# Question 84ead

Aug 11, 2017

Whether the system is at equilibrium or not can be judged by the closeness of $Q$, the reaction quotient, with $K$, the equilibrium constant.

They are both written almost the same way, but they use concentrations from different moments during the reaction (not at equilibrium or at equilibrium, respectively).

In this case, we are apparently not even close to equilibrium. It's way skewed towards the product.

(That's a good thing. Industrial synthesis processes try to push the reaction forward to minimize cost and maximize product yield!)

And the basic summary of this idea is:

• If $Q < K$, the reaction shall shift to the right, and was previously skewed towards the reactants.
• If $Q > K$, the reaction shall shift to the left, and was previously skewed towards the products.
• If $Q = K$, we're good to go, the reaction is already at equilibrium...

You were given the reaction

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

And apparently, you were given the change in Gibbs' free energy at $\text{700 K}$ at some standard pressure, i.e.

DeltaG_("700 K")^@ = "13.5 kJ/mol"

This is not entirely clear from your question, so you should check your question later... If this is correct, however, recall the equation for relating $\Delta G$ at nonstandard conditions:

$\textcolor{g r e e n}{\Delta G = \Delta {G}^{\circ} + R T \ln Q}$

In this case, your reference temperature is $\text{700 K}$ (it doesn't have to be $\text{298.15 K}$), and your reference pressure is probably $\text{1 atm}$.

Either way, $\Delta G$ is zero at chemical equilibrium, so we automatically get

${\cancel{\Delta G}}^{0} - R T \ln K = \Delta {G}_{\text{700 K}}^{\circ}$

$\implies \Delta {G}_{\text{700 K}}^{\circ} = - R T \ln K$

since $Q = K$ at equilibrium. Furthermore, since the reaction contains all gases, this $K$ is ${K}_{P}$, the pressure-based equilibrium constant, which uses pressure in $\text{atm}$.

This gives you an equilibrium constant of:

K_P = "exp"(-(DeltaG_"700 K"^@)/(RT))

$= {e}^{- \text{13.5 kJ/mol"//"0.008314472 kJ/mol"cdot"K"//"700 K}}$

$= \underline{0.098}$ (units omitted)

Now that you have an equilibrium constant to compare with, calculate the reaction quotient $Q$.

$\textcolor{red}{{Q}_{P}} = \frac{{P}_{C {H}_{3} O H}}{{P}_{C O} \cdot {P}_{{H}_{2}}^{2}}$

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

$= \frac{1}{2.0 \times {10}^{- 9} {\text{atm}}^{2}}$

$= \underline{\textcolor{red}{5.0 \times {10}^{8}}}$ (units omitted)

Here, $Q$ is much larger than $K$, so the reaction is apparently not even close to equilibrium. It's way skewed towards the product, methanol.