Question

Given the decomposition reaction:2SI3(g)->2SO2(g) + O2(g)According to Le Châtelier’s principle, what will happen when the volume of the container is increased for the chemical reaction that had reached equilibrium? A)Increasing volume, increases pressure and favors the products. B) Increasing volume, decreases pressure and favors the products.C)Increasing volume, increases pressure and favors the reactants.D) Increasing volume, decreases pressure and favors the reactants.

Answer 1

To analyze the equilibrium and how it shifts aaccording to Le Châtelier’s principle, we have to see how the system reacts to the change.

The change is an increase in the volume of the container. Since all the compounds on equilibrium are in gaseous state, their volume is the same as its container, so an increase in the volume of the container increase the volume of the compounds.

The reactant is SI₃, but we have 2 of them for each reaction.

The products are 2 molecules of SO₂ and one molecule of O₂.

In total for each reaction, we have 2 molecules in the reactant part and 3 molecules in the product part.

Since they are all in gaseous form, this means that the products occupy more space than the reactants, that is, an increase in the volume will favor the products, because this increase will left more space for them to occupy.

Thinking in preassure, an increase in the volume will decrease pressure, because, by the Boyle's Law, they are inversely proportional (assuming ideal gas). Since there are more molecules per reaction on the products side, this will favor the products, since more molecules make more pressure and now it has been decreased.

Thus:

Increase in volume -> decrease in pressure -> favors the products.

This matches alternative B.