At #627^@"C"# and #"1 atm"#, #"SO"_3# is partially dissociated into #"SO"_2"# and #"O"_2# . If the density of the equilibrium mixture is #"0.925 g/L"#, what is the degree of dissociation?
1 Answer
Explanation:
The idea here is that you need to calculate the average molar mass of the mixture, as this will help you calculate how many moles of sulfur trioxide dissociated to produce sulfur dioxide and oxygen gas.
Your tool of choice will be this equation--I will not derive the equation here!
#M_M = (rho * R * T)/P#
Here
#M_M# is the molar mass of an ideal gas#rho# is the density of the gas at a pressure#P# and an absolute temperature#T# #R# is the universal gas constant, equal to#0.0821 quad ("atm" * "L")/("mol" * "K")#
In your case, this equation will help you calculate the average molar mass of the mixture. Plug in your values to find--do not forget to convert the temperature of the mixture to Kelvin!
#M_M = ("0.925 g" color(red)(cancel(color(black)("L"^(-1)))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (627 + 273.15) color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))#
#M_M = "68.36 g mol"^(-1)#
Now, it's important to realize that this actually represents the weighted average of the molar masses of each individual gas. In other words, each gas will contribute to the average molar mass in proportion to its mole fraction in the mixture.
The balanced chemical equation that describes this equilibrium looks like this
#2"SO"_ (3(g)) rightleftharpoons 2"SO"_ (2(g)) + "O"_ (2(g))#
Notice that for every
To make the calculations easier, let's assume that you start with
#(1 - x) quad "moles SO"_3# #x quad "moles SO"_2# #x/2 quad "moles O"_2# This is the case because the number of moles of oxygen gas produced will always be half the number of moles of sulfur trioxide that dissociate.
This means that. at equilibrium, the total number of moles of gas present in the reaction vessel is equal to
#(1 - color(red)(cancel(color(black)(x))) + color(red)(cancel(color(black)(x))) + x/2) quad "moles" = (1 + x/2) quad "moles"#
The mole fraction of sulfur trioxide will be
#chi_ ("SO"_ 3) = ((1-x) quad color(red)(cancel(color(black)("moles"))))/((1+x/2)color(red)(cancel(color(black)("moles")))) = (1-x)/(1+x/2)#
Similarly, the mole fractions of sulfur dioxide and of oxygen gas will be
#chi_ ("SO"_ 2) = (x color(red)(cancel(color(black)("moles"))))/((1 + x/2)color(red)(cancel(color(black)("moles")))) = x/(1+x/2)#
#chi_ ("O"_ 2) = (x/2 color(red)(cancel(color(black)("moles"))))/((1+x/2)color(red)(cancel(color(black)("moles")))) = (x/2)/(1+x/2)#
This means that the average molar mas of the mixture can be written as
#68.36 color(red)(cancel(color(black)("g mol"^(-1))))= overbrace( ((1-x)/(1+x/2)) * 80.066 color(red)(cancel(color(black)("g mol"^(-1)))))^(color(blue)("the contribution of SO"_3)) + #
#color(white)(68.36 color(white)(cancel(color(white)("g mol"^(-1)))) = ) + overbrace((x/(1+x/2)) * 64.066 color(red)(cancel(color(black)("g mol"^(-1)))))^(color(blue)("the contribution of SO"_2)) + #
#color(white)(68.36 color(white)(cancel(color(white)("g mol"^(-1)))) = ) + overbrace( ((x/2)/(1+x/2)) * 32.0 color(red)(cancel(color(black)("g mol"^(-1)))))^(color(blue)("the contribution of O"_2))#
This means that you have
#68.36 * (1 + x/2) = (1-x) * 80.066 + x * 64.066 + x/2 * 32.0#
This is equivalent to
#136.72 + 68.36 * x = 160.132 - 160.132 * x + 128.132 * x + 32.0 * x#
which gets you
#68.36 * x = 23.96 implies x= 23.96/68.36 = 0.35#
You can thus say that the degree of dissociation is equal to
#alpha = (0.35 color(red)(cancel(color(black)("moles"))))/(1color(red)(cancel(color(black)("mole")))) = 0.35# This basically means that for every
#1# mole of sulfur trioxide present in the mixture,#0.35# moles will dissociate to produce sulfur dioxide and oxygen gas.
I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the pressure of the mixture.
