Which of the following equilibria will shift toward formation of more products if the volume of a reaction mixture at equilibrium increases by a factor of 2? a. 2 SO2(g)+ O2(g) 2 SO3(g) b. NO(g) + O3(g) NO2(g) + O2(g) c. 2N2O5(g) 4 NO2(g)+ O2(g) d. N2O4(g) 2 NO2(g)

As per Le Chatelier's principle, the reaction will shift toward the side that permits to overcome or minimize the force that disturbs the equilibrium.

With those concepts you can predict which equilibria will shift toward formation of more products if the volume of a reaction mixture at equilibrium increases by a factor of 2

Let's dig into each option.

a. 2 SO₂(g)+ O₂(g) ⇄ 2 SO3(g)

Incorrect.

There are three molecules in the reactant side and two in the product side. so the increase in volume will favor the reverse reaction. This is, the equilibrium will shift to the formation of more reactants, and this is an incorrect choice.

b. NO(g) + O₃(g) ⇄ NO₂(g) + O₂(g)

Incorrect.

There are the same number of molecules in the reactant side and the product side. Hence, the increase of volume will not produce a change in the equlibrium.

c. 2N₂O₅(g) ⇄ 4 NO₂(g)+ O₂(g)

Correct.

There are 2 molecules in the reactant side (left) and four molecules in the product side (right). So, the increase in volume in this system will produce a shift toward the product side.

d. N₂O₄(g) ⇄ 2 NO₂(g)

Correct.

There are more molecules in the product side than in the reactant side, so you predict that the equilibrium will shift toward the formation of more product to overcome the increase of volume.

aliasger2709 aliasger2709 · Answer 2 · 2022-03-02T06:27:18+01:00

Equilibrium shift is the shift of the reaction towards the stressed conditions in the reactions. The shift towards the formation of the products will occur in [tex]\rm 2N_{2}O_{5}(g) \rightleftharpoons 4 NO_{2}(g)+ O_{2}(g)[/tex] and [tex]\rm N_{2}O_{4}(g) \rightleftharpoons 2 NO_{2}(g)[/tex].

What is the equilibrium shift?

According to the Le principle the reaction shift towards the site or the reactants and products where the disturbance or the stress is present so that it can be overcome.

In the third equation reaction, [tex]\rm 2N_{2}O_{5}(g) \rightleftharpoons 4 NO_{2}(g)+ O_{2}(g)[/tex] the number of the molecules on the left side is 2 and on the right side of the product have 4 molecules. When the volume is increased then the reaction shifts towards the right or the product side.

In the fourth reaction, [tex]\rm N_{2}O_{4}(g) \rightleftharpoons 2 NO_{2}(g)[/tex] the number of the products are more compared to the reactants and hence the increase in the volume will shift the reaction towards the formation of the product.

Therefore, option c. [tex]\rm 2N_{2}O_{5}(g) \rightleftharpoons 4 NO_{2}(g)+ O_{2}(g)[/tex] and option d. [tex]\rm N_{2}O_{4}(g) \rightleftharpoons 2 NO_{2}(g)[/tex] are the reactions that shift towards product formation.

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