Select your language

Suggested languages for you:
Log In Start studying!
StudySmarter - The all-in-one study app.
4.8 • +11k Ratings
More than 3 Million Downloads
Free
|
|

All-in-one learning app

  • Flashcards
  • NotesNotes
  • ExplanationsExplanations
  • Study Planner
  • Textbook solutions
Start studying

Reaction Quotient and Le Chatelier's Principle

Save Save
Print Print
Edit Edit
Sign up to use all features for free. Sign up now
Reaction Quotient and Le Chatelier's Principle

Le Chatelier's principle tells us that when a system at equilibrium is disturbed, the position of the equilibrium shifts to counteract the disturbance. But we can also deduce this by looking at the effect of a disturbance on the reaction quotient and the equilibrium constant. By considering the relative magnitudes of these values, we can predict the overall direction of a reaction in response to a change in conditions.

  • This article is about the reaction quotient and Le Chatelier's principle in chemistry.
  • We'll compare the reaction quotient with the equilibrium constant before providing you with a recap of Le Chatelier's principle.
  • We'll then look at how changing reaction conditions changes the reaction quotient and the equilibrium constant, and how this supports Le Chatelier's principle.

What is the reaction quotient?

In a system at equilibrium, the relative amounts of products and reactants remain constant. We can represent the ratio between them using a value known as the equilibrium constant, Keq. There are a few different forms of Keq, but you'll most commonly work with Kc and Kp:

  • Kc measures the concentrations of aqueous or gaseous species in a system at equilibrium.
  • Kp measures the partial pressures of gaseous species in a system at equilibrium.

For the reaction \(aA(g)+bB(g) \rightleftharpoons cC(g)+dD(g)\), the equilibrium constants take the following expressions:

$$K_c=\frac{{[C]_{eqm}}^c\space {[D]_{eqm}}^d}{{[A]_{eqm}}^a\space {[B]_{eqm}}^b}\qquad K_p=\frac{{{(P_C)}_{eqm}}^c\space {{(P_D)}_{eqm}}^d}{{{(P_A)}_{eqm}}^a\space {{(P_B)}_{eqm}}^b}$$

But systems aren't always at equilibrium. In this case, we turn to the reaction quotient. The reaction quotient Q is similar to Keq, but instead of measuring relative amounts of reactants and products at equilibrium, it measures them at one particular moment at any point in the reaction.

The reaction quotient Q is a value that tells us the relative amounts of products and reactants in a system at a particular moment, at any point in the reaction.

Like Kc, Qc measures concentration, and like Kp, Qp measures partial pressure.

Reversible reactions in a closed system always head towards a point of equilibrium. This leads us on to an important point: as the reaction progresses, Q tends towards Keq. Furthermore, if Q equals Keq, then the reaction is at equilibrium. This will come in handy later on.

Reaction quotient expression

The expressions for the reaction quotient are much like the expressions for the equilibrium constant. But instead of taking their measurements at equilibrium, they measure the concentration or partial pressure at any one particular moment in the reaction:

$$Q_c=\frac{[C]^c\space [D]^d}{[A]^a\space [B]^b}\qquad Q_p=\frac{(P_C)^c\space (P_D)^d}{(P_A)^a\space (P_B)^b}$$

Reaction quotient vs equilibrium constant

Confused about how the reaction quotient and equilibrium constant compare? Here's a handy table to help summarize your learning:

Reaction Quotient and Le Chatelier's Principle equilibrium constant reaction quotient Keq C comparison table StudySmarterA table comparing the equilibrium constant and the reaction quotient.StudySmarter Originals

Find out more about Keq in "Equilibrium Constant", and discover Q in more depth in Reaction Quotient. You can also see how we use both values in Calculating Equilibrium Concentrations, Calculating Equilibrium Constant, and Using the Reaction Quotient.

Moving on - let's remind ourselves about Le Chatelier's principle.

What is Le Chatelier's principle?

We learned in the article Le Chatelier's Principle that changing the conditions of an equilibrium alters the position of that equilibrium. This temporarily favors one reaction or the other.

Le Chatelier's principle states that when a system at equilibrium is disturbed, the position of the equilibrium shifts to counteract the disturbance.

