Question

In which direction will the position of the equilibrium

$\mathbf{2}\mathit{H}\mathit{I}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{I}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

be shifted for each of the following changes?

a. H₂(g) is added.

b. l₂ is removed.

c. HI(g) is removed.

d. In a rigid reaction container, some Ar(g) is added.

e. The volume of the container is doubled.

f. The temperature is decreased (the reaction is exothermic).

Short answer

a. It will shift the reaction to the left to reestablish the equilibrium.

b. The reaction will shift the equilibrium to the right.

c. The equilibrium favor the formation of HI(g) when HI(g) is removed. Thus, the equilibrium will shift towards left side.

d. There will be no change when argon gas is added.

e. There will be no effect observed on equilibrium.

f. The reaction will shift the equilibrium to the right to reestablish equilibrium.

Step-by-step solution

Subpart a)

Adding more hydrogen gas will move the reaction to the left to consume some of the added hydrogen gas. Hence, this will shift the reaction to the left to reestablish the equilibrium.

Subpart b)

When iodine gas is removed, the reaction will shift the equilibrium to the right in order to generate more iodine gas. Hence, the reaction will shift the equilibrium to the right. Hence, the formation of HI(g) is favored.

Subpart c)

Hydrogen iodide is present in the reactant side of the equilibrium reaction and the removal of HI gas will shift the equilibrium towards left side. Hence, it will shift the reaction to the left.

Subpart d)

At constant volume, the addition of inert gas (argon gas) has no effect because the concentrations/partial pressures of all the species in the reaction are unchanged. Hence, there will be no change on the equilibrium.

Subpart e)

In the reaction, the number of reactant gas molecules is equal to the number of product gas molecules. So, doubling the volume will have no effect on the equilibrium. Hence, there will be no change on the equilibrium.

Subpart f)

As the temperature is decreased for the given exothermic reaction, the reaction will shift the equilibrium to the right. For this case, heat is a product. When the product is removed, the reaction will shift to the right. Hence, the reaction will shift the equilibrium to the right.