# Determine whether the system is at equilibrium?

## One of the methods for synthesizing Methanol $\left(C {H}_{3} O H\right)$ involves reacting CO with ${H}_{2}$. the equilibrium is $C O \left(g\right) + 2 {H}_{2} \left(g\right) \leftrightarrow C {H}_{3} O H \left(g\right)$. At 427 degrees Celsius a mixture of CO, ${H}_{2}$, and $C {H}_{3} O H$ having the following Partial pressures; ${P}_{C O} = 2.0 \cdot {10}^{-} 3 a t m$, ${P}_{{H}_{2}} = 1.0 \cdot {10}^{-} 6 a t m$ and ${P}_{C {H}_{3} O H} = 1.0 \cdot {10}^{-} 6$. For this reaction $\Delta {G}_{700 K} = 13.5 k J$. Determine whether the system is at equilibrium? I know that when K=Q the system is at equilibrium, but where does $\Delta {G}_{700 K}$ come in? Is there a specific difference between the formulas for working out Q and K? Thank you very much in advance?

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 $\Delta {G}^{\circ}$ for the reaction at $\text{700 K}$, because if it IS at equilibrium, then $\Delta {G}_{\text{700 K}} = 0$ and $\Delta {G}_{\text{700 K}}^{\circ} \ne 0$.

So, I ASSUME that $\Delta {G}_{\text{700 K"^@ = "13.5 kJ/mol}}$ for this reaction. We know that

$\Delta G = \Delta {G}^{\circ} + R T \ln Q$

where $Q$ is the reaction quotient, $R$ is the universal gas constant, $T$ is temperature in $\text{K}$, and ""^@ indicates the reference state for the change in the Gibbs' free energy $\Delta G$.

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

At equilibrium, by definition, $\Delta G = 0$ and $Q = K$, so...

$\Delta {G}^{\circ} = - R T \ln K$

Therefore, at equilibrium, for a gas-phase reaction (where $K \equiv {K}_{p}$), we can calculate ${K}_{p}^{\circ}$ (since $\Delta {G}^{\circ}$ is supposedly known):

${K}_{p}^{\circ} = {e}^{- \Delta {G}^{\circ} / R T}$

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

$= 0.0983$

in implied units of $\text{atm}$. Now we compare this with ${Q}_{p}^{\circ}$ calculated from the initial states given for the partial pressures in the reaction, each divided by the standard pressure ${P}^{\circ} \equiv \text{1 atm}$:

$\text{CO"(g) + 2"H"_2(g) rightleftharpoons "CH"_3"OH} \left(g\right)$

${Q}_{p}^{\circ} = \frac{{P}_{C {H}_{3} O H} / {P}^{\circ}}{\left({P}_{C O} / {P}^{\circ}\right) {\left({P}_{{H}_{2}} / {P}^{\circ}\right)}^{2}}$

$= \left({\left(1.0 \times {10}^{- 6} \text{atm"//"1 atm"))/((2.0 xx 10^(-3) "atm"//"1 atm")(1.0 xx 10^(-6) "atm"//"1 atm}\right)}^{2}\right)$

$= 5.0 \times {10}^{8}$

This would definitely not be at equilibrium. ${Q}_{P}^{\circ}$ $\text{>>}$ $0.0983$.