Calculation of K_{eq }and Concentrations 
After you have finished this you should be able to:
1) Calculate the equilibrium constant. 2) Calculate the equilibrium concentration of a participant when the value of K_{eq }is known as well as the concentrations of the other participants. 3) Determination of the net direction of a reaction prior to establishing an equilibrium. 4) Calculation of equilibrium []'s when initial []'s and the equilibrium constant are known. 5) Calculation of the % dissociation and the % yield of a reaction. 
Example Problems 
Problem #1 
When 0.40 moles of PCl_{5 }is heated in a 10.0 L container, an equilibrium is established is which 0.25 moles of Cl_{2 }is present. 

a) What is the number of moles of PCl_{5 }and PCl_{3}
at equilibrium?
b) What are the equilibrium concentration of all three components?
PCl_{5
} <=====> _{
}PCl_{3}
+
Cl_{2}

From the balanced equation and the coefficients it should
be obvious
that everything works on a onetoone basis. So there should be 0.25
moles
of PCl_{3} and 0.25 moles of PCl_{5} should have been
consumed,
so there are 0.400.25 moles = 0.15_{ }moles of PCl_{5 }left.

Therefore [PCl_{5}] = 0.15 moles = 0.015
moles/L
10.0 L 
[PCl_{3}] and [Cl_{2}] both are 0.25 moles
each.
Therefore their concentrations are identical as well.

[PCl_{3}] = [Cl_{2}] = 0.25 moles =
0.025
moles/L
10.0 L 
Problem #2 
When 1.00 mole of NH_{3 }gas and 0.40 moles of N_{2
}gas
are placed in a 5.0 L container and allowed to reach an equilibrium at
a certain temperature, it is found that 0.78 moles of NH3 is present.
The
reaction is:
2 NH_{3}(g)
<======>
3 H_{2}(g) + N_{2}(g)

a) What are the number of moles of H_{2} and N_{2}
at equilibrium?
b) What is the concentration in moles/L of each species?

Problem #3 
A mixture of H_{2} and I_{2} is allowed to
react
st 448^{o}C. When equilibrium is established, the
concentrations
of the participants are found to be [H_{2}] = 0.46 mol/L, [I_{2}]
= 0.39 mol/L, and [HI] = 3.0 mol/L. Calculate the value of the K_{eq
}at
448^{o}C.
H_{2}(g) + I_{2}(g) <=====> 2 HI(g)


Problem #4 
Assume that in the analysis of another equilibrium mixture
of the
same reaction as above, at the same temperature of 448^{o}C,
the
equilibrium concentrations of I_{2} and H_{2} are both
0.50 mol/L. What is the equilibrium concentrations of HI?

Problem #5
The equilibrium constant for the reaction represented below is 50 at 448^{o}C H_{2}(g) + I_{2}(g) <=====> 2 HI(g) a) How many moles of HI are present at equilibrium when 1.0 moles of H_{2} is mixed with 1.0 moles of I_{2} in a 0.50 L container and allowed to react at 448^{o}C? b) How many moles of H_{2} and I_{2} are left unreacted? c) If the conversion of H_{2} and I_{2} to HI is essentially complete, how many moles of HI would be present? d) What is the percent yield of the equilibrium mixture?

Solution
a) First write the K_{eq} equation based on the balanced reaction. You should always make sure that the reaction is balanced first. The concentrations at the start are: [H_{2}] = 1.0 mol/0.50 L = 2.0 mol/L [I_{2}] = 1.0 mol/0.50 L = 2.0 mol/L [HI] = 0 moles therefore 0 mol/L Fill in the numbers under the reaction for the starting concentrations.
H_{2 }+ I_{2}
<=====> 2 HI
Let 'x' be the number of moles of H_{2 }per litre consumed. finish 2.0  x
2.0 
x
2x

Stop and confirm the above. If 'x' amount of H_{2}
is consumed
then 'x' amount of I_{2} will also be consumed because the H_{2
}and
I_{2} react on a 1:1 basis. Also if 'x' amount of H_{2}
(or for that matter I_{2}) reacts then '2x' of HI will be
produced
because the reactants and product react on a 1:2 basis.

Fill in the finish concentrations into the K_{eq }equation.
and solve for 'x'.


50 =
(2x)^{2}
(2.0  x)(2.0  x) 
50 = (2x)^{2}
___
(2.0  x)^{2} 
This is going to work nicely because both the top and
bottom on
the right are squares. There square root the entire right hand side,
along
with the left.


