Question #a750a

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Question #a750a

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Answer 1 · 2016-06-20T22:02:02

Here's what's going on here.

Explanation:

The trick here is to realize that because you know the equilibrium concentration of sulfur trioxide, ${SO}_{3}$ $"SO"_3$ , you practically know the value of $x$ $x$ , so you don't need to use $x$ $x$ anymore.

In fact, we only use $x$ $x$ because we don't know the equilibrium concentration of one or more chemical species that take part in the reaction.

So, your equilibrium reaction looks like this

$2 {SO}_{2 (g)} + O_{2 (g)} ⇌ 2 {SO}_{3 (g)}$ $color(red)(2)"SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons color(red)(2)"SO"_ (3(g))$

Now, you know that the equilibrium concentration of ${SO}_{3}$ $"SO"_3$ is equal to $0.04 M$ $"0.04 M"$ .

Assuming that you started with $0 M$ $"0 M"$ of ${SO}_{3}$ $"SO"_3$ in the reaction vessel, you can say that the reaction produced twice as many moles of ${SO}_{3}$ $"SO"_3$ than moles of $O_{2}$ $"O"_2$ consumed $\to$ $->$ $O_{2}$ $"O"_2$ and ${SO}_{3}$ $"SO"_3$ are in a $1 : 2$ $1:color(red)(2)$ mole ratio.

So if the reaction produced $0.04 M$ $"0.04 M"$ of ${SO}_{3}$ $"SO"_3$ , it follows that it must have consumed $0.02 M$ $"0.02 M"$ of $O_{2}$ $"O"_2$ .

Likewise, the reaction produced $0.04 M$ $"0.04 M"$ of ${SO}_{3}$ $"SO"_3$ , so it must have consumed $0.04 M$ $"0.04 M"$ of ${SO}_{2}$ $"SO"_2$ $\to$ $->$ ${SO}_{2}$ $"SO"_2$ and ${SO}_{3}$ $"SO"_3$ are in a $2 : 2$ $color(red)(2):color(red)(2)$ mole ratio.

Therefore, the ICE table would look like this

$2 {SO}_{2 (g)} + O_{2 (g)} ⇌ 2 {SO}_{3 (g)}$ $" " color(red)(2)"SO"_ (2(g)) " "+" " "O"_ (2(g)) " "rightleftharpoons " " color(red)(2)"SO"_ (3(g))$

$I a a a a a a 0.06 a a a a a a a a 0.05 a a a a a a a a a a a 0$ $color(purple)("I")color(white)(aaaaaacolor(black)(0.06)aaaaaaaacolor(black)(0.05)aaaaaaaaaaacolor(black)(0)$
$C a a a a - 0.04 a a a a a a - 0.02 a a a a a a a a a + 0.04$ $color(purple)("C")color(white)(aaaacolor(black)(-0.04)aaaaaacolor(black)(-0.02)aaaaaaaaacolor(black)(+0.04)$
$E a a a a a 0.02 a a a a a a a a 0.03 a a a a a a a a a a a 0.04$ $color(purple)("E")color(white)(aaaaacolor(black)(0.02)aaaaaaaacolor(black)(0.03)aaaaaaaaaaacolor(black)(0.04)$

This is exactly what you get if you use $x$ $x$ to represent the number of moles of oxygen gas that are consumed by the reaction

$2 {SO}_{2 (g)} + O_{2 (g)} ⇌ 2 {SO}_{3 (g)}$ $" " color(red)(2)"SO"_ (2(g)) " "+" " "O"_ (2(g)) " "rightleftharpoons " " color(red)(2)"SO"_ (3(g))$

$I a a a a a a 0.06 a a a a a a a a 0.05 a a a a a a a a a a a 0$ $color(purple)("I")color(white)(aaaaaacolor(black)(0.06)aaaaaaaacolor(black)(0.05)aaaaaaaaaaacolor(black)(0)$
$C a a a a a - 2 x a a a a a a a a - x a a a a a a a a a a + 2 x$ $color(purple)("C")color(white)(aaaaacolor(black)(-color(red)(2)x)aaaaaaaacolor(black)(-x)aaaaaaaaaacolor(black)(+color(red)(2)x)$
$E a a a 0.06 - 2 x a a a a a 0.05 - x a a a a a a a a 2 x$ $color(purple)("E")color(white)(aaacolor(black)(0.06-color(red)(2)x)aaaaacolor(black)(0.05-x)aaaaaaaacolor(black)(color(red)(2)x)$

But since the equilibrium concentration of ${SO}_{3}$ $"SO"_3$ is $0.04 M$ $"0.04 M"$ , it follows that

$2 x = 0.04 M \Rightarrow x = 0.02 M$ $color(red)(2)x = "0.04 M" implies x = "0.02 M"$

Once again, the ICE table takes the form given to you by your teacher.

So remember, don't get too hung up on $x$ $x$ , that's just a way we have of figuring out the equilibrium concentrations of the chemical species when we don't know what they are.

When you do know what they are, there's no need to use $x$ $x$ , just use the stoichiometric coefficients of the reaction, i.e. the mole ratios.

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Question #a750a

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