# Gibbs Free Energy and Equlibrium Constant

I have some problems understanding the relationship between $\Delta$$G° and K. For example, in the reaction of$$\ce{N2 + H2 <=> 2NH3}$$where \Delta$$G°= - 33.5$ kJ/mol.

$\Delta$$G°=- 33.5 kJ/mol is obtained from the \sum G^°_{\text{f, products}} – \sum G^°_{\text{f, reactants}} . This means that the complete conversion of 1 molar of N_2 gas and 1 molar of H_2 gas gives 2 molars of NH_3 gives -33.5 kJ/mol of free energy. This is assuming that the reaction goes fully to completion, yet in reality it doesn't. From this formula below \Delta{G^°} = –RT \ln K_p \tag{5-6} K_p can be obtained as 7* 10^5, assuming at 298K and 1 atm. My question is, since the complete reaction corresponds to \Delta$$G°= - 33.5$ kJ/mol of free energy, why should there be $K$ in the first place? $- 33.5$ kJ/mol tells me this value is for the complete reaction, so why should we insert this value into the equation just to get $K$ which tells me otherwise that the reaction doesn't go to completion? It doesn't make sense to me? In fact, shouldn't have K be infinity since the value that we had inserted is one obtained for a complete reaction?

Also, from this equation $\Delta{G} = \Delta{G^°} + RT \ln Q \tag{5-5}$

does having 1 mol of $N_2$ , 1 mol of $H_2$ and 1 mols $NH_3$ of in the reaction correspond to $- 33.5$ kJ/mol, since that makes $Q = 1$? And since $Q$ slowly reaches the value of $K$, $\Delta{G} = 0$, then $- 33.5$ kJ of free energy must have corresponded from the initial state to the equlibrium state. Yet again to me this doesn't make sense, because my understanding is that $- 33.5$ kJ is a value obtained only for the ideal scenario that the reaction goes to completion, and not from its initial standard states to equilibrium.

How does this all work out?

• The $\Delta G$ value doesn't need to correspond to the reaction going to completion. It only tells us what the free energy released per mole of reaction is. At standard conditions, this reaction releases $-33.5$ kJ of free energy per mole of reaction, but that doesn't necessarily mean that if we have $1$ mol of $\ce{N2} \text{ and } 3\text{mols of } \ce{H2}$ that the reaction will proceed to completion. – Tyberius Mar 29 '17 at 16:08

Q is supposed to be calculated in terms of partial pressures, not moles. $\Delta G^0$ is based on a reversible process you have devised to take the pure reactants in stoichiometric proportions and 1 atm (actually, 1 bar) and 298 K and convert them to pure products in stoichiometric proportions, also at 1 atm. and 298 K.

You can't really measure $\Delta G^\circ$ directly, rather you measure $K$ and convert it to $\Delta G^\circ$. Put another way, $K$ is the quantity you measure to get $\Delta G^\circ$, just like heat exchanged ($q$) is the quantity you measure to get $\Delta H^\circ$.

To measure $\Delta H^\circ$ for the reaction you would, for example, measure $q$ under conditions (excess $\ce{H2}$) where a known quantity of $\ce{N2}$ is completely converted to $\ce{NH3}$.

To predict $q$ for an arbitrary set of reaction conditions you compute how much reactant/product is consumed/produced from $K$ and multiply that change in partial pressure by by $\Delta H^\circ$.

First of all, it confuses many when $\Delta$ in thermodynamics is used for both real changes (as in $\Delta U = q + w$, unit e.g. kJ) and for derivatives, i.e. infinitesimal changes (as in the $\Delta G$ in the question, unit kJ/mol). In my opinion, to reduce this confusion it is better to use $\Delta_r G$ and $\Delta_r G^0$ than $\Delta G$ and $\Delta G^0$, as (some) textbooks do.

Next one needs to understand that $\Delta_r G$ and $\Delta_r G^0$ are not changes from beginning to end of the reaction, and not at all to completion of the reaction, but they describe infinitesimal changes (change in $G$ per mole reaction: $\Delta_r G$ for the actual experimental composition and $\Delta_r G^0$ for the standard reference composition). Experimentally the mathematical term "infinitesimal" means so small a change that the molecular composition in the reaction vessel does not change significantly. Provided the change in composition is negligible, one gets the actual change by multiplying with the actual number of moles that has reacted. This number does not have to be 1 mol. Just as if you drive on the highway with, say, 120 km/h, this does not mean that you must drive for exactly 1 hour and move 120 km, in fact it may be only for one minute and you have moved 2 km. Your speed was still 120 km/h. And it is easier to keep a constant speed for 1 min than for 1 hour.

So, $\Delta G_r^0$ = -33.7 kJ/mol is not for the reaction running to completion, but for the special case where the partial pressures of $\ce{N2}$, $\ce{H2}$, and $\ce{NH3}$ all are 1 bar (or 1 atm if old standard), both before and after the reaction. In this very special case, after 1 mol reaction, i.e. after 1 mol $\ce{N2}$ has reacted with 1 mol $\ce{H2}$ and formed additional 2 mol $\ce{NH3}$, you would then get $\Delta G = \Delta_r G^0 \times 1\ {\rm mol} = -33.7\ {\rm kJ}$. Experimentally this three digit number would only be correct to the three digits if one had a very big container with many moles of the three gases, such that adding or removing a mole would not significantly change the partial pressures of the three gases. Of course, keeping the change of composition negligible would be easier to achieve after e.g. only 1 $\mu$mol reacted instead of 1 mol; cf. my comparison of 1 hour and 1 minute driving on the highway. (Note that it is not important if you are able to measure this $\Delta G$ experimentally or not, it is still the correct number if the measurement could be done!)

In your eq. 5-5 ( $\Delta_r G = \Delta_r G^0 + RT \ln Q$ ) the term "$RT \ln Q$" describes the change in $\Delta_r G$ when changing from the reference standard composition to the actual composition. In this case, changing from 1 bar partial pressure of each of the 3 gases to the actual partial pressures in the experiment. We can then ask ourselves: when will there be equilibrium? The mathematics of thermodynamics tells us that we have equilibrium at constant pressure and temperature when $G$ has its minimum value, and minimum means that the derivative of $G$ with respect to further reaction is zero: $\Delta_r G = 0$. Inserting this in eq. 5-5 gives $0 = \Delta_r G^0 + RT \ln Q_{eq}$, which is easily rearranged to eq. 5-6. Conventionally we call the numerical value of $Q_{eq}$ the equilibrium constant $K$.