Question

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)

Answer 1

Answer:

• Rightwardshift: (a), (b), (f) and (h)

• Leftwardshift: (c), (d), and (e)

• No shift: (g)

Explanation:

1. Balanced chemical equation (given):

      $2A(aq)+2B(aq)\rightleftharpoons 5C(aq)$

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.

          $K_{c}=\frac{[C(aq)]^5}{[A(aq)]^2\cdot [B(aq)]^2}$

a. Increase [B]

• Rightward shift

Since, by assumption, the temperature of the reaction is the same, the equilibrium constant   $K_{c}$ 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]

• Rightward shift.

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]

• Leftward shift.

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  $K_{c}$ : by increasing the denominator the numerator has to increase to keep the same value of  $K_{c}$

d. Decrease [A]

• Leftward shift.

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

e. Decrease[B]

• Leftward shift.

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

f. Decrease [C]

• Rightward shift.

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

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

• No shift

You need to perform some calculations and determine the reaction coefficient, $Q_c$ to compare with the equilibrium constant $K_{c}$.

The expression for $Q_c$ has the same form of the equation for  $K_{c}$. but the it uses the inital concentrations instead of the equilibrium concentrations.

            $Q{c}=\frac{[C(aq)]^5}{[A(aq)]^2\cdot [B(aq)]^2}$

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

When  $Q_c$  = $K_{c}$ the reaction is at equilibrium, so no shift will occur.

h. Double both [B] and [C]

• Rightward shift.

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

When   $Q_c$  > $K_{c}$ 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:

• Rightwardshift: (a), (b), (f) and (h)

• Leftwardshift: (c), (d), and (e)

• No shift: (g)