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?

Question

Get Advice on Live Video Call
Earn $ Cash $ with
consultations on Bissoy App

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?

Share with your friends

Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

$"8.5 mols of gas"$

Write the reaction:

$2"SO"_2(g) + "O"_2(g) -> 2"SO"_3(g)$

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:

$2"SO"_2(g) + "O"_2(g) -> 2"SO"_3(g)$

$"I"" "5" "" "" "" "5" "" "" "0$
$"C"" "-2x" "" "-x" "" "+2x$
$"E"" "5 - 2x" "5-x" "" "2x$

Remember to include the stoichiometric coefficients in the change in concentration, i.e. $-2x$ 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 $"SO"_2$ was consumed, so the fraction of dissociation $alpha$ is given as $alpha = 1 - 0.60$:

$5 - 2x = alpha xx 5$

$= (1 - 0.60)(5) = 2$

$5 - 2 = 2x$

$=> x = "1.5 mols gas"$

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

$color(blue)(n_(eq,t ot)) = (5 - 2x) + (5 - x) + (2x)$

$= 10 - x$

$= 10 - 1.5$

$=$ $color(blue)("8.5 mols gas total")$

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?

1 Answers