# 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 themolar massof an ideal gas#rho# is thedensityof the gas at apressure#P# and anabsolute temperature#T# #R# is theuniversal 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** **moles** of sulfur trioxide **that dissociate**, the reaction produces **moles** of sulfur dioxide and **mole** of oxygen gas.

To make the calculations easier, let's assume that you start with **mole** of sulfur trioxide. if you take **moles** to be the number of moles of sulfur trioxide **that dissociate**, you can say that, at equilibrium, the reaction vessel will contain

#(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

halfthe number of moles of sulfur trioxidethat 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# moleof sulfur trioxide present in the mixture,#0.35# moleswill 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.