# Question 32a47

Apr 11, 2016

Increasing the pressure.

#### Explanation:

You're dealing with a gaseous equilibrium that features sulfur dioxide, ${\text{SO}}_{2}$, and oxygen gas, ${\text{O}}_{2}$, as the reactants and sulfur trioxide, ${\text{SO}}_{3}$, as the product for the forward reaction.

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

Moreover, notice that the forward reaction is exothermic, i.e. it gives off heat.

This tells you that you can actually treat heat as a product of the forward reaction and as a reactant for the reverse reaction.

Now, equilibrium reactions are governed by Le Chatelier's Principle, which states that any stress placed on the position of the equilibrium will cause the equilibrium to shift in such as way as to reduce that stress.

$1. \textcolor{b l u e}{\underline{\text{Increasing the temperature}}}$

The forward reaction is exothermic, which of course implies that the reverse reaction is endothermic. Heat can be treated as a product in the forward reaction, which means that increasing the temperature will cause the equilibrium to shift to the left, since the forward reaction "consumes" the excess heat.

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

color(white)(aaaaaa)stackrel(color(green)(larr))(color(white)(aaaa)color(red)("Shift to the left")color(white)(aaaaa)

The equilibrium will thus remove the added heat by favoring the reaction that requires heat, i.e. the reverse reaction.

$2. \textcolor{b l u e}{\underline{\text{Increasing the pressure}}}$

When pressure is increased, the equilibrium will shift in such a way as to reduce this stress on its position.

As you know, gas pressure is caused by the collisions that take place between the molecules of gas and the walls of the reaction vessel.

Since volume is being kept constant, the only way to decrease pressure is by reducing the number of molecules of gas present in the reaction vessel.

This means that the reaction that produces the fewest number of moles of gas particles will be favored. In this case, you have

{: ("2 moles SO"_2), ("1 mole O"_2) :}} -> three moles of gas on the reactants' side

${\text{2 moles SO}}_{3} \to$ two moles of gas on the products' side

Since the product side of the forward reaction contains fewer moles of gas, the equilibrium will shift to the right.

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

color(white)(aaaaaa)stackrel(color(green)(->))(color(white)(aaaa)color(red)("Shift to the right")color(white)(aaaaa)

$3. \textcolor{b l u e}{\underline{{\text{Decreasing the amount of SO}}_{2}}}$

Sulfur dioxide is a reactant in the forward reaction, so decreasing its concentration will cause the equilibrium to shift in such a way as to increase the amount of sulfur dioxide present.

This of course implies that the reverse reaction, which consumes sulfur trioxide and produces sulfur dioxide and oxygen, will be favored.

As a result, the equilibrium will shift to the left.

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

color(white)(aaaaaa)stackrel(color(green)(larr))(color(white)(aaaa)color(red)("Shift to the left")color(white)(aaaaa)

$4. \textcolor{b l u e}{\underline{{\text{Decreasing the amount of O}}_{2}}}$

The exact same thing will happen when you decrease the concentration of oxygen gas.

In order to increase the amount of oxygen gas present, the equilibrium will once again favor the reverse reaction, i.e. a shift to the left occurs.

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

color(white)(aaaaaa)stackrel(color(green)(larr))(color(white)(aaaa)color(red)("Shift to the left")color(white)(aaaaa)#