What happens if stress is applied to a system at equilibrium

Le Chatelier's Principle states that when a stress is applied to a system at equilibrium, the equilibrium will shift in a direction to partially counteract the stress and once again reach equilibrium.

If a system at equilibrium is subjected to an external stress, the equilibrium will shift to minimize the effects of that stress. Equilibrium is all about rates—the rate of the forward reaction is equal to the rate of the reverse reaction. External stresses are factors that will cause the rate of either the forward or reverse reaction to change, throwing the system out of balance. Le Chatelier's Principle allows us to predict how this will affect our system.

In our unit on Kinetics we examined factors that influenced reaction rates. Recall these factors:. For instance, if a stress is applied by increasing the concentration of a reactant, the reaction will adjust in such a way that the reactants and products can get back to equilibrium. In the case of too much reactant, the reaction will use up some of the reactant to make more product.

It is said in this scenario that the reaction "shifts to the products" or "shifts to the right". If the concentration of a product is increased, there is an opposite effect. The reaction will use up some of the product to make more reactant. The reaction "shifts to the reactants" or "shifts to the left". To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Donate Login Sign up Search for courses, skills, and videos.

Worked example: Calculating the equilibrium total pressure after a change in volume. Next lesson. Current timeTotal duration Google Classroom Facebook Twitter. Video transcript - [Instructor] Le Chatelier's principle says, if a stress is applied to a reaction mixture at equilibrium, the net reaction goes in the direction that relieves the stress.

Change in the concentration of a reactant or product is one way to place a stress on a reaction at equilibrium. For example, let's consider the hypothetical reaction where gas A turns into gas B. And let's say the reaction is at equilibrium. And we suddenly introduce a stress such as we increase the concentration of reactant A. According to Le Chatelier's principle, the net reaction is gonna go in the direction that relieves the stress. And since we increase the concentration of A, the net reaction is gonna go to the right to decrease the concentration of A.

Let's use some particular diagrams so we can get into the details of how the reaction goes to the right. So we're gonna symbolize gas A by red particles and gas B by blue particles.

And for this hypothetical reaction, the equilibrium constant is equal to three at 25 degrees Celsius. Let's start by writing the reaction quotient. Qc is equal to, and we get that from our balanced equation. That would be the concentration of B to the first power divided by the concentration of A, also to the first power.

Let's calculate the concentrations of B and A from our first particular diagram. So B is represented by the blue spheres and there are three blue spheres. If each particle represents 0. So the concentration of B is 0. For A, we have one particles, that's 0. So the concentration of A is 0. And 0. So Qc at this moment in time is equal to three. Notice we could have just counted our particles, three blues and one red and said three over one.

That would have been a little bit faster. So Qc is equal to three and Kc is also equal to three. When synthesizing an ester, for example, how can a chemist control the reaction conditions to obtain the maximum amount of the desired product?

Only three types of stresses can change the composition of an equilibrium mixture: 1 a change in the concentrations or partial pressures of the components by adding or removing reactants or products, 2 a change in the total pressure or volume, and 3 a change in the temperature of the system. In this section, we explore how changes in reaction conditions can affect the equilibrium composition of a system. We will explore each of these possibilities in turn. If we add a small volume of carbon tetrachloride CCl 4 solvent to a flask containing crystals of iodine, we obtain a saturated solution of I 2 in CCl 4 , along with undissolved crystals:.

If we add more CCl 4 , thereby diluting the solution, Q is now less than K. Adding solvent stressed the system by decreasing the concentration of dissolved I 2. Hence more crystals will dissolve, thereby increasing the concentration of dissolved I 2 until the system again reaches equilibrium if enough solid I 2 is available Figure By adding solvent, we drove the reaction shown in Equation Figure The concentration of I 2 decreases initially due to dilution but returns to its original value as long as solid I 2 is present.

We encounter a more complex system in the reaction of hydrogen and nitrogen to form ammonia:. The K p for this reaction is 2. Because the stress is an increase in P H 2 , the system must respond in some way that decreases the partial pressure of hydrogen to counteract the stress.

The reaction will therefore proceed to the right as written, consuming H 2 and N 2 and forming additional NH 3. Initially, the partial pressures of H 2 and N 2 will decrease, and the partial pressure of NH 3 will increase until the system eventually reaches a new equilibrium composition, which will have a net increase in P H 2.

We can confirm that this is indeed what will happen by evaluating Q p under the new conditions and comparing its value with K p. The equations used to evaluate K p and Q p have the same form: substituting the values after adding hydrogen into the expression for