A disturbance can be a change in concentration, pressure, or temperature. The system always responds by trying to undo the change:

  • Increasing the concentration of a species causes the position of the equilibrium to shift to use up some of that species, thereby reducing its concentration. For example, increasing the concentration of a reactant shifts the equilibrium to the right, favoring the forward reaction.
  • Increasing the pressure of a gaseous equilibrium causes the position of the equilibrium to shift to decrease the pressure. This means favoring the reaction that produces the fewest moles of gas.
  • Increasing the temperature of an equilibrium causes the position of the equilibrium to shift to use up some of the extra heat energy. This means favoring the endothermic reaction.

However, it is important to note that shifting the position of the equilibrium doesn't necessarily mean changing the equilibrium constant, Keq:

  • Changing the concentration of an equilibrium doesn't affect Keq.
  • Changing the pressure of an equilibrium doesn't affect Keq either.
  • However, changing the temperature of an equilibrium does affect Keq.

But in all three cases, changing the conditions of an equilibrium, and thus shifting its position, does affect the reaction quotient Q. We can use the difference between the reaction quotient and the equilibrium constant to predict the direction a reaction will move in. In the next section, we'll explore how.

Reaction Quotient and Le Chatelier's Principle

As we've already explored, changing the conditions of an equilibrium shifts its position in order to counteract the change. This doesn't necessarily affect the equilibrium constant Keq, but it does affect the reaction quotient Q. We can use this to determine the direction an equilibrium will shift in. Remember that as a reaction progresses, Q tends towards Keq. This tells us which reaction is favored in order to reach equilibrium:

  • If Q is less than Keq, the forward reaction is favored. This increases the relative amount of products until Q equals Keq.
  • If Q is greater than Keq, the backward reaction is favored. This increases the relative amount of reactants until Q equals Keq.

By disturbing the system's equilibrium, we change the value of Q. The system, therefore, responds by shifting the position of the equilibrium until Q and Keq are equal again. This follows what Le Chatelier's principle tells us about how a system reacts to a disturbance. Let's now use some examples to understand how the process works for changes in concentration, pressure, and temperature.

Concentration

According to Le Chatelier's principle, increasing the concentration of a species causes the position of the equilibrium to shift to use up some of that species, thereby reducing its concentration. For example, increasing the concentration of a reactant shifts the equilibrium to the right, favoring the forward reaction. This is supported by the reaction quotient and the equilibrium constant.

A system at equilibrium is represented by the following equation:

$$A(g)\rightleftharpoons 2B(g)\qquad K_c=2.0$$

The system contains 2.0 M A and 2.0 M B. Calculate Qc at this point and use it to predict the shift in the equilibrium's position.

First of all, let's check that the system is initially at equilibrium. We can do this by calculating the reaction quotient Qc and comparing it to the equilibrium constant Kc. If Qc = Kc, then the system is at equilibrium. For this reaction, Qc takes the following expression:

$$Q_c=\frac{{[B]}^2}{[A]}$$

If we substitute the data given in the question into the expression, we find that Qc does in fact equal Kc. The system is at equilibrium:

$$Q_c=\frac{{(2.0)}^2}{(2.0)}$$ $$Q_c=\frac{4.0}{2.0}=2.0$$ $$Q_c=K_c$$

Now let's consider what happens when the concentration of A is increased by 2.0 M, giving a total concentration of 4.0 M. Changing concentration doesn't affect Kc, but it does change the value of Qc:

$$Q_c=\frac{{(2.0)}^2}{(4.0)}$$ $$Q_c=\frac{4.0}{4.0}=1.0$$

Thanks to increasing the concentration of A, Qc is now less than Kc. But we know that Qc will try to tend towards Kc, in order to reach equilibrium again. This means that the position of the equilibrium shifts to the right to favor the forward reaction, increasing the concentration of B and thus increasing the value of Qc.

Le Chatelier's principle agrees with the above example: increasing the concentration of a species favors the reaction that uses some of that species up. In this case, we increased the concentration of the reactant, and so the forward reaction is favored.

Pressure

According to Le Chatelier's principle, increasing the pressure of a gaseous equilibrium causes the position of the equilibrium to shift to decrease the pressure. This means favoring the reaction that produces the fewest moles of gas. To understand how this affects the reaction quotient, we need to look at the system in terms of partial pressure.

A system at equilibrium is represented by the following equation:

$$A(g)+2B(g)\rightleftharpoons 2C(g)\qquad K_p=1.0$$

The system contains 1.0 moles of A, 2.0 moles of B, and 2.0 moles of C, and initially has a pressure of 5.0 kPa. The pressure is then increased to 6.0 kPa. Calculate Qp at the point of pressure increase and use it to predict the shift in the equilibrium's position.

