Consider the following system at equilibrium: 2A(aq)+2B(aq)⇌5C(aq) Classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the net reaction. a. increase (b)
b. increase(a)
c. increase(c)
d. decrease(a)
e. decrease(b)
f. decrease(c)
g. double(a) and reduce (b) to one half
h. double both (b) and (c)

Explanation:

1. Balanced chemical equation (given):

2. Equilibrium constant

The equilibrium constant is the ratio of the product of the concentrations of the products, at equilibrium, each raised to its stoichiometric coefficient, to the product of the concentrations of the reactants, at the equilibrium, each raised to its stoichiometric coefficient.

a. Increase [B]

Since, by assumption, the temperature of the reaction is the same, the equilibrium constant   is the same, meaning that an increase in the concentration of the species B must cause a rightward shift to increase the concentration of the species C, such that the ratio expressed by the equilibrium constant remains unchanged.

b. Increase [A]

This is exactly the same case for the increase of [B], since it is in the same side of the equilibrium chemical equation.

c. Increase [C]

C is on the right side of the equilibrium equation, thus, following Le Chatelier's principle, an increase of its concentration must shift the reaction to the left to restore the equilibrium. Of course, same conclusion is drawn by analyzing the expression for   : by increasing the denominator the numerator has to increase to keep the same value of  

d. Decrease [A]

This is the opposite change to the case {b), thus it will cause the opposite effect.

e. Decrease[B]

This is the opposite to case (a), thus it will cause the opposite change.

f. Decrease [C]

This is the opposite to case (c), thus it will cause the opposite change.

g. Double [A] and reduce [B] to one half

You need to perform some calculations and determine the reaction coefficient, to compare with the equilibrium constant .

The expression for has the same form of the equation for  . but the it uses the inital concentrations instead of the equilibrium concentrations.

Doubling [A] and reducing  [B] to one half would leave the product of [A]² by [B]² unchanged, thus  will be equal to .

When    = the reaction is at equilibrium, so no shift will occur.

h. Double both [B] and [C]

Again, using the expression for , you will realize that the [C] is raised to the fifth power (5) while [B] is squared (power 2). That means that will be greater than  ..

When    > the equilibrium must be displaced to the left some of the reactants will need to become products, causing the reaction to shift to the right.

Summarizing:

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