Calculate #P_{SO_2}# at equilibrium at 298 K?

For the following reaction run at 298 K; a mixture of equilibrium contains #P_{O_2}=0.50# atm, #P_{SO_2}=2.0# atm. Calculate #P_{SO_2}# at equilibrium at 298 K?
#2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)#
#\Delta G°_{f,SO_3(g)}=-371\frac{kJ}{mol}#
#\Delta G°_{f,SO_2(g)}=-300\frac{kJ}{mol}#

1 Answer
Truong-Son N.
Jul 28, 2018

Well, apparently all the #"SO"_2# was consumed...

First, write the reaction and equilibrium constant:

#2"SO"_2(g) + "O"_2(g) rightleftharpoons 2"SO"_3(g)#

#K_P = P_(SO_3)^2/(P_(SO_2)^2P_(O_2))#

What we want to solve for is the (always positive) partial pressure of #"SO"_2# (not #"SO"_3#, apparently):

#P_(SO_2) = sqrt(P_(SO_3)^2/(K_PP_(O_2))#

I assume you gave #P_(O_2)# and #ulul(P_(SO_3))#...

We know all of these quantities except for #K_P#. Remember that gas-phase reactions have #K -= K_P#, so:

#DeltaG_(rxn)^@("gas-phase") = -RTlnK_P#

We already have enough information to find #DeltaG^@#, so we can find #K_P# afterwards. These #DeltaG_f^@# are correct and the reaction is balanced.

#DeltaG_(rxn)^@ = ["2 SO"_3(g) cdot DeltaG_f("SO"_3(g))^@] - ["2 SO"_2(g) cdot DeltaG_f("SO"_2(g))^@ + "1 O"_2(g) cdot "0 kJ/mol O"_2(g)]#

#= [-"742 kJ/mol"] - [-"600 kJ/mol"]#

#= -"142 kJ/mol"#

Therefore,

#K_P = e^(-DeltaG_(rxn)^@//RT)#

#= e^(-(-"142 kJ/mol")//("0.008314 kJ/mol"cdot"K" cdot "298 K")#

#= e^(57.3)#

As a result, we get the nearly zero partial pressure of #"SO"_2#:

#P_(SO_2) = sqrt(("2.0 atm")^2/(e^(57.3)("0.50 atm"))#

#=# #ul(1.02 xx 10^(-12) "atm")#

i.e. basically zero...

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