# Determine whether the system is at equilibrium?

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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?

One of the methods for synthesizing Methanol

I know that when K=Q the system is at equilibrium, but where does

Is there a specific difference between the formulas for working out Q and K?

Thank you very much in advance?

##### 1 Answer

I think the question has an error... In order to determine whether the system is at equilibrium or not, we require the value of

So, I ASSUME 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

At equilibrium, by definition,

#DeltaG^@ = -RTlnK#

Therefore, at equilibrium, for a gas-phase reaction (where

#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

#"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.