expression:
Step 3: The initial concentrations are:
[ NH 3 ] 0 = [ H 2 ] 0 = [ N 2 ] 0 =
Step 4: There is no need to calculate Q, since no H 2 and N 2 is initially present, AND we know the system will shift to
the right (form products) to reach equilibrium.
Step 5: I.C.E., If we let x represent the number of moles per liter of NH 3 consumed to reach equilibrium.
( Concentration ) 2 [ NH 3 ] 3 [ H 2 ] + [ N 2 ]
Initial: 2.75 0 0
Change: - 2x + 3x + x
Equilibrium: 2.75 - 2x 3x x
Step 6: Substituting the equilibrium concentrations into the equilibrium expression gives:
3
H
N
NH
3
3x x
2.75 2x
K 5.7810
2 2 5
2 2
3
Q < K
Okay, now for the really cool part. Since K is so small, taking the difference of 2.75 – x would be very close to a
value of 2.75 (more like, 2.69 something), we can IGNORE the “x” in the equation for ammonia. This type of problem
is an algebra NIGHTMARE !!! Therefore, we can use the 5% rule. The 5% rule is a common practice (and on every
AP exam (question number one)) to help determine concentrations at equilibrium. You will need to know how to use
the quadratic equation when you get to college (they will expect you know how to do this)…

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).