Equilibrium: Calculations with the equilibrium constant, Kc
 
Calculating Percent Reaction 

Percent reaction is based on the idea of product yield.
Where

 percent yield =  x 100 %
than:

Percent reaction =  x 100 %
A high percent reaction indicates that lots of products are formed while a low percent reaction indicates that few products are formed.

Hydrogen and iodine combine to form hydrogen iodide according to the following reaction:

H2 (g) + I2 (g)  2 HI (g)

If 0.100 mol of both H2  and I2 were placed in a 1.00 L container and allowed to come to equilibrium, and it was found that the equilibrium concentration of HI was 4.56 x 10-3 mol / L, what is the percent reaction?

Percent reaction =  x 100 %
[HI]eqm = 4.56 x 10-3 mol / L

[HI]theor comes from stoichiometry

Since there are equal amounts and they combine in a 1:1 ratio there is no need to worry about limiting reagent conditions.

[HI]theor = moles H2 x mole ratio   =  0.100 mol H2 x 2 mol HI  x     1       = 0.200 mol HI / L
		volume                                                 1 mol H2     1.00 L

Percent reaction = [HI]eqm    x 100 % =    4.56 x 10-3 mol / L  x 100 %   = 2.28 %
                               [HI]theor                          0.200 mol / L  

Calculating Kc from equilibrium Concentrations

If the concentration of the various species is known at equilibrium, the equilibrium constant can be calculated.  Equilibrium concentrations can be either given directly, calculated from percent reaction, or from determining the change and applying it in an ICE chart. 

At equilibrium, the concentration of N2 was 0.105 mol / L, H2 was 0.230 mol / L, and NH3 was 0.011 mol / L at a certain temperature. Calculate the Kc.

N2 (g) + 3 H2 (g)     2 NH3 (g)

Kc =      [NH3]2  =         (0.011)2         = 0.0947
          [N2][H2]3     (0.105)(0.230)3

Calculation Equilibrium concentration of one species form Kc

In this case, the initial concentration of one species is given as is the equilibrium concentration of at least one species in the equilibrium expression. They do not have to be the same, as an ICE chart will be used to determine both the change and the equilibrium concentrations of the other entities.

Chlorine gas will react with ammonia according to the following reaction:
3 Cl2 (g) + NH3 (g)    NCl3 (g) + HCl (g)
0.300 mol of Cl2 and 0.200 mol of NH3 were placed into a 1.00 L container and allowed to come to equilibrium. At equilibrium it was found that the concentration of HCl was 0.0214 mol/L. Using this information determine the Kc for this reaction.

We can use an ICE chart to determine the change in reactants and products and also their equilibrium concentrations.

[Cl2]initial = 0.300 mool / L		[NH3]initial = 0.200 mol / L	[NCl3]initial = [HCl}initial = 0 mol / L

Some of NH3 will be consumed, as it is a reactant, and we can designate the number of moles of NH3 lost as x. In addition, some Cl2 has to be lost as well and since for every mol o NH3 consumed 3 moles of CL2 will be consumed, we can designate the loss of Cl2 as 3x, where x is the minimum number of moles that will change in the expression. Since there was no NCl3 or HCl initially, we will gain some. We can designate the gain for both of them as x since for every one mol of NH3 consumed 1 mole of each of them will be produced.

	3 Cl2 (g) +	NH3 (g)		NCl3 (g) +	HCl (g)
	
I	0.300	0.200		0	0	
C	– 3x	– x		+x	+x	
E	0.300 – 3x	0.200 - x		x	x	

Since we know that the equilibrium concentration of HCl is 0.0214 mol / L, it tells us what x is
[HCl]eqm = x = 0.0214

Therefore we can determine the equilibrium concentrations of the other species.

[NCl3] = x = 0.0214 mol/L	   
[NH3] = 0.200 – x = 0.200 – 0.0214 = 0.1786 mol/L 	
[Cl2] = 0.300 – 3x = 0.300 – 3(0.0214) = 0.2358 mol/L

Now that the equilibrium concentrations are know, you can calculate Kc

Kc = [NCl3][HCl] =  (0.0214) (0.0214) = 0.196
        [Cl2]3[NH3]       (0.2358)3(0.1786)

Calculating Equilibrium Concentrations from Initial Concentration and Kc

If the initial concentration of one of the species is known, can use an ‘ICE’ chart to solve these problems. This is where we   identify the Initial concentration, the Change in concentration, and the equilibrium concentration. Since the reaction occurs in the same vessel (volume) can use mole ratios as the change – represented with a variable x.

A 2.00 L container had 0.200 mol of phosphorus pentachloride, PCl5, added to it and the system was allowed to come to equilibrium. If the equilibrium constant Kc at a particular temperature is 0.2016, determine the equilibrium concentration of all of the species. 

PCl5(g)   PCl3(g) + Cl2(g)
Answer
The initial concentration of PCl5 is known as we are told the amount of gas and the volume of the container.
[PCl5]initial = 0.200 mol / 2.00 L = 0.100 mol / L
Since there was no PCl3 or Cl2 initial placed into the contained, their initial concentrations are 0
[PCl3]initial = [Cl2]initial = 0 mol / L

Some of PCl5 has to be lost we designate any lose as x. Since there was no PCl3 or Cl2 initially, we will gain some. We can designate the gain for both of them as x as well since the three compounds are in a 1:1:1 ratio.

