Calculation of Keq 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 Keq 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 PCl5 is heated in a 10.0 L container, an equilibrium is established is which 0.25 moles of Cl2 is present.
PCl5(g) <====>  PCl3(g) + Cl2(g)

a) What is the number of moles of PCl5 and PCl3 at equilibrium?
b) What are the equilibrium concentration of all three components?

                PCl5   <=====>       PCl3     +      Cl2
start   0.40 moles                    ------          -------
after   0.15 moles               0.25 moles    0.25 moles

From the balanced equation and the coefficients it should be obvious that everything works on a one-to-one basis. So there should be 0.25 moles of PCl3 and 0.25 moles of PCl5 should have been consumed, so there are 0.40-0.25 moles = 0.15 moles of PCl5 left.
Therefore [PCl5] = 0.15 moles = 0.015 moles/L
                                 10.0 L
[PCl3] and [Cl2] both are 0.25 moles each. Therefore their concentrations are identical as well.
[PCl3] = [Cl2] = 0.25 moles = 0.025 moles/L
                             10.0 L
Problem #2
When 1.00 mole of NH3 gas and 0.40 moles of N2 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 NH3(g)  <======>   3 H2(g) + N2(g)

a) What are the number of moles of H2 and N2 at equilibrium?

b) What is the concentration in moles/L of each species?

Problem #3
A mixture of H2 and I2 is allowed to react st 448oC. When equilibrium is established, the concentrations of the participants are found to be [H2] = 0.46 mol/L, [I2] = 0.39 mol/L, and [HI] = 3.0 mol/L. Calculate the value of the Keq at 448oC.             H2(g) + I2(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 448oC, the equilibrium concentrations of I2 and H2 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 448oC

       H2(g) + I2(g) <=====>  2 HI(g)

a) How many moles of HI are present at equilibrium when 1.0 moles of H2 is mixed with 1.0 moles of I2 in a 0.50 L container and allowed to react at 448oC?

b) How many moles of H2 and I2 are left unreacted?

c) If the conversion of H2 and I2 to HI is essentially complete, how many moles of HI would be present?

d) What is the percent yield of the equilibrium mixture?

a) First write the Keq equation based on the balanced reaction. You should always make sure that the reaction is balanced first.

The concentrations at the start are:

[H2] = 1.0 mol/0.50 L = 2.0 mol/L

[I2] = 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.

                 H2     +     I2      <=====> 2 HI
  start    2.0 M      2.0 M                    0 M

Let 'x' be the number of moles of H2 per litre consumed.

finish    2.0 - x     2.0 - x                      2x

Stop and confirm the above. If 'x' amount of H2 is consumed then 'x' amount of I2 will also be consumed because the H2 and I2 react on a 1:1 basis. Also if 'x' amount of H2 (or for that matter I2) 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 Keq 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?

[H2] = 2.0 mols/ 1 - x = 2.0 mols/L - 1.56 mols/L = 0.44 mols/L

[I2] = 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 H2 and I2 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

[H2] = [I2] = 0.5 L * 0.44 mols/L = 0.22 moles

Therefore 0.22 moles of H2 and 0.22 moles of I2 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 H2 to react with I2. Theoretically if both the H2 and I2 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
                               Theoretical Yield

1.56 moles/L * 100
    2.00 moles/L

= 78% yield (note that the moles/L units cancel)


Percentage Yield = 3.12 moles * 100
                              4.00 moles

= 78% yield (note that the units still cancel)

Problem #6
How many moles of HI are present at equilibrium when 2.0 moles of H2 is mixed with 1.0 moles of I2 in a 0.50 L container and allowed to react at 448oC. At this temperature Keq = 50.
First the equation: H2 + I2 <=====>  2 HI

Then the concentrations:

[H2] = 2.0 moles/0.50 L = 4.0 moles/L

[I2] = 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.

               H2    +      I2   <====> 2 HI
start    4.0 M        2.0 M             0 M
finish   4.0-x          2.0-x               2x

We will subtract 'x' from both H2 and I2 because this is the amount of reactant that will convert into product '2x'.
Now substitute these 'x' factors into the Keq 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 =        4x2
         8.0 - 6.0x + x2

400 - 300x + 50x2 = 4x2

46x2 - 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 H2 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.

                H2    +      I  <======>  2  HI
start     4.0 M      2.0 M                      0 M
finish    4.0-x       2.0-x                         2x

          4.0-1.9    2.0-1.9                    2(1.9)
            2.1 M     0.1 M                     3.8 M

To finally answer the question:
The moles of HI present at equilibrium would be:

moles = concentration * volume
         = 3.8 moles/L * 0.5 L
         = 1.9 moles

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 H2, and 1.5 moles of I2 are placed in a 1.0 L container at 448oC, will a reaction occur?   If so, which reaction will take place?
Calculate the concentrations and substitute them into an expression and compute the "experimental concentration quotient" or "Q".
If Q is equal to Keq (50 in this case) then the system is already at equilibrium and stable.

If Q > Keq then the [product] is too high and must decrease. Therefore the reverse reaction is favoured.

If Q < Keq then the [product] is still to small and must increase, therefore the forward reaction is favoured.

            H2 + I2 <=======>   2 HI

therefore Q =          (3.0 mol/L)2
                      (2.0 mol/L)(1.5 mol/L)

                  = 9.0 = 3

Therefore Q < Keq since 3 < 50. The forward reaction is favoured to occur until the ratio of product:reactants = 50.

In-Class Follow-up Problems
1.   When 0.040 moles of PCl5 is heated to 250oC in a 1.0 L vessel, an equilibrium is established in which the equilibrium concentrations of Cl2 is 0.025 mol/L. Find the equilibrium constant, Keq , at 250oC for the reaction:

PCl5(g) <======> PCl3(g) + Cl2(g)

Note that the initial, not the equilibrium concentration for PCl3 is given. The amount of PCl5 reacted is the same as the amount of Cl2 formed.

2. Assume that the analysis of another equilibrium mixture of the system from above shows that the equilibrium concentration of PCl5 is 0.012 mol/L and that of Cl2 is 0.049 mol/L. What is the equilibrium concentration of PCl3 at 250oC?

3. How many moles of PCl5 must be heated in a 1.0 L flask at 250oC 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 PCl5, 0.60 mole of Cl2, and 0.60 mole of PCl3 are placed in a 1.0 L flask and heated to 250oC. If so, which reaction takes place?

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