A chemical equilibrium between gaseous reactants and products is shown.

N2(g) + 3H2(g) ⇌ 2NH3(g)

How will the reaction be affected if the pressure on the system is decreased? (3 points)

It will shift toward the reactant side as there is lower pressure on the reactant side.

It will shift toward the product side as there is higher pressure on the product side.

It will shift toward the reactant side as there are a greater number of moles of gas on the reactant side.

It will shift toward the product side as there are a fewer number of moles of gas on the product side.

Explanation:

In the chemical equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g), two molecules of N₂(g) react with 3 molecules of H₂(g) to form 2 molecules of NH₃(g).

Hence, in the reactant (left) side there are a total of 4 molecules reacting, while in the product (right) side there are a total of 2 molecules produced.

Following Le Chatelier's principle, an increase of the pressure on the system must be followed by a shift  of the equilibrium to counteract that disturbance.

Hence, if the pressure on the system is decreased, to oppose this distrubance, the reaction will shift in the direction that increases the pressure, and that is to the side that more molecules are produced (because the pressure is proportional to the number of molecules).

As stated, the stoichiometry shows that there are more molecules on the left side. Then, you conclude that the reaction will shift toward the reactant side as there are a greater number of moles of gas on the reactant side.

just wanted to confirm that it is the third statement

Explanation:

just took the quiz :)

the answer to your question is c