Question

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?

Answer 1

`"8.5 mols of gas"`

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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")`