Determine whether the system is at equilibrium?

One of the methods for synthesizing Methanol #(CH_3OH)# involves reacting CO with #H_2#. the equilibrium is #CO(g) + 2H_2(g) harr CH_3OH(g)#. At 427 degrees Celsius a mixture of CO, #H_2#, and #CH_3OH# having the following Partial pressures; #P_(CO)=2.0*10^-3 atm#, #P_(H_2)=1.0*10^-6 atm# and #P_(CH_3OH)=1.0*10^-6#. For this reaction #DeltaG_(700K)=13.5kJ#. Determine whether the system is at equilibrium?
I know that when K=Q the system is at equilibrium, but where does #DeltaG_(700K)# come in?
Is there a specific difference between the formulas for working out Q and K?
Thank you very much in advance?

1 Answer
Truong-Son N.
Jul 25, 2018

I think the question has an error... In order to determine whether the system is at equilibrium or not, we require the value of #DeltaG^@# for the reaction at #"700 K"#, because if it IS at equilibrium, then #DeltaG_"700 K" = 0# and #DeltaG_"700 K"^@ ne 0#.

So, I ASSUME that #DeltaG_"700 K"^@ = "13.5 kJ/mol"# for this reaction. We know that

#DeltaG = DeltaG^@ + RTlnQ#

where #Q# is the reaction quotient, #R# is the universal gas constant, #T# is temperature in #"K"#, and #""^@# indicates the reference state for the change in the Gibbs' free energy #DeltaG#.

(Furthermore, the Gibbs' free energy is a state function, so it should AT LEAST be #DeltaG#, and not #G#...)

At equilibrium, by definition, #DeltaG = 0# and #Q = K#, so...

#DeltaG^@ = -RTlnK#

Therefore, at equilibrium, for a gas-phase reaction (where #K -= K_p#), we can calculate #K_p^@# (since #DeltaG^@# is supposedly known):

#K_p^@ = e^(-DeltaG^@//RT)#

#= e^(-"13.5 kJ/mol"//("0.008314 kJ/mol"cdot"K" cdot "700 K"))#

#= 0.0983#

in implied units of #"atm"#. Now we compare this with #Q_p^@# calculated from the initial states given for the partial pressures in the reaction, each divided by the standard pressure #P^@ -= "1 atm"#:

#"CO"(g) + 2"H"_2(g) rightleftharpoons "CH"_3"OH"(g)#

#Q_p^@ = (P_(CH_3OH)//P^@)/((P_(CO)//P^@)(P_(H_2)//P^@)^2)#

#= ((1.0 xx 10^(-6) "atm"//"1 atm"))/((2.0 xx 10^(-3) "atm"//"1 atm")(1.0 xx 10^(-6) "atm"//"1 atm")^2)#

#= 5.0 xx 10^8#

This would definitely not be at equilibrium. #Q_P^@# #">>"# #0.0983#.

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