7.1 = 2x
2.0  x 14.2  7.1x = 2x x = 14.2 = 1.56 mol/L

You are solving for 'x' which is in moles/L therefore you
can automatically
fill in the units for any value of 'x'.
Now to answer the question. What are the final reactant and product concentrations? [H_{2}] = 2.0 mols/ 1  x = 2.0 mols/L  1.56 mols/L = 0.44 mols/L [I_{2}] = 2.0 mols/ l  x = 2.0 mols/L  1.56 mols/L = 0.44 mols/L [HI] = 2 x = 2 * 1.56 mols/L = 3.12 moles/L

b) From a) you get the number of moles/L of H_{2}
and I_{2} left unreacted. This is the concentration not the
answer
to the question. You are asked for the number of moles left unreacted
therefore
moles = concentration * volume
[H_{2}] = [I_{2}] = 0.5 L * 0.44 mols/L = 0.22 moles Therefore 0.22 moles of H_{2} and 0.22 moles of I_{2}
are left unreacted. Note that the L's unit cancels out.

c) If you look at the equation again you can see that
there is exactly
the right amount of H_{2} to react with I_{2}.
Theoretically
if both the H_{2} and I_{2} where all used up then we
should
make 2 moles of HI. This is the theoretical yield because it is what
could
be made in theory if everything went to completion.

d) The percentage yield is calculated by using the actual
yield
from part a) and the theoretical yield from part b). You may use either
moles or moles/L but make sure that they are both the same.
Percentage Yield = Actual
Yield
* 100
= 1.56 moles/L * 100
= 78% yield (note that the moles/L units cancel) OR Percentage Yield = 3.12 moles * 100
= 78% yield (note that the units still cancel)

Problem #6 
How many moles of HI are present at equilibrium when 2.0
moles of
H_{2} is mixed with 1.0 moles of I_{2} in a 0.50 L
container
and allowed to react at 448^{o}C. At this temperature K_{eq}
= 50.

Solution
First the equation: H_{2} + I_{2} <=====> 2 HI Then the concentrations: [H_{2}] = 2.0 moles/0.50 L = 4.0 moles/L [I_{2}] = 1.0 moles/0.50 L = 2.0 moles/L [HI] = 0 moles therefore 0 moles/L Then the Keq equation is:

Write down the equation and the start concentrations. Let
'x' be
the finish concentrations.
H_{2} + I_{2 }
<====> 2 HI

We will subtract 'x' from both H_{2} and I_{2}
because
this is the amount of reactant that will convert into product '2x'.

Now substitute these 'x' factors into the K_{eq }equations
and solve for 'x'.


50 =
(2x)^{2}
(4.0  x)(2.0  x) We're not so lucky this time. Multiply this out and get the
equation
into standard form.

50 = 4x^{2}
8.0  6.0x + x^{2} 400  300x + 50x^{2 }= 4x^{2} 46x^{2 } 300x + 400 = 0 (standard form) Use the quadratic equation to solve for 'x'.
In this case a = +46, b = 300 and c = +400.

The quadratic equation is a mathematical relationship for
solving
a line to find it's roots. In chemistry the 'roots' represent real life
values. One will be realistic and the other will not be.


4.7 is to large since we only started with 4 mols/L of H_{2
}to
begin with. It is therefore the unreal root. So let x = 1.9 mol/L.
Substitute 'x =1.9 mol/L' back up into the original concentration calculations to see how much is left after reaction.
H_{2} + I_{2 }
<======> 2 HI
4.01.9
2.01.9
2(1.9)
To finally answer the question:
moles = concentration * volume
There would be 1.9 moles of HI in the reaction vessel at
equilibrium.

Problem #7 
When 3.0 moles of HI, 20 moles of H_{2}, and 1.5
moles of
I_{2} are placed in a 1.0 L container at 448^{o}C, will
a reaction occur? If so, which reaction will take place?

Solution
Calculate the concentrations and substitute them into an expression and compute the "experimental concentration quotient" or "Q". 
If Q is equal to K_{eq} (50 in this case) then the
system
is already at equilibrium and stable.
If Q > K_{eq} then the [product] is too high and must decrease. Therefore the reverse reaction is favoured. If Q < K_{eq} then the [product] is still to small and must increase, therefore the forward reaction is favoured. H_{2} + I_{2 }<=======> 2 HI therefore Q =
(3.0 mol/L)^{2}
=
9.0 = 3
Therefore Q < K_{eq} since 3 < 50. The
forward reaction
is favoured to occur until the ratio of product:reactants = 50.

InClass Followup Problems 
1. When 0.040 moles of PCl_{5 }is
heated to
250^{o}C in a 1.0 L vessel, an equilibrium is established in
which
the equilibrium concentrations of Cl_{2 }is 0.025 mol/L. Find
the
equilibrium constant, K_{eq }, at 250^{o}C for the
reaction:
PCl_{5}(g) <======> PCl_{3}(g) + Cl_{2}(g) Note that the initial, not the equilibrium concentration for PCl_{3} is given. The amount of PCl_{5 }reacted is the same as the amount of Cl_{2} formed. 2. Assume that the analysis of another equilibrium mixture of the system from above shows that the equilibrium concentration of PCl_{5} is 0.012 mol/L and that of Cl_{2 }is 0.049 mol/L. What is the equilibrium concentration of PCl_{3} at 250^{o}C? 3. How many moles of PCl_{5} must be heated in a 1.0 L flask at 250^{o}C in order to produce enough chlorine to give an equilibrium concentration of 0.10 mol/L? 4. Will there be a net reaction when 2.5 moles of PCl_{5}, 0.60 mole of Cl_{2}, and 0.60 mole of PCl_{3} are placed in a 1.0 L flask and heated to 250^{o}C. If so, which reaction takes place? 