Question #81435

1 Answer
Mar 1, 2017

The forward reaction will be favored.


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.

In your case, you have

#2"SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"SO"_ (3(g))#

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, #"SO"_2#, and oxygen gas, #"O"_2#, will be consumed and sulfur trioxide, #"SO"_3#, will be produced.

#2"SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"SO"_ (3(g))#

#color(white)(aa)stackrel(color(red)(->))(color(white)(aacolor(blue)("shift to the right")aaaa))#

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

#"2 moles SO"_2 + "1 mole O"_2 = "3 moles gas"#

on the reactants' side and

#"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.