Think about a chemical reaction at equilibrium. In the generic reaction

for example, how is the equilibrium affected if we:

remove some product D ?

add some product A ?

increase or decrease the temperature ?

increase or decrease the pressure ?

**Le Châtelier's principle** will help us to predict the direction of the equilibrium shift when changes like this are made to the reaction mixture.

In this section we'll be using phrases like "shifts to the right" or "shifts to the left." When an equilibrium shifts to the right, it produces more product than before whatever caused it to shift. The opposite is true for shifting to the left (toward reactants).

Think about a seesaw that is in balance. Now put some extra weight on one side. You'll have to add weight to the other side to regain balance. Chemical reactions make the same sorts of adjustments automatically, and Le Châtelier's principle will help us to understand it.

Le Châtelier's principle is named for Henry Louis Le Châtelier, a French chemist who died in 1936 at the age of 85. Versions of Le Châtelier's principal are used in economics to predict the direction in which economic indicators will shift upon certain changes.

Any change that occurs in a system at chemical equilibrium causes an adjustment in the concentrations of reactant(s) or product(s) that re-establishes the equilibrium.

Think about a mass suspended from a spring. When we stretch the spring away from its **equilibrium position** (a perfect balance between the gravitational force which pulls the mass down and the pulling force of the spring), a restoring force tries to pull it back toward equilibrium. As the spring is compressed, another restoring force tries to extend it toward equilibrium (**EQ** in the animation).

I've put this spring on its side so gravity will have to work in the left-right direction, but you get the idea. Chemical reactions behave in much the same way: Depending on the disturbance from equilibrium, the system will adjust in the direction of regaining equilibrium.

What we want to understand here is what the effect of changing the concentration of *one* reaction component is on all of the others. Let's start very simply with a reversible reaction that converts some molecule $A$ to $B$ and $B$ back to $A$:

$$A \leftrightharpoons B$$

The equilibrium constant expression is easy to write:

$$K_{eq} = \frac{[B]}{[A]}$$

Now let's consider making some changes to the reaction mixture once it's at equilibrium.

Imagine that once the mixture has reached equilibrium, we could remove some product, $B$. Doing that would upset the equilibrium, and it would readjust, more $A$ converting to $B$ until the equilibrium was re-established.

Adding reactant has the same effect as removing product; both **move the reaction to the right**, toward product.

Removing reactant, $A$, causes the reverse reaction to increase. Enough $B$ converts back to $A$ so that equilibrium is re-established.

Adding product also causes an upset in the equilibrium that will drive the reaction **to the left**, back toward reactants.

Generalizing to more complicated reactions

These principles are the same for more complicated reactions like

$$aA + bB \leftrightharpoons cC + dD$$

Adding reactant(s) or removing product(s) causes the equilibrium to shift to the right.

Removing reactant(s) or adding product(s) causes the equilibrium to shift to the left.

These ideas can be very important in chemistry of all kinds. Consider, for example, a reaction in which you'd like (and are able) to continuously remove a product, something you desire to have. Then removing product and adding reactants is a way to keep the reaction continuously moving to the right, toward the product(s) you'd like to form.

For any chemical reaction at equilibrium,

Adding reactant(s) or removing product(s) causes the equilibrium to shift to the right, toward products.

Removing reactant(s) or adding product(s) causes the equilibrium to shift to the left, toward reactants.

We should think about what actually happens over time as a system in chemical equilibrium is disturbed, then moves to re-establish equilibrium.

Consider changing the concentration of a reactant, $A$. If $[A]_o$ is the initial equilibrium concentration of $A$, then adding $A$ rapidly will rapidly increase the concentration of $A$. But that will upset the equilibrium, causing more $A$ to react to form products, thus reducing its concentration.

The new concentration of $A$ will be $[A]'$. Notice that the new concentration won't (can't be the same as the old concentration. It will go up by a little.

