# Given the reaction at equilibrium 2SO_2(g) + O_2(g) rightleftharpoons 2SO_3(g), as the pressure is increased at a constant temperature, will the number of moles of SO_3(g) produced decrease, increase, or stay the same? Why?

Apr 5, 2017

The number of moles of $S {O}_{3} \left(g\right)$ produced will increase, as according to the comparison between pressure & moles via the Ideal Gas Law. (See explanation)

#### Explanation:

We start by finding the relationship between pressure and number of moles. Since temperature is constant, we can consider this:(derived from the Ideal Gas Law, $P V = n R T$:
${P}_{1} / {n}_{1} = {P}_{2} / {n}_{2}$, so pressure and moles are inversely proportional.

Since we increased the pressure, the number of moles of gas must decrease.

Now we look at the reaction:
On the reactants side, $2 S {O}_{2} \left(g\right) + {O}_{2} \left(g\right)$, there's a total of 3 moles of gaseous molecules.
On the products side, $2 S {O}_{3} \left(g\right)$, there's a total of 2 moles of gaseous molecules.

Since the number of moles of gaseous molecules must decrease, we would convert the reactants (3 moles of gaseous molecules) to products (2 moles of gaseous molecules). So, the number of moles of $S {O}_{3} \left(g\right)$ produced will increase.