# Question 8b160

May 14, 2016

${P}_{{O}_{2}} = \text{0.41 atm}$

#### Explanation:

Start by writing a balanced chemical equation for this synthesis reaction

$\textcolor{red}{2} {\text{SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons color(blue)(2)"SO}}_{3 \left(g\right)}$

The idea here is that you need to determine

• the number of moles of oxygen gas that remain unreacted after the reaction takes place
• the total number of moles present in the reaction vessel after the reaction takes place

Notice that the reaction consumes $\textcolor{red}{2}$ moles of sulfur dioxide for every $1$ mole of oxygen gas that takes part in the reaction.

Now, you are told that the reaction consumes 60% of the initial amount of sulfur dioxide before reaching equilibrium. Since you know that you're starting with $5$ moles of sulfur dioxide, you can say that the reaction consumed

5 color(red)(cancel(color(black)("moles SO"_2))) * overbrace(("60 moles SO"_2color(white)(a)"used up")/(100color(red)(cancel(color(black)("moles SO"_2)))))^(color(purple)("= 60% SO"_2color(white)(a)"consumed")) = "3 moles SO"_2

So, the reaction consumes $3$ moles of sulfur dioxide, which means that after the reaction is complete, the reaction vessel will contain

n_("SO"_2color(white)(a)"remaining") = "5 moles" - "3 moles" = "2 moles SO"_2

Use the mole ratios that exist between sulfur dioxide and oxygen gas to determine how many moles of the latter are consumed by the reaction

3color(red)(cancel(color(black)("moles SO"_2))) * "1 mole O"_2/(color(red)(2)color(red)(cancel(color(black)("moles SO"_2)))) = "1.5 moles O"_2

The reaction consumes $1.5$ moles of oxygen gas, which means that after the reaction is complete, the reaction vessel will contain

n_("O"_2color(white)(a)"remaining") = "5 moles" - "1.5 moles" = "3.5 moles O"_2

Finally, the reaction will produce

3color(red)(cancel(color(black)("moles SO"_2))) * (color(blue)(2)color(white)(a)"moles SO"_3)/(color(red)(2)color(red)(cancel(color(black)("moles SO"_2)))) = "3 moles SO"_3

The total number of moles present in the reaction vessel after the reaction is complete will be

${n}_{\text{total" = overbrace("2 moles")^(color(purple)("moles SO"_2)) + overbrace("3.5 moles")^(color(brown)("moles O"_2)) + overbrace("3 moles")^(color(orange)("moles SO"_3)) = "8.5 moles gas}}$

Now, the partial pressure of oxygen gas will depend on its mole fraction -- think Dalton's Law of Partial Pressures here.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{{O}_{2}} = {\chi}_{{O}_{2}} \times {P}_{\text{total}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The mole fraction of oxygen gas is equal to the number of moles of oxygen gas divided by the total number of moles present in the reaction vessel

${\chi}_{{O}_{2}} = \left(3.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(8.5color(red)(cancel(color(black)("moles}}}}\right) = 0.4118$

The partial pressure of oxygen gas will thus be

P_(O_2) = 0.4118 xx "1 atm" = color(green)(|bar(ul(color(white)(a/a)"0.41 atm"color(white)(a/a)|)))#