Let's do a sample calculation and see if we can follow the changes in all components of the equilibrium

$$A + B \leftrightharpoons C + D$$

Let's let $K_{eq} = 0.25$ (picked at random, just for this generic example), then the equilibrium constant expression is:

$$K_{eq} = \frac{[C][D]}{[A][B]} = 0.25$$

Now for simplicity, let's let

$$[A] = [B] = 1$$

Next we construct an ICE table:

$$ \begin{array}{lcrc} & \underline{I} & \underline{C} & \underline{E} \\[3pt] [A] & 1.0 & -x & 1-x \\[3pt] [B] & 1.0 & -x & 1-x \\[3pt] [C] & 0 & +x & x \\[3pt] [D] & 0 & +x & x \end{array}$$

Then

$$\frac{x^2}{(1 - x)^2} = 0.25$$

Both sides are squares, so taking the square root of both sides gives:

$$\frac{x}{1 - x} = 0.5$$

A little rearrangement and some easy algebra gives us

$$ \begin{align} x &= 0.5 - 0.5 x \\ 1.5 x &= 0.5 \\ x &= 0.33 \end{align}$$

and our final set of equilibrium concentrations is

$$ \begin{align} [A] &= 0.67 \\ [B] &= 0.67 \\ [C] &= 0.33 \\ [D] &= 0.33 \end{align}$$

Now let's make a change. We'll rapidly increase the concentration of $A$ from 1 to 2 (I'm not using units here just to keep things simple). The new ICE table is

$$ \begin{array}{lcrc} & \underline{I} & \underline{C} & \underline{E} \\[3pt] [A] & \color{#E90F89}{2.0} & -x & 2-x \\[3pt] [B] & 1.0 & -x & 1-x \\[3pt] [C] & 0 & +x & x \\[3pt] [D] & 0 & +x & x \end{array}$$

Plugging the equilibrium values into the equilibrium constant expression gives us a slightly more complicated expression, but it's still do-able:

$$\frac{x^2}{(1 - x)(2 - x)} = 0.25$$

Expanding the denominator gives

$$\frac{x^2}{x^2 - 3x + 2} = 0.25$$

Now we multiply both sides by that denominator

$$x^2 = 0.25 x^2 - 0.75 x + 0.5$$

and gather terms to get a nice quadratic equation:

$$0.75 x^2 + 0.75 x - 0.5 = 0$$

It can be solved to find

$$x = 0.457$$

And our new set of equilibrium concentrations is

$$ \begin{align} [A] &= 1.54 \\ [B] &= 0.54 \\ [C] &= 0.457 \\ [D] &= 0.457 \end{align}$$

Now we can construct a diagram that models what happens as the equilibrium is re-established after the addition of more $A$. It might look like this:

See if you can follow the course of the reaction from the point of view of each of the components. Notice that the concentration of $A$ is greater than before the addition of more $A$, but that some of it has reacted with more $B$ to produce increases in the concentrations of $C$ and $D$.

Things can get a little more complicated when the reaction coefficients are bigger than 1, but the general idea is the same. There is a short time when the reaction is out-of-balance, until it achieves equilibrium, and at that point, all components will have a different equilibrium concentration than before.

Predict the direction of the equilibrium shift. Roll over/tap on the question to see the answer.

5. |
The equilibrium $$H_3O^+_{(aq)} + HCO_{3 \, (aq)}^- \leftrightharpoons 2 H_2O_{(l)} + CO_{(l)}$$ occurs when a carbonated soda bottle is opened. In which direction is the equilibrium shifted when the bottle is opened, and why? ## SolutionWhen the bottle is opened, CO If we take the equilibrium expression, $$K_{eq} = \frac{[CO_2]}{[H_3O^+][HCO_3^-]},$$ and rearrange it like this: $$[CO_2] = K_{eq} [H_3O^+] [HCO_3^+],$$ it's easy to see that as the concentration of CO |

5. |
The synthesis of ammonia (NH $$N_{2 \, (g)} + 3 H_{2 \, (g)} \leftrightharpoons 2 NH_{3 \, (g)} \phantom{000} K_{eq} = 9.6 \text{ at } T = 300^{\circ}C$$ (a) Calculate the equilibrium partial pressures of all components if P (b) Calculate new equilibrium partial pressures if half of the ammonia is removed from the equilibrium mixture. ## SolutionOur equilibrium expression for this reaction is $$\frac{P_{NH_3}^2}{P_{N_2} P_{H_2}^3} = 9.6$$ Our equilibrium ICE table looks like this: So we're left to solve this equation for x: $$\frac{(2x)^2}{(1 - x)(1 - 3x)^3} = 9.6$$ We can do that numerically on a calculator by making this simple rearrangement, $$9.6(1 - x)(1 - 3x)^3 = 4x^2,$$ and simply finding the intersection(s). This graph from a TI84 calculator shows two intersections. The rightmost gives x > 1, which can't happen, so we discard that solution. The left solution gives x = 0.231, so our equilibrium amounts are: $$ \begin{align} P_{NH_3} &= 0.462 \; \text{atm} \\[5pt] P_{N_2} &= 0.769 \; \text{atm} \\[5pt] P_{H_2} &= 0.307 \; \text{atm} \end{align}$$ Now if we remove half of the equilibrium pressure of ammonia (NH Similarly, we now need to solve this equation: $$\frac{(0.231 + 2x)^2}{(0.796 - x)(0.307 - 3x)^3} = 9.6$$ A similar calculation gives x = 0.032. Using our table, we obtain the new equilibrium values: $$ \begin{align} P_{NH_3} &= 0.231 + 2(0.032) = 0.295 \; \text{atm} \\[5pt] P_{N_2} &= 0.769 - 0.032 = 0.737 \; \text{atm} \\[5pt] P_{H_2} &= 0.307 - 3(0.032) = 0.275 \; \text{atm} \end{align}$$ You can see that removal of product led to formation of more ammonia. The Haber process works by continuous removal of ammonia and resupply of H |