Once again, we'll start by confirming that the system is indeed at equilibrium. We do this by working out the partial pressure of each gas and substituting the values into an expression for Qp. If Qp equals Kp, then the reaction is at equilibrium. Remember that the partial pressure of each gas is simply its molar fraction multiplied by the total pressure of the system. You should get the following results:

$$(P_A)=\frac{1}{5}\times 5.0=1.0\space kPa$$ $$(P_B)=\frac{2}{5}\times 5.0=2.0\space kPa$$ $$(P_C)=\frac{2}{5}\times 5.0=2.0\space kPa$$ $$Q_p=\frac{(P_C)^2}{(P_A)\space (P_B)^2}$$ $$Q_p=\frac{(2.0)^2}{(1.0)\space (2.0)^2}$$ $$Q_p=\frac{4.0}{4.0}=1.0

Here, Qp does in fact equal Kp, and so the system is at equilibrium.

We then increase the total pressure. This means that the partial pressure of each species increases:

$$(P_A)=\frac{1}{5}\times 6.0=1.2\space kPa$$ $$(P_B)=\frac{2}{5}\times 6.0=2.4\space kPa$$ $$(P_C)=\frac{2}{5}\times 6.0=2.4\space kPa$$

Putting these values into the expression for Qp, we find that Qp has decreased:

$$Q_p=\frac{(2.4)^2}{(1.2)\space (2.4)^2}=0.83$$

We know that Qp will try to tend towards Kp, in order to reach equilibrium again. Therefore, the position of the equilibrium shifts to the right to favor the forward reaction and increase the value of Qp.

Le Chatelier's principle agrees with the above example: increasing the pressure of a gaseous equilibrium favors the reaction that produces the fewest moles of gas. In this case, the forward reaction only produces 2 moles of gas, whilst the backward reaction produces 3. That means that the system favors the forward reaction.

Ready for a more confusing example? Consider what would happen if we added an inert gas to a gaseous equilibrium.

A system is represented by the following equation:

$$A(g)+B(g)\rightleftharpoons C(g)\qquad K_p=0.50$$

At equilibrium, the system has a total pressure of 6.0 kPa, and equal quantities of A, B, and C. By calculating the reaction quotient for each of the following cases, predict the shift in the equilibrium's position if:

  1. An equal quantity of an inert gas is added, but the system increases in volume so the total pressure remains at 6.0 kPa.
  2. An equal quantity of an inert gas is added, but the volume of the system stays the same so the overall pressure increases to 8.0 kPa.

Let's start by looking at the partial pressures of each of the gases in the original system. We have equal amounts of A, B and C, and so they each have a molar fraction of a third. This gives them all a partial pressure of 2 kPa:

$$(P_A)=\frac{1}{3}\times 6.0=2.0\space kPa$$ $$(P_B)=\frac{1}{3}\times 6.0=2.0\space kPa$$ $$(P_C)=\frac{1}{3}\times 6.0=2.0\space kPa$$

Putting them into the expression for Qp, we find that the system is indeed at equilibrium:

$$Q_p=\frac{(P_C)}{(P_A)\space (P_B)}$$ $$Q_p=\frac{(2.0)}{2.0)\space (2.0)}=0.50$$

We'll now consider the first case. The total pressure is the same, but we have added an equal quantity of another gas. A, B, and C now each only have a mole fraction of a quarter. This reduces their partial pressures:

$$(P_A)=\frac{1}{4}\times 6.0=1.5\space kPa$$ $$(P_B)=\frac{1}{4}\times 6.0=1.5\space kPa$$ $$(P_C)=\frac{1}{4}\times 6.0=1.5\space kPa$$

Putting them into the expression for Qp, we find that Qp has increased. It is now greater than Kp:

$$Q_p=\frac{(1.5)}{1.5)\space (1.5)}=0.67$$

The position of the equilibrium, therefore, shifts to the left, favoring the backward reaction in order to reduce Qp.

But if we look at the second case, we see something a little different. Once again, we've added an equal quantity of an inert gas, but this time the volume has stayed the same and so the overall pressure has increased to 8 kPa. Like before, A, B, and C have a mole fraction of a quarter, but because the overall pressure has increased, their partial pressures stay the same:

$$(P_A)=\frac{1}{4}\times 8.0=2.0\space kPa$$ $$(P_B)=\frac{1}{4}\times 8.0=2.0\space kPa$$ $$(P_C)=\frac{1}{4}\times 8.0=2.0\space kPa$$

This means that Qp doesn't change. Qp still equals Kp, and so the position of the equilibrium doesn't shift.

