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Unit 7

7.10 Reaction Quotient and Le Chatelier's Principle

4 min readjune 1, 2021

fiveable-dylan

Dylan Black


7.10: Reaction Quotient and Le Chatelier’s Principle

In the last section, we discussed the concept of Le Chatelier’s Principle and described how it can be used to predict changes in concentrations as a result of an external stress being placed upon a system. However, why does Le Chatelier’s Principle actually work? We learned some rules last time and sort of “logic'd” 💪 it out but in this guide we’ll take a look at how Le Chatelier’s Principle is justified by using Q, the reaction quotient.

Review of Using Q

In Unit 7.3 we discussed the value Q, a number called the reaction quotient and how it could tell us how equilibrium would shift based on conditions not at equilibrium. For example, if we found that Q was less than the assigned K value for a reaction our reaction would respond by increasing the concentration of products in order for Q to become K. Remember that when Q = K our system is at equilibrium and thus there will be no more changes in concentrations all else held equal. The formulas for Q and K look exactly the same but remember that with Q we can calculate the value at any concentration whereas at K it is only equilibrium concentrations
This way of thinking about Q vs. K will help guide our mathematical justification of Le Chatelier’s Principle for concentration and pressure, but we’ll also see how temperature stands out as an exception to using Q to justify.

Applying Q To Le Chatelier’s Principle

Concentration

As we learned in the last section, increasing the concentration of products or reactants will cause the system to respond by creating more of the other side. For example in the reaction A ⇌ B, if the concentration of B was increased the system would respond by increasing the production of A and vice versa. Let’s think about how Q might help us understand why this occurs.
We know that Q is the ratio of products over reactants raised to their stoichiometric coefficients, so if we increase a concentration on either side, we’re either increasing Q in the case of adding products or decreasing Q in the case of adding reactants. This will change the value of Q to be greater than or less than equilibrium and thus will cause the reaction to respond by shifting back to equilibrium.
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-2E9gvsTXy8xh.png?alt=media&token=b9ece397-063c-4c0a-9512-aabe632451a5

Image From LibreTexts

As seen in the above image, when the concentration of products increases from equilibrium we will find that Q > K and thus there will be a net reaction towards the products and vice versa.

Pressure

Similar logic can be applied to pressure, however in this case we want to give special attention to the exponents on the reactants and products when calculating Q. When learning about Le Chatelier’s Principle we found that when pressure was increased on a system, we shift towards the side with fewer moles of gas and vice versa.
To understand how Q relates to pressure let’s look at the value of Q when dealing with partial pressures:
We can define a value Qp similar to the way we define Q by being the ratio of partial pressures raised to their stoichiometric coefficients. Let’s suppose our reaction is     A + B ⇌ C.
Qp = P(C)/P(A)*P(B)
The same rules between Q and Kc apply with Qp and Kp
If our overall pressure increases, our partial pressures will also increase proportionally to the overall increase (remember, PA = XA * P where XA is the mole fraction of A). Let’s suppose our pressure increased by a factor of 2. This means that our new Qp will be:
2P(C) / 2P(A) * 2P(B)
However, we can cancel out our 2 in the numerator and one 2 in the denominator to find that at the end, only the denominator will be multiplied by a factor of 2 meaning that our reactant pressure has gone up making Qp < Kp shifting the reaction towards the reactants. The same logic could be applied if our pressure went down by a certain factor. 
However, let’s suppose our reaction was instead 3A + B ⇌ 2C. Same as before we can write out Qp = P(C)^2 / P(A)^3 * P(B) and then imagine that our pressure increased by a factor of 2:
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-sig9riXErSp9.png?alt=media&token=d3d557d9-837c-476e-8992-7cc1eda572e7
We can then cancel out the 2^2 to find that our denominator is increased by a factor of 4 and the numerator’s increase has been canceled out. This means that Qp has decreased. Therefore Qp < Kp meaning our reaction will proceed by creating more products in response to an increase in pressure. The opposite would occur with a pressure decrease.
This shows us that when the pressure increases or decreases, it is the number of moles of gas on either side of the reaction that has an impact on the direction of equilibrium.

Temperature as the Exception

Unlike concentration and temperature, Q is not used to explain Le Chatelier’s Principle when it comes to 
In the last section, we addressed temperature and Le Chatelier’s Principle by describing heat as either a reactant or a product of a reaction based on whether a reaction is exothermic or endothermic. However, that isn’t quite what actually happens. While this way of thinking is valid and will get correct answers, what is happening, in reality, is that the equilibrium constant is changing! Although we never discuss K changing, remember that in reality, it’s a temperature-dependent value. This means that K is only constant when the temperature is constant. Based on your reaction, K will either increase or decrease when the temperature is increased. This is when thinking about heat as a reactant or products will help figure out the direction the reaction will go. However, remember that this part of Le Chatelier’s Principle is not justified by using Q. 

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