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
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.