To conclude, if we add an inert gas to an equilibrium and the pressure stays the same, the equilibrium shifts to favor the reaction producing the greatest number of moles of gas. This same concept works for diluting an aqueous equilibrium - all of the concentrations decrease equally, and so the position of the equilibrium shifts to favor the reaction that produces more aqueous species. But if we add an inert gas to an equilibrium and the volume stays the same, the position of the equilibrium doesn't move.

Temperature

The final factor that we'll explore today is temperature. When it comes to equilibria, temperature works a little differently. Unlike changing concentration or pressure, changing the temperature of a system at equilibrium does change the equilibrium constant Keq:

  • If the forward reaction is endothermic, then increasing the temperature increases the value of Keq.
  • If the forward reaction is exothermic, then increasing the temperature decreases the value of Keq.

Le Chatelier's principle also tells us increasing the temperature of a system causes the position of the equilibrium to shift to use up some of the extra heat energy. This means favoring the endothermic reaction. We can still compare Q to the new Keq in order to confirm this shift in position.

A system is represented by the following equation:

$$A(g)\rightleftharpoons B(g)\qquad\Delta ^\uptheta =+67.8\space kJ\space mol^{-1}\qquad K_c=3.0$$

At temperature X, it contains 1.0 M A and 3.0 M B at equilibrium. The temperature is then increased to temperature Y. Using your knowledge of Kc and Qc, predict the shift in the equilibrium's position .

Let's start by calculating the reaction quotient for the reaction:

$$Q_c=\frac{[B]}{[A]}$$ $$Q_c=\frac{[3.0]}{[1.0]}=3.0$$

At this temperature, Qc equals Kc and so the system is at equilibrium.

We're told that the temperature increases. Because the reaction is endothermic, we know that this increases the value of Kc. But the concentrations of A and B haven't changed, and so Qc hasn't changed. That means that Qc is now less than Kc, and the reaction isn't at equilibrium. The position of the equilibrium, therefore, shifts to the right, favoring the forward reaction to increase the value of Qc so that it tends towards Kc.

This agrees with Le Chatelier's principle: Increasing the temperature of a system at equilibrium favors the endothermic reaction. In this case, that is the forward reaction.

We've reached the end of this article. You should now be able to define the reaction quotient Q and compare it to the equilibrium constant Keq. In addition, you should be able to recall Le Chatelier's principle and explain how changing concentration, pressure, or temperature affects the position of an equilibrium. You should also be able to state their effect on the equilibrium constant Keq. Finally, you should be able to use the reaction quotient Q to predict how a system will respond to a change in conditions and relate this to Le Chatelier's principle.

Reaction Quotient and Le Châtelier's Principle - Key takeaways

  • The reaction quotient Q is a value that tells us the relative amounts of products and reactants in a system at a particular moment, at any point in the reaction. In contrast, the equilibrium constant Keq tells us the relative amounts of products and reactants in a system at equilibrium.
  • Q always tends towards Keq.
  • Le Chatelier's principle states that when a system at equilibrium is disturbed, the position of the equilibrium shifts to counteract the disturbance. Disturbances include changes in concentration, pressure, and temperature.
  • Changes in concentration and pressure don't change Keq, but they do change Q. We can use the relative magnitudes of Q and Keq to predict the shift in the equilibrium's position, which supports Le Chatelier's principle.
  • A change in temperature does change Keq. We can still compare Q to the new value of Keq in order to predict the shift in the equilibrium's position.

Frequently Asked Questions about Reaction Quotient and Le Chatelier's Principle

Le Chatelier's principle states that the position of an equilibrium is affected by changes in concentration, pressure and temperature.

The position of an equilibrium is affected by changes in concentration, pressure and temperature. However, only changes in temperature affect the value of the equilibrium constant Keq - changes in concentration or pressure only affect the reaction quotient Q.

Le Chatelier's principle states that when a system at equilibrium is disturbed, the position of the equilibrium will shift to counteract the disturbance. Simply put, it means that if you change the conditions of a system at equilibrium, one of the reactions will be temporarily favored to try and undo the change and restore equilibrium. For example, if you increase the concentration of a species, the system will favor the reaction that uses up that species in order to reduce its concentration again.

