# If a reaction of sulfur dioxide with dioxygen gas to form sulfur trioxide starts with "5 mols SO"_2 and "5 mols O"_2, and 60% of the "SO"_2 was consumed in the reaction, what is the total mols of gas at equilibrium?

Aug 24, 2017

$\text{8.5 mols of gas}$

Write the reaction:

$2 {\text{SO"_2(g) + "O"_2(g) -> 2"SO}}_{3} \left(g\right)$

For this reaction, you know that the starting mols are:

n_("SO"_2,i) = "5 mols SO"_2

n_("O"_2,i) = "5 mols O"_2

We construct the ICE table to show the changes in mols at constant temperature and total pressure:

$2 {\text{SO"_2(g) + "O"_2(g) -> 2"SO}}_{3} \left(g\right)$

$\text{I"" "5" "" "" "" "5" "" "" } 0$
$\text{C"" "-2x" "" "-x" "" } + 2 x$
$\text{E"" "5 - 2x" "5-x" "" } 2 x$

Remember to include the stoichiometric coefficients in the change in concentration, i.e. $- 2 x$ for a reactant written with a coefficient of $2$, $+ x$ for a product with a coefficient of $1$, etc.

In this case, we know that 60% of the ${\text{SO}}_{2}$ was consumed, so the fraction of dissociation $\alpha$ is given as $\alpha = 1 - 0.60$:

$5 - 2 x = \alpha \times 5$

$= \left(1 - 0.60\right) \left(5\right) = 2$

$5 - 2 = 2 x$

$\implies x = \text{1.5 mols gas}$

Therefore, at equilibrium, the mols of gas in the vessel are given by:

$\textcolor{b l u e}{{n}_{e q , t o t}} = \left(5 - 2 x\right) + \left(5 - x\right) + \left(2 x\right)$

$= 10 - x$

$= 10 - 1.5$

$=$ $\textcolor{b l u e}{\text{8.5 mols gas total}}$