Le Châtelier’s principle - a generalization that states that:
chemical systems at equilibrium shift to restore equilibrium when a change occurs that disturbs the equilibrium.
An adjustment by a system at equilibrium that results in a change in the concentrations of reactants and products is called an equilibrium shift.
Le Châtelier’s principle allows chemists to predict the qualitative effects of changes in concentration, pressure, and temperature on a chemical reaction system at equilibrium.
We can tell a reaction is at equilibrium if the reaction quotient (Q) is equal to the equilibrium constant (K).
If a system at equilibrium is subjected to a perturbance or stress (such as a change in concentration) the position of equilibrium changes.
Since this stress affects the concentrations of the reactants and the products, the value of Q will no longer equal the value of K.
To re-establish equilibrium, the system will either shift toward the products (if Q < K) or the reactants (if Q > K) until Q returns to the same value as K.
The Relationship Between the Equilibrium Constant and the Reaction Quotient
If reaction quotient (Q_{c}) is greater than equilibrium constant (K_{c}), the numerator of the reaction quotient expression must be very large.
The concentrations of the chemicals on the right side of the equation
(mA + nB + ⇌ xC + yD)
must be more than their concentrations at equilibrium.
In this situation, the system attains equilibrium by moving to the left.
Conversely, if Q_{c} is less than K_{c} , the system attains equilibrium by moving to the right.
Effect of Change in Concentration on Equilibrium
(a) The test tube contains 0.1 M Fe^{3+}.
(b) Thiocyanate ion has been added to solution in (a), forming the red Fe(SCN)^{2+} ion.
Fe^{3+} (aq) + SCN^{−} (aq) ⇌ Fe(SCN)^{2+} (aq).
(c) Silver nitrate has been added to the solution in (b), precipitating some of the SCN − as the white solid AgSCN.
Ag^{+} (aq) + SCN^{−} (aq) ⇌ AgSCN(s).
The decrease in the SCN^{−} concentration shifts the first equilibrium in the solution to the left, decreasing the concentration (and lightening color) of the Fe(SCN)^{2+}.
The stress on the system in the image above is the reduction of the equilibrium concentration of SCN− (lowering the concentration of one of the reactants would cause Q to be larger than K).
As a consequence, Le Châtelier's principle leads us to predict that the concentration of Fe(SCN)2+ should decrease, increasing the concentration of SCN− part way back to its original concentration, and increasing the concentration of Fe3+ above its initial equilibrium concentration.
Collision Theory and Concentration Changes in an Equilibrium System
According to collision theory, entities in a chemical system must collide to react.
When the concentration of an entity in a chemical reaction system is increased, it is more likely that that entity will collide with other entities. There are simply more of them present. However, only collisions between reactant entities can potentially contribute to a chemical reaction. Even so, the more frequently collisions occur overall, the more likely it is that a chemical reaction will take place.
Collision theory explains the response of a chemical reaction system at equilibrium to a change in concentration as the result of random collisions and probability.
When we add more reactant entities in an equilibrium system, the equilibrium shifts to the right because the number of successful collisions for the forward reaction increases.
If, instead, we add more product entities, then the number of successive collisions for the reverse reaction will increase and the equilibrium will shift to the left.
Le Châtelier’s Principle and Changes in Energy
For a given system at equilibrium, the value of the equilibrium constant depends only on temperature.
Changing the temperature of a reacting mixture changes the rate of the forward and reverse reactions by different amounts, because the forward and reverse reactions have different activation energies.
A reacting mixture at one temperature has an equilibrium constant whose value changes if the mixture is allowed to reach equilibrium at a different temperature.
To apply Le Châtelier’s principle and predict how a change in energy will affect a chemical system at equilibrium, we can think of energy as a reactant or a product.
Energy is absorbed in an endothermic reaction. If we consider energy to be a reactant, we can write the word equation:
reactants + energy ⇌ products
Since energy is released during an exothermic reaction, we can consider energy to be a reactant and write
reactants ⇌ products + energy
To use Le Châtelier’s principle to predict how an equilibrium will shift in response to a change in energy, consider how the system can counteract this shift.
Endothermic Reactions
If an endothermic reaction is cooled (thermal energy removed), we can consider that the quantity of one of the reactants has been decreased.
We can therefore predict that the equilibrium will shift to the left (toward the reactants), and energy will be released.