To find the reaction quotient Q, you need to know the partial pressures or concentrations of all of the species in an equilibrium at one particular moment. You then substitute these values into the expression for Q. The expression is given by a fraction. To find the numerator of the fraction, you raise the concentrations or partial pressures of each of the products to the power of their coefficients in the balanced chemical equation, and then multiply them together. To find the denominator, you raise the concentrations or partial pressures of each of the reactants to the power of their coefficients in the balanced chemical equation, and then multiply them together. You then calculate Q by simply dividing the numerator by the denominator.

The reaction quotient Q is a is a value that tells us the relative amounts of products and reactants in a system at a particular moment, at any point in the reaction. In contrast, the equilibrium constant Keq measures the relative amounts of products and reactants in a system at equilibrium.

Final Reaction Quotient and Le Chatelier's Principle Quiz

Question

What is the reaction quotient?

Show answer

Answer

The reaction quotient is a value that tells us the relative amounts of products and reactants in a system at a particular moment, at any point in the reaction. 

Show question

Question

What is the symbol for the reaction quotient?

Show answer

Answer

Q

Show question

Question

Compare the reaction quotient and the equilibrium constant.

Show answer

Answer

  • Both measure the relative amounts of products and reactants in a system of reversible reactions.
  • Both measure either concentration or partial pressure.
  • The reaction quotient Q takes these measurements at any one particular point in the reaction, whilst the equilibrium constant Keq takes these measurements at equilbrium.
  • For a particular reaction at a specific temperature, Q can vary, but Keq is always the same.

Show question

Question

True or false? At equilibrium, Q = Keq.

Show answer

Answer

True

Show question

Question

True or false? In a closed system of reversible reactions, Q always tends towards Keq.

Show answer

Answer

True

Show question

Question

Describe how increasing the concentration of one of the reactants in an equilibrium affects its values of Qc and Kc. Use this to predict the shift in the equilibrium's position.

Show answer

Answer

Increasing the concentration of the reactants decreases the value of Qc but doesn't change the value of Kc. In a closed system, Qc always tends towards Kc. The position of the equilibrium therefore shifts to the right, favoring the forward reaction in order to increase the value of Qc.

Show question

Question

Increasing the pressure of a gaseous equilibrium favors ____.

Show answer

Answer

The reaction that produces the fewest moles of gas.

Show question

Question

Increasing the concentration of the products of a system at equilibrium favors ____.

Show answer

Answer

The backward reaction.

Show question

Question

Decreasing the temperature of a system at equilibrium favors ____.

Show answer

Answer

The exothermic reaction.

Show question

Question

The forward reaction of a reversible reaction is endothermic. Increasing the temperature ____ its value of Keq.

Show answer

Answer

Increases

Show question

Question

Describe how increasing the temperature of an equilibrium with an exothermic forward reaction affects its values of Q and Keq. Use this to predict the shift in the equilibrium's position.

Show answer

Answer

Because the forward reaction is exothermic, increasing the temperature causes Keq to decrease. However, Q doesn't change. In a closed system, Q always tends towards Keq. The system therefore shifts to the left, favoring the backward reaction in order to decrease the value of Q.

Show question

Question

Adding an inert gas to a gaseous equilibrium at constant pressure favors ____.

Show answer

Answer

The reaction that produces the greatest number of moles of gas.

Show question

Question

Adding an inert gas to a gaseous equilibrium at constant volume favors ____.

Show answer

Answer

Neither reaction - the position of the equilibrium stays the same.

Show question

More about Reaction Quotient and Le Chatelier's Principle
60%

of the users don't pass the Reaction Quotient and Le Chatelier's Principle quiz! Will you pass the quiz?

Start Quiz

Discover the right content for your subjects

No need to cheat if you have everything you need to succeed! Packed into one app!

Study Plan

Be perfectly prepared on time with an individual plan.

Quizzes

Test your knowledge with gamified quizzes.

Flashcards

Create and find flashcards in record time.

Notes

Create beautiful notes faster than ever before.

Study Sets

Have all your study materials in one place.

Documents

Upload unlimited documents and save them online.

Study Analytics

Identify your study strength and weaknesses.

Weekly Goals

Set individual study goals and earn points reaching them.

Smart Reminders

Stop procrastinating with our study reminders.

Rewards

Earn points, unlock badges and level up while studying.

Magic Marker

Create flashcards in notes completely automatically.

Smart Formatting

Create the most beautiful study materials using our templates.

Sign up to highlight and take notes. It’s 100% free.