	PCl5 (g)		PCl3 (g)	Cl2 (g)
	
I	0.100		0	0	
C	-x		+x	+x	
E	0.100 - x		x	x	

Now that the E line in the ICE chart is known, can solve using the Kc value and the equilibrium expression.

Kc =   [PCl3][Cl2]   =    (x)   (x)      = 0.2016     solve  
            [PCl5]             0.100 – x

Isolate and solve for x

x2 = (0.100 – x) (0.2016)
x2 = 0.02016 – 0.2016x
0 = x2 + 0.2016x – 0.02016

have to use quadratic to solve

sub x into the E line of the ice chart

[PCl5]eqm = 0.100 – x = 0.100 – 0.0733 = 0.0267 mol / L

[PCl3]eqm = [Cl2]eqm = 0.0733 mol / L

x = 0.0733 or x = -0.275

cannot have a negative concentration change, therefore x = 0.0733

Calculating Equilibrium Concentrations from Initial Concentration and Kc: 
      Perfect Squares

Sometimes a root can be taken to solve for the equilibrium concentrations rather than having to use the quadratic equation.

[CO]eqm = [H2O]eqm = 0.100 – x = 0.100 – 0.0668 = 0.033 mol / L

[CO2]eqm = [H2]eqm = 0.0668 mol / L

The reaction
CO (g) + H2O (g)  CO2 (g) + H2 (g)
has a Kc = 4.06 at 500oC. If 0.100 mol of CO and 0.100 mol of H2 are placed into a 1.oo L container, and allowed to come to equilibrium, what would the concentration of each of the species be?

Can use an ICE chart
   [CO]initial = [H2O]initial = 0.100 mol / L
   [CO2]initial = [H2] initial = 0 mol / L

	CO (g) +	H2O (g)		CO2 (g) +	H2 (g)
	
I	0.100	0.100		0	0	
C	-x	-x		+x	+x	
E	0.100 – x	0.100 – x		x	x	

Kc = [CO2][H2]   =              (x)(x)                 = 4.06   
        [CO][H2O]     (0.100 – x)(0.100 – x)

This expression can be rewritten as 

        x2        = 4.06
(0.100 – x)2

can take the square root of both sides

         x        =  = 2.01
(0.100 – x)

Isolate and solve for x

x = (0.100 – x)(2.01) = 0.201 – 2.01x
3.01x = 0.201
x = 0.0668

Calculating Equilibrium Concentrations from Initial Concentration and Kc:
      when Kc is very small

As we have seen, if the Kc is really small than very little of the reactant will change into product. As a result we can say that there will be approximately no change in the concentration of the reactant, especially when we take into account significant figures. 

Check

Change in [N2O]  x 100 % < 5 %
    [N2O]initial

    x        x 100%   = 6.38 x 10-4    x 100 %
(0.05)                            0.05

1.28 %  <  5 %, therefore assumption is valid

[N2O]eqm = 0.0500 – x = 0.0500 – 0.000638 = 0.0494 mol / L 

[NO2]eqm = 0.100 – 2x = 0.100 – 2(0.000638) = 0.099 mol / L

[NO]eqm = 3x = 3 (6.38 x 10-4) = 1.91 x 10-3 mol / L 

Check
[N2O]  =   0.05         = 3571 > 500 
  Kc         1.4 x 10-5 

Assumption is safe to use

At 200oC, Kc = 1.4 x 10-5 for the reaction:
N2O (g) + 2 NO2 (g)  3 NO (g)
If 0.200 mol of N2O and 0.400 mol of NO2 are placed into a 4.00 L container, what would the NO concentration be at equilibrium if this equilibrium was allowed to establish?

[N2O]initial = 0.200 mol / 4.00 L = 0.0500 mol/L
[NO2]initial = 0.400 mol / 4.00 L = 0.100 mol/L

	N2O (g) +	2 NO2 (g)		3 NO (g)
	
I	0.0500	0.100		0	
C	-x	-2x		+3x	
E	0.0500 - x	0.100 – 2x		3x	

Since the kc for this system is small, it is expected that there will be little change to the [N2O] or [NO2] when the system reaches equilibrium. Therefore, [N2O]eqm  will be approximately [N2O]initial and [NO2]eqm will also be approximately [NO2]initial, this allows for the simplification of the ‘E’ line to:
 
	N2O (g) +	2 NO2 (g)		3 NO (g)
	
E	0.0500	0.100		3x	

To test that this assumption is valid, we conduct the 500 rule test where we take the initial concentration of the entity that we are assuming will not have a significant change and dividing it by the kc value. If this results in a number greater than 500 it assumption is probably valid and we are safe to us it in the calculation.

Kc =         [NO]3         =       (3x)3        =  1.4 x 10-5
          [N2O][NO2]2	  (0.05)(0.1)2 
 
27x3  = 7.00 x 10-9  

x3 = 2.59 x 10-10

x = 6.38 x 10-4

after solving for x, you have to check that the assumption is 
valid by making sure that less than 5 % of the reactant 
would change