If thermal energy were added to this equilibrium system by heating, the equilibrium would likely shift to the right.
Exothermic Reactions
If thermal energy is removed from an exothermic reaction—where energy is a product—then the equilibrium will shift to the right (toward the products), and energy will be released to counteract the change.
If energy is added to an exothermic reaction, the equilibrium will shift to the left to compensate for the change, and the energy will be used as products are converted to reactants.
Le Châtelier’s Principle and Changes in Gas Volume
Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system.
An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium.
While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for K_{c}) or partial pressure (for K_{p}).
Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression.
Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.
As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components.
In accordance with Le Châtelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress.
The reverse reaction would be favored by a decrease in pressure.
Consider what happens when we increase the pressure on a system in which N_{2}, H_{2}, and NH_{3} are at equilibrium:
N_{2 }(g) + 3H_{2} (g) ⇌ 2NH_{3} (g)
The formation of additional amounts of NH_{3} decreases the total number of molecules in the system because each time two molecules of NH_{3} form, a total of four molecules of N_{2} and H_{2} are consumed.
This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure.
On the other hand, a decrease in the pressure on the system favors decomposition of NH_{3} into N_{2} and H_{2} , which tends to restore the pressure.
Le Châtelier’s Principle and Changes in Gas Volume
(a) The container holds the equilibrium reaction N_{2 }(g) + 3H_{2} (g) ⇌ 2NH_{3} (g) .
(b) The volume of the container is decreased by half, which doubles the total pressure of the system.
(c) The reaction shifts to the right, which decreases the total number of particles in the container and the total pressure by producing more ammonia, NH_{3} (g).
The Effect of a Catalyst on Equilibrium
A catalyst speeds up the rate of a reaction, either by allowing a different reaction mechanism or by providing additional mechanisms.
The overall effect is to lower the activation energy, which increases the rate of reaction.
The activation energy is lowered the same amount for the forward and reverse reactions, however. There is the same increase in reaction rates for both reactions.
As a result, a catalyst does not affect the position of equilibrium. It only affects the time that is taken to achieve equilibrium.
The Effects of Changing Conditions on a System at Equilibrium
Type of reaction | Change to system | Effect on K_{c} | Direction of change |
all reactions | increasing any reactant concentration, or decreasing any product concentration | no effect | toward products |
decreasing any reactant concentration, or increasing any product concentration | no effect | toward reactants | |
using a catalyst | no effect | no change | |
exothermic | increasing temperature | decreases | toward reactants |
decreasing temperature | increases | toward products | |
endothermic | increasing temperature | increases | toward products |
decreasing temperature | decreases | toward reactants | |
equal number of reactant and product gas molecules | changing the volume of the container, or adding a non-reacting gas | no effect | no change |
more gaseous product molecules than reactant gaseous molecules | decreasing the volume of the container at constant temperature | no effect | toward reactants |
increasing the volume of the container at constant temperature, or adding a non-reacting gas at contstant pressure | no effect | toward products | |
fewer gaseous product molecules than reactant gaseous molecules | decreasing the volume of the container at constant temperature | no effect | toward products |
increasing the volume of the container at constant temperature | no effect | toward reactants |
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Calculation of Concentration Changes as a Reaction Goes to Equilibrium
Under certain conditions, the equilibrium constant for the decomposition of PCl_{5} (g) into PCl_{3} (g) and Cl_{2} (g) is 0.0211. What are the equilibrium concentrations of PCl_{5}, PCl_{3}, and Cl_{2} if the initial concentration of PCl_{5} was 1.00 M?
Solution:
Step 1:
Determine the direction the reaction proceeds.
The balanced equation for the decomposition of PCl_{5} is
PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g)
Because we have no products initially, Q_{c} = 0 and the reaction will proceed to the right.
Step 2:
Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.
Let us represent the increase in concentration of PCl_{3} by the symbol x. The other changes may be written in terms of x by considering the coefficients in the chemical equation.
The changes in concentration and the expressions for the equilibrium concentrations are (ICE Table):
PCl_{5} (g) | PCl_{3} (g) | Cl_{2} (g) | |
Initial concentration | 1.00 | 0 | 0 |
Change | -x | +x | +x |
Equilibrium concentration | 1.00-x | 0+x=x | 0+x=x |
Step 3:
Solve for the change and the equilibrium concentrations.
