# Chemical Equilibrium - Le Chatelier's principle

If the concentration of reactants is increased such that one is a limiting agent, wouldn't the concentration of products also increase (because excess of a reactant will get carried to the other(product) side of the equation considering that the limiting agent is in aqueous medium?

e.g. (Consider 1 mole of each reactant) $$\ce{H2O + NH3 <=> NH4+ + OH-}$$

Now, let us take another scenario in which there are two moles of NH3 and 1 mole of H20. Then concentration of one reactant has increased (from the previous scenario). Wouldn't the concentration of the products also increase because one mole of NH3 does not react and gets carried to the products' side of the equation?

I am confused with how Le Chatelier's principle is applied in the 2nd scenario. Why does the equilibrium shift to the right if the concentration of the reactants and products increase (if my assumption is right)

If my assumption is wrong (concentration of product increases to balance excess reactant), my concern is such an equation is not balanced. I am unable to reconcile how the Le Chatelier's principle is applied in the 2nd scenario.

(Based on the answers below, I would like to restate my question) I understand that

$$\ce{2H2O <=> H3O+ + OH-}$$

Now if I add HCl, I have

$$\ce{H2O + HCl <=> H3O+ + Cl-}$$

The way I interpret this is : the concentration of reactants is 1 + 1 = 2 moles and the concentration of products is also 1 + 1 = 2 moles. Since the concentration on both sides of the equation is same, why do we say equilibrium shifts left. What am I missing?

• The reaction may be described an equilibrium constant (actually: two constants, one for the forward reaction and one for the reverse) which is/are independent of the concentrations of the reactants. If you change the concentration of one of the molecules involved in the equilibrium, the concentrations of the others will follow. May 10 '20 at 7:41
• The equation for chemical reactions is typically written for the balanced equation. Typically the excess reactants are not carried over to the right side as products. Doing that would make creating the mathematical expression for the equilibrium from the chemical formula confusing. You could write for example $$\ce{NH3 + H2O <=>[excess NH3] NH4+ + OH-}$$
– MaxW
May 10 '20 at 8:02
• It does not make any sense to consider $1$ mole $\ce{H_2O}$ ($18$ g) and $1$ mole $\ce{NH_3}$ ($17$ g). This mixture will not be homogeneous. $18$ g water cannot dissolve more than $6$ g $\ce{NH_3}$ at room temperature and pressure. For the same reason, your second system, with $2$ moles $\ce{NH_3}$ and $1$ mole $\ce{H_2O}$ is even more impossible to reach. May 10 '20 at 10:41
• @MaxW: Thanks for your response. I have a follow-up question to improve my understanding. What happens when HCl is added to water? Literature says equilibrium shifts to the left because the concentration of products increases. However I don't understand why concentration of the reactants doesn't increase due to HCl (on the side of reactants) May 10 '20 at 12:27
• A bit late to this discussion, but it appears to me that your big confusion is that Le Châtelier's Principle applies when the system is at equilibrium. In your first example, when you have 1 mol of each reactant, how much of each product is present? Now if you adjust the composition of reactants, what happens to the total balance of species in terms of both reactants and products?
– Zhe
Jun 10 '20 at 14:34

For the case you're describing, having left-over product species will likely not cause the reverse reaction to increase.

When you define a reaction equation like: $$\ce{aA + bB <=> cC + dD}$$, you are explicitly defining A and B as 'reactants' and C and D as 'products'. In the scenario you've described, just because you have excess reactants because you have a limiting concentration of either A or B, doesn't mean the concentration of products will increase. This is because we can think of concentration of products as $$[Products](t) = [C](t) + [D](t)$$ and not $$[Products](t) = [C](t) + [D](t) + [A](t) + [B](t)$$.

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A nice definition of Le Chatlier's principle is the following: "A change in one of the variables that describe a system at equilibrium produces a shift in the position of the equilibrium that counteracts the effect of this change."

If one is to get a little more technical with this, equilibrium is a function of many factors, primarily Temperature (T), Pressure (P), and mixture species (n). This is typically described by thermodynamics by the state property, Gibbs Free Energy, or $$\Delta G$$.

For the purposes of an example to quantify Le Chatlier's principle, let's look at the differential form of Gibbs Free Energy: $$dG = -SdT +VdP + \sum^{n}_{i=1}\mu_idn_i$$. If we consider a case where temperature isn't changing (isothermal), and where pressure isn't changing (isobaric), then the Gibbs Free Energy equation is affected by changes in system composition: $$dG = \sum^{n}_{i=1}\mu_idn_i$$. For an equilibrium case, which means that $$\Delta G = 0$$, it can be shown that if you expand this expression out for a reaction mechanism (like the one mentioned above), you end up with an expression like: $$\frac{\Delta G^o_{T_o}}{RT_o} = -ln(K_{eq})$$. This particular equation is defined for standard conditions (0 degC, 1 atm).

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We're almost to an answer, let's look a little closer at $$K_{eq}$$. Let's use a gas phase system $$\ce{aA + bB <=> cC + dD}$$ as an example. If you look up the derivation of the $$\frac{\Delta G^o_{T_o}}{RT_o} = -ln(K_{eq})$$ in any standard thermodynamics textbook, you will see that $$K_{eq}$$ is actually equal to the ratio of the mol fractions of each species multiplied by the system pressure: $$K_{eq} = \frac{y_C^c y_D^d}{y_A^a y_B^b} P^{c + d - a - b}$$. If one assumes that the pressure is fixed and that $$a = b = c = d = 1$$, then this expression would reduce down to: $$K_{eq} = \frac{y_C y_D}{y_A y_B}$$.

You can further add detail to this expression by defining the mol fraction in terms of the extent of reaction $$\xi$$ for a reaction system. Where, $$n_i = n_{io} + v_in_{io}\xi$$--which essentially says that the mols of a given species is equal to inital mols +/- the extent to which that chemical is generated or reacted away (based on the specified chemical of reference $$n_{io}$$). If you plug this into the expression for $$K_{eq}$$ you can see that now the equilibrium conversion of a reaction system is dependent on the amount of mols of each species. For the system described above, the total mols $$N_{total}$$ will cancel out of the expression leaving:

$$K_{eq} = \frac{(n_{Ao} + v_An_{io})(n_{Bo} + v_Bn_{io})}{(n_{Co} + v_CN_{io})(n_{Do} + v_Dn_{io})} = exp(-\frac{\Delta G^o_{T_o}}{RT_o})$$

For a given temperature and pressure, the Gibbs Free Energy ($$\Delta G$$) will be a fixed value. Therefore the conditions in the equation (composition of the mixture of reaction system) must balance in such a way that the equilibrium conversion $$\xi$$ will satisfy the condition. You can see here that as one either varies products (C,D) or reactants (A,B) that the solution for $$\xi$$ will change. You can generally show that an excess in either products or reactants will shift the system to maximum equilibrium conversion $$\xi$$. And as such, if one suddenly added or removed products or reactants you would cause a shift in the equilibrium conversion $$\xi$$.

• While you probably mean good, please don't finish your posts with a tagline. I also recommend our guides on formatting, please have a look here and here. May 10 '20 at 23:55
• Thank you for your feedback, I will definitely look into these.
– samp
May 11 '20 at 0:06