Question #8b160

1 Answer
May 14, 2016

Answer:

#P_(O_2) = "0.41 atm"#

Explanation:

Start by writing a balanced chemical equation for this synthesis reaction

#color(red)(2)"SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons color(blue)(2)"SO"_(3(g))#

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 #color(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_"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.

#color(blue)(|bar(ul(color(white)(a/a)P_(O_2) = chi_(O_2) xx P_"total"color(white)(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) = (3.5 color(red)(cancel(color(black)("moles"))))/(8.5color(red)(cancel(color(black)("moles")))) = 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)|)))#