Substituting the equilibrium concentrations into the equilibrium constant equation gives
K_{c} = [PCl 3 ][Cl 2 ] / [PCl 5] = 0.0211
(x*x) / (1.00-x) = 0.0211
This equation contains only one variable, x, the change in concentration. We can write the equation as a quadratic equation and solve for x using the quadratic formula.
0.0211(1.00 − x) = x 2
x 2 + 0.0211x − 0.0211 = 0
An equation of the form ax 2 + bx + c = 0 can be rearranged to solve for x:
x = (−b ± √(b − 4ac)) / 2a
In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a, b, and c yields:
x = (−0.0211 ± √((0.0211) 2 − 4(1)(−0.0211))) / 2(1)
Hence
x = 0.135
or
x = −0.156
Quadratic equations often have two different solutions, one that is physically possible and one that is physically impossible (an extraneous root). In this case, the second solution (−0.156) is physically impossible because we know the change must be a positive number (otherwise we would end up with negative values for concentrations of the products). Thus, x = 0.135 M.
The equilibrium concentrations are:
[PCl_{5}] = 1.00 − 0.135 = 0.87 M
[PCl_{3}] = x = 0.135 M
[Cl_{2}] = x = 0.135 M
{slider title="Approximate Solution Starting Close to Equilibrium"}
Approximate Solution Starting Close to Equilibrium
The Meaning of a Small Equilibrium Constant
Understanding the meaning of a small equilibrium constant can sometimes help to simplify a calculation that would otherwise involve a quadratic equation.
When K_{c} is small compared with the initial concentration, the value of the initial concentration minus x is approximately equal to the initial concentration. Thus, you can ignore x.
Of course, if the initial concentration of a substance is zero, any equilibrium concentration of the substance, no matter how small, is significant.
In general, values of K_{c} are not measured with accuracy better than 5%.
Therefore, making the approximation is justified if the calculation error you introduce is less than 5%.
To help you decide whether or not the approximation is justified, divide the initial concentration by the value of K_{c}.
- If the answer is greater than 500, the approximation is justified.
- If the answer is between 100 and 500, it may be justified.
- If the answer is less than 100, it is not justified. The equilibrium expression must be solved in full.
Problem:
The atmosphere contains large amounts of oxygen and nitrogen. The two gases do not react, however, at ordinary temperatures. They do react at high temperatures, such as the temperatures produced by a lightning flash or a running car engine.
In fact, nitrogen oxides from exhaust gases are a serious pollution problem.
A chemist is studying the following equilibrium reaction.
N_{2} (g) + O_{2} (g) ⇌ 2NO (g)
The chemist puts 0.085 mol of N_{2} (g) and 0.038 mol of O_{2} (g) in a 1.0 L rigid cylinder. At the temperature of the exhaust gases from a particular engine, the value of K c is 4.2 × 10 −8 . What is the concentration of NO (g) in the mixture at equilibrium?
Required:
You need to find the concentration of NO at equilibrium.
Given:
You have the balanced chemical equation. You know the value of K c and the following concentrations: [N_{2}] = 0.085 mol/L and [O_{2}] = 0.038 mol/L.
Analysis:
Step 1 Divide the smallest initial concentration by K_{c} to determine whether you can ignore the change in concentration.
Step 2 Set up an ICE table. Let x represent the change in [N_{2}] and [O_{2}].
Step 3 Write the equilibrium expression. Substitute the equilibrium concentrations into the equilibrium expression. Solve the equilibrium expression for x.
Step 4 Calculate [NO] at equilibrium.
Solution:
Step 1
Smallest initial concentration / K c = 0.038 / 4.2 × 10 −8 = 9.0 × 10^{5}
Because this is well above 500, you can ignore the changes in [N_{2}] and [O_{2}].
Step 2
ICE Table
N_{2} (g) | O_{2} (g) | 2NO (g) | |
Initial concentration | 0.085 | 0.038 | 0 |
Change | -x | -x | +2x |
Equilibrium concentration | 0.085 − x ≈ 0.085 | 0.038 − x ≈ 0.038 | 2x |
Step 3
K_{c} = [NO]^{2} / ([N_{2}][O_{2}])
4.2 × 10 −8 = (2x) 2 / (0.085 × 0.038)
x = √3.39 × 10^{−11} = 5.82 × 10^{−6}
Step 4
[NO] = 2x
Therefore, the concentration of NO (g) at equilibrium is 1.2 × 10^{−5} mol/L.
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