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)

The correct answers are:

Explanation:

Following Le Chatelier's principle, when a chemical equilibrium is disturbed by an external force, the reaction will shift toward the side that permits to overcome or minimize the disturbance and comeback to a new equilibrium.

Since, the pressure is inversely related to the volume (the greater the volume the lower the pressure and viceversa), ff the volume of a reaction mixture at equilibrium increases by a factor of 2, the pressure of the system will decrease.

Since, the pressure is directly related to the number of particles (atoms,or molecules), you can predict that the reaction will shift to the side where more molecules are produced to overcome the decrease in the pressure generated by the increase on the volume.

Let's see each option.

a. 2 SO₂(g)+ O₂(g) ⇄ 2 SO3(g)

Incorrect.

There are more molecules on the left side (reactant side) so the reverse reaction is favored. This is the equilibrium will shift to the formation of more reactants. So, this is an incorrect option.

b. NO(g) + O₃(g) ⇄ NO₂(g) + O₂(g)

Incorrect.

There are two molecules in the reactant side and two molecules in the product side. Hence, in this equilibrium the increase of volume will not yield a change in the equlibrium.

c. 2N₂O₅(g) ⇄ 4 NO₂(g)+ O₂(g)

Correct.

In this reaction, there are four molecules in the product side (right) against 2 molecules in the reactant side (left). So, the increase in volume 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 a shift toward the formation of more product to overcome the increase of volume.