Question

Consider the following reversible reaction:

A(g) + 3B(g)  ⇌ AB3(g)   \DeltaΔH = -245 kJ/mol

On your Calcs sheet, please do the following:

a) write the expression for the equilibrium constant for this reaction.

b) discuss what effect (if any) would EACH of the following procedures have on the equilibrium, and explain the reason(s.)

In here, choose the option that would cause a change in the equilibrium constant.

Increasing the volume of the reaction container.

Increasing the concentration of product AB₃  

Increasing the concentration of reactant B

Increasing the temperature.

Adding a catalyst.

Answer 1

Solution :

For the equilibrium reaction,

A(g) + 3B(g) == AB3(g) , ΔH = -245 kJ/mol

Part A) The equilibrium expression of the given reaction is written as,

Keq = [AB3] / [A] [B]^3

Part B)

I)  Increasing the volume of container

>> Increasing the volume causes a decrease in pressure. According to Le-Chatelier principle, decrease in pressure causes an equilibrium to shift in that direction where number of moles are maximum. Hence, equilibrium shifted in reactants direction. Thus, equilibrium constant decreases.

II) Increasing the concentration of AB3:

Increasing the concentration of product shifted equilibrium towards reactant direction, hence equilibrium constant will decrease.

III) Increasing the concentration of B:

Increasing the concentration of reactant shifted equilibrium towards forward direction, hence equilibrium constant will increase.

IV) Increasing the temperature :

The given reaction is an example of exothermic reaction, hence increase in temperature shifted equilibrium towards reactant side, hence equilibrium constant will decrease.

V) Addition of catalyst :

Addition of catalyst does not affect the equilibrium constant because it increases the rate of both forward and reverse reactions.