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}#
For the following reaction run at 298 K; a mixture of equilibrium contains
1 Answer
Well, apparently all the
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
#P_(SO_2) = sqrt(P_(SO_3)^2/(K_PP_(O_2))#
I assume you gave
We know all of these quantities except for
#DeltaG_(rxn)^@("gas-phase") = -RTlnK_P#
We already have enough information to find
#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
#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...