Now that you're familiar with how changes in concentration affect equilibria, it's fairly simple to figure out how temperature changes will affect them.

The key is to treat *heat* as a reactant or product, whichever is appropriate for the thermodynamics of the process. Here's what's meant by that.

Consider the reaction

$$CH_{4 \, (g)} + 2 H_2S_{(g)} \leftrightharpoons CS_{2 \, (g)} + 4 H_{2 \, (g)}$$

This reaction is known to be exothermic, so we could treat heat as another product, and write the reaction like this:

$$CH_{4 \, (g)} + 2 H_2S_{(g)} \leftrightharpoons CS_{2 \, (g)} + 4 H_{2 \, (g)} + \color{red}{\text{heat}}$$

Now consider what happens when we heat up this reaction mixture. The equilibrium will shift to the left, just as if *heat* were a chemical product and we added more.

On the other hand, if we cooled down this reaction, which means removing heat, we'd be removing a product, and we'd expect the reaction equilibrium to move to the right. That's pretty handy to know if your goal is to make as much CS_{2} as possible – just cool the reaction down a little.

Now consider the endothermic reaction, the dissolving of ammonium chloride in water:

$$NH_4Cl_{(s)} + \color{red}{\text{heat}} \leftrightharpoons NH^+_{4 \, (g)} + Cl^-_{(aq)}$$

This time if we add heat, we predict that the reaction will shift to the right in order to re-equilibrate. And if we cool the mixture, removing a "reactant," we'd expect the equilibrium to shift to the left, toward reactants.

In this case, to dissolve more ammonium chloride, we'd want to heat the mixture up.

**reactant** or a **product**, as appropriate. Then increasing temperature is adding to heat and decreasing it is reducing the amount of added heat.

1. |
Consider the exothermic reaction $$CO_{(g)} + H_2O_{(g)} \leftrightharpoons CO_{2 \, (g)} + H_{2 \, (g)}$$ What would happen to the equilibrium if - The reaction mixture was heated up?
- Steam was added?
- The temperature was reduced?
## Solution(a) Because the reaction is exothermic, heat is a product. Adding products causes the reaction to move to the left, so heating this reaction up will cause a shift toward reactants. (b) Steam can be a mixture of gaseous and liquid water. If it's hot enough, it's all H (c) Reducing the temperature means removing heat. Because heat is a product of an exothermic reaction, cooling the reaction should move the equilibrium toward products. |

2. |
Consider the endothermic dissolution of urea in water $$CH_4N_2O_{(s)} \leftrightharpoons CH_4N_2O_{(aq)}$$ - What would happen if the temperature was elevated?
- If the temperature was reduced?
## Solution(a) This reaction is endothermic, which means that heat can be considered to be a reactant. Adding heat, then, should move the equilibrium toward the right, toward the product. (b) Likewise, if the temperature is reduced, the equilibrium should shift to the left. |

The effect of changing pressure on an equilibrium is a little more tricky. We'll divide it into two parts,

(1) adding a non-reactive gas to raise the pressure, and

(2) raising the pressure of an equilibrium mixture by reducing its volume.

Think about a gaseous mixture at equilibrium. The concentration of each gas can be expressed using the ideal gas law,

$$PV = nRT$$

by the rearrangement

$$\frac{n}{V} = \frac{P}{RT}$$

$n$ is the number of moles, so $n/v$ is the molar concentration, and $P$ is the *partial* pressure of the gas of interest. Recall that $R$ is the molar gas constant $(R = 0.0821 \, \text{ J/mol·K}$ and $T$ is the Kelvin temperature.

