Question cfc3a

May 28, 2017

["O"_2] = "0.5 mol dm"^(-3)

Explanation:

For starters, you know that

${K}_{c} > 1$

This tells you that the equilibrium lies to the right, meaning that the forward reaction is favored. This implies that, at equilibrium, the reaction vessel will contain more products than reactants.

So even without doing any calculations, you can expect to find

["O"_2] < "3 mol dm"^(-3)

Now, the equilibrium reaction looks like this

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

Sulfur takes part in the reaction as a solid, which means that it will not be included in the expression of the equilibrium constant--remember, we always assume that the concentration of a solid is constant.

The equilibrium constant that describes this reaction will look like this

${K}_{c} = \left(\left[{\text{SO"_2])/(["O}}_{2}\right]\right)$

Rearrange to solve for the equilibrium concentration of oxygen gas

["O"_2] = (["SO"_2])/K_c

Plug in your values to find

["O"_2] = "3 mol dm"^(-3)/6 = color(darkgreen)(ul(color(black)("0.5 mol dm"^(-3))))#

The answer is rounded to one significant figure.

As predicted, the equilibrium concentration of oxygen gas came out to be $<$ than the equilibrium concentration of sulfur dioxide.