# Question #81435

Mar 1, 2017

The forward reaction will be favored.

#### Explanation:

Equilibrium reactions are governed by Le Chatelier's Principle, which states that a system at equilibrium will respond to a stress on the current position of the equilibrium in such a way as to counteract that stress and reestablish the equilibrium.

$2 {\text{SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"SO}}_{3 \left(g\right)}$

Now, when you increase the amount of oxygen, you're placing a stress on the equilibrium.

In order to counteract this stress, the equilibrium will shift in such a way as to decrease the amount of oxygen gas.

This implies that the equilibrium shift to the right. As a result, the forward reaction will be favored, i.e. sulfur dioxide, ${\text{SO}}_{2}$, and oxygen gas, ${\text{O}}_{2}$, will be consumed and sulfur trioxide, ${\text{SO}}_{3}$, will be produced.

$2 {\text{SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"SO}}_{3 \left(g\right)}$

$\textcolor{w h i t e}{a a} \stackrel{\textcolor{red}{\to}}{\textcolor{w h i t e}{a a \textcolor{b l u e}{\text{shift to the right}} a a a a}}$

Now, another important thing to notice here is that when the equilibrium shifts to the right, the pressure in the reaction vessel decreases.

This is the case because you have

$\text{2 moles SO"_2 + "1 mole O"_2 = "3 moles gas}$

on the reactants' side and

${\text{2 moles SO}}_{3}$

on the products' side. Since pressure is caused by the collisions that take place between the gas molecules and the walls of the reaction vessel, a decrease in the total number of moles of gas present will cause the pressure to decrease as well.