The concentration of each gas is related to its own partial pressure, and not the *total pressure*, so raising the total pressure by simply introducing other gases (so long as they don't interfere with the equilibrium by reacting with any of the constituents) shouldn't affect the equilibrium.

There are limits to this, of course. Sometimes adding an inert gas can increase the probability of three-body collisions, in which the inert particle can carry away excess energy that allows a reaction to happen, so rates of reaction can be affected, sometimes differentially (more forward than reverse, for example).

The partial pressure of each component of a gas mixture is

$$P = \frac{nRT}{V}$$

An increase in volume reduces the partial pressure of each component, and an decrease in volume increases the partial pressure.

Consider the simple reaction

$$A + B \leftrightharpoons C + D$$

One mole of $A$ and $B$ produce one mole each of $C$ and $D$. In this case, a change in volume changes each partial pressure, $P_A$, $P_B$, $P_C$ and $P_D$, in the same proportion, so there won't be a change in the equilibrium.

But consider another reaction

$$A + B \leftrightharpoons 3C$$

In this case two moles of reactants produce three moles of product, and there will be a difference in partial pressures with a change in volume because the number of moles, $n$ is different.

On the left side of this equation there are two moles of gas, but on the right there are three. A change in volume will affect the partial pressure of each gas in proportion to its coefficient – the number of moles present in the reaction. When the volume of this reaction is reduced, the partial pressure of $C$ will increase more rapidly than those of $A$ or $B$, thus the reaction equilibrium will tend toward the left, toward products and toward fewer moles.

Conversely, when the volume is increased, the partial pressure on the right will decrease faster than those on the left, forcing the reaction to the right.

Let's do an example to further illustrate this idea.

Consider the synthesis of ammonia,

$$N_{2 \, (g)} + 3 H_{2 \, (g)} \leftrightharpoons 2 NH_{3 \, (g)}$$

If the volume is reduced ten-fold, in which direction will the reaction adjust in order to maintain equilibrium?

**Solution**

The equilibrium constant expression is

$$K_{eq} = \frac{P_{NH_3}^2}{P_{N_2}\cdot P_{H_2}^3}$$

where

$$K_{eq} = 4.34 \times 10^{-3} \; \text{at T = 300˚C}$$

Now according to the ideal gas law, the partial pressure of a component of a gas mixture is

$$P = \frac{nRT}{V}$$

$R$ is a constant and if we keep the temperature constant, the partial pressure is proportional to the number of moles divided by the volume (the concentration):

$$P \propto \frac{n}{V}$$

Now as the volume is decreased, the partial pressures of components of the reaction mixture will increase or decrease in proportion to the number of moles present. In the case of the ammonia synthesis, $P_{NH_3}$ and $P_{{H_2}$ will rise faster than $P_{N_2}$.

Here is a look at the partial pressures of our three components as the volume is decreased from 1L (right side of the graph) to 0.1L (left side). Notice that around 0.2L of volume, $P_{NH_3}$ crosses the $P_{N_2}$ curve. So at high pressures, the product is favored over the reactants, or the equilibrium shifts to the right, toward the side with fewer total moles.

There is a simple set of rules for predicting how equilibria with different total numbers of moles of reactants and products will behave upon changes in volume (and thus pressure):

When the volume is *decreased*, the equilibrium moves toward the side with fewer moles.

When the volume is *increased*, the equilibrium moves toward the side with more total moles.

For a gaseous mixture at equilibrium, in which the total number of moles of reactant(s) is *different* than the total number of moles of product(s),

When the volume is **decreased**, the equilibrium moves in the direction of the side with **fewer** total moles, and

when the volume is **increased**, the equilibrium moves in the direction of the side with **more** total moles.

A catalyst is something that increases the rate of a chemical reaction. Catalysts have no effect on equilibrium, however, because they increase both the forward and reverse rates of chemical reaction. It doesn't sound like there's much to gain from using a catalyst, but consider most of the chemical reactions that take place in cells.

Left to their own without catalysts (which we call enzymes in biology), most biochemical processes would never happen. That's how slow they are. So catalysts speed up reactions, sometimes allowing them even to occur at all.

Equilibrium is a fact of reversible chemical reactions with which we always have to cope.

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