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Reaction Quotient and Le Chatelier's Principle

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**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.

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, K**_{eq}. There are a few different forms of K_{eq}, but you'll most commonly work with **K**_{c} and **K**_{p}:

**K**_{c}measures the**concentrations of aqueous or gaseous species**in a system at equilibrium.**K**_{p}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 K_{eq}, 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 K_{c}, **Q**_{c} measures **concentration**, and like K_{p}, **Q**_{p} 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 K**_{eq}. Furthermore, **if** **Q equals K**_{eq}, **then the reaction is at equilibrium**. This will come in handy later on.

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**:

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

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

Find out more about K_{eq} 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.

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, K_{eq}:

- Changing the
**concentration**of an equilibrium**doesn't affect K**_{eq}. - Changing the
**pressure**of an equilibrium**doesn't affect K**_{eq}either. - However, changing the
**temperature**of an equilibrium*does*affect K_{eq}.

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.

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 K**_{eq}, 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,

- If
**Q is less than K**_{eq}, the**forward reaction**is favored. This increases the relative amount of products until Q equals K_{eq}. - If
**Q is greater than K**_{eq}, the**backward reaction**is favored. This increases the relative amount of reactants until Q equals K_{eq}.

**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 K _{eq} 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.

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 Q _{c} 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 Q_{c} and comparing it to the equilibrium constant K_{c}. If Q_{c} = K_{c}, then the system is at equilibrium. For this reaction, Q_{c} 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 Q_{c} does in fact equal K_{c}. 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 K_{c}, but it does change the value of Q_{c}:

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

Thanks to increasing the concentration of A, Q_{c} is now less than K_{c}. But we know that Q_{c} will try to tend towards K_{c}, 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 Q_{c}.

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.

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 Q _{p} 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 Q_{p}. If Q_{p} equals K_{p}, 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, Q_{p} does in fact equal K_{p}, 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 Q_{p}, we find that Q_{p} has decreased:

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

We know that Q_{p} will try to tend towards K_{p}, 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 Q_{p}.

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:**

**An equal quantity of an inert gas is added, but the system increases in volume so the total pressure remains at 6.0 kPa.****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 Q_{p}, 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 Q_{p}, we find that Q_{p} has increased. It is now greater than K_{p}:

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

The position of the equilibrium, therefore, shifts to the left, favoring the backward reaction in order to reduce Q_{p}.

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 Q_{p} doesn't change. Q_{p} still equals K_{p}, and so the position of the equilibrium doesn't shift.

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 K_{eq}:

- If the forward reaction is
**endothermic**, then increasing the temperature**increases the value of K**_{eq}. - If the forward reaction is
**exothermic**, then increasing the temperature**decreases the value of K**_{eq}.

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 K_{eq} 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 K _{c} and Q_{c}, **

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, Q_{c} equals K_{c} 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 K_{c}. But the concentrations of A and B haven't changed, and so Q_{c} hasn't changed. That means that Q_{c} is now less than K_{c}, 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 Q_{c} so that it tends towards K_{c}.

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 K**_{eq}. 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 K**_{eq}. 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**.

- The
**reaction quotient Q****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 K**_{eq }tells us the relative amounts of products and reactants in a system.*at equilibrium* **Q always tends towards K**_{eq}.**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 K**_{eq}, but they**do change Q**. We can use the relative magnitudes of Q and K_{eq}to predict the shift in the equilibrium's position, which supports Le Chatelier's principle. - A change in
**temperature does change K**_{eq}. We can still compare Q to the new value of K_{eq}in order to predict the shift in the equilibrium's position.

_{eq}_{ }- changes in concentration or pressure only affect the reaction quotient Q.

_{eq} measures the relative amounts of products and reactants in a system at equilibrium.

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