Humboldt State University ® Department of Chemistry

Richard A. Paselk
Chem 107

Fundamentals of Chemistry

Fall 2009
Lecture Notes: Equilibrium Examples

© R. Paselk 2005
  
 

Worked Examples
  

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Example: Consider the reaction

CO + NO2 NO + CO2 + heat (226 kJ)

Note that heat appears on the product side - the system is giving up heat, \ DH is negative, DH = - 226 kJ/mol

So, what will happen to [CO2] if:

    • CO is added?
      • [CO2] will increase. The increased [CO] will react more rapidly with NO2, driving the reaction to the right.
    • NO2 is added?
      • [CO2] will increase. The increased [NO2] will react more rapidly with CO, driving the reaction to the right.
    • NO is added?
      • [CO2] will decrease. The increased [NO] will react with CO2, driving the reaction to the left.
    • T is increased?
      • [CO2] will decrease. If T increases heat must have been added, therefore the reaction will use up some of the extra heat, driving the reaction to the left.
    • V is increased?
      • No change in relative concentrations. Although a smaller volume will result in a higher pressure (more collisions) and a higher concentration, the concentrations of all species will increase in proportion. Since we have equal numbers of reactants on both sides, neither side will be favored.
    • Ar is added?
      • No change. Ar is unreactive and will have no effect.

 

 

 

Example: Consider the gas phase reaction:

PCl5 PCl3 + Cl2

Keq = 5.0 x 10-2 @ 150 °C

Find [Cl2] if [PCl5] = 0.40 M and [PCl3] = 0.20 M.

Know that Keq = [PCl3] [Cl2] / [PCl5]

Substituting, Keq = (0.20)[Cl2] / (.040) = 5.0 x 10-2

Rearranging, [Cl2] = (5.0 x 10-2)(0.40)/(0.20) = 1.0 x 10-1 M

[Cl2] = 0.10 M

 

 

 

 

 

Example: Consider the reaction of carbon monoxide and water to give carbon dioxide and hydrogen. Calculate the concentrations of all species at equilibrium if we start with 0.341 moles each of carbon monoxide and water in a 2.71 L container @ 600 K. Keq = 302.

  CO + H2O CO2 + H2
Before reaction 0.341/2.71 = 0.126 M   0.341/2.71 = 0.126 M   0 0
@ Equilibrium 0.126 - x   0.126 - x   x   x

So Keq = [CO2] [H2] / [CO] [H2O]

Substituting, K = (x)(x)/(0.126 - x)(0.126 - x) = 302

Taking the square roots of both sides: x/(0.126 - x) = 17.4

x + 17.4 x = (0.126)(17.4); and x = 0.119, so

[CO2] = [H2] = 0.119 M

[CO] = [H2O] = 0.126 - 0.119 = 0.007 M

 

 

 

 

Example: Consider the gas phase dissociation of carbonyl chloride to carbon monoxide and chlorine @ 100 °C.

If 0.20 moles of carbonyl chloride (COCl2) is placed in a 2.5 L container at 100 °C calculate the concentrations of all species at equilibrium. Keq = 2.6 x 10-10 @ 100 °C.

  COCl2 CO + Cl2
Before reaction 0.20/2.5 = 0.080   0 0
@ Equilibrium 0.080 - x   x   x

Keq = [CO] [Cl2] / [COCl2]

Keq = (x)(x)/(0.080 -x) = 2.6 x 10-10

Notice that this will give a quadratic equation - yuk! This is to be avoided if possible, if for no other reason than more steps means more opportunities to make a mistake! So we will take advantage of the fact that we are working with experimental work - there is always an error, and if x is smaller than our error, then we can simplify.

Thus the trick is to assume x<< 0.080 and if we look at K, its value (2.6 x 10-10) is small enough that it is likely that our assumptionis correct. So:

Keq = x2/ 0.080 = 2.6 x 10-10

x2 = (0.080)(2.6 x 10-10) = 2.08x 10-11

and x = 4.6 x 10-6

Note that x = 4.6 x 10-6<< 0.080, so our assumption is correct!

\ [CO] = [Cl2] = 4.6 x 10-6 M

[COCl2] = 0.080 M

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Last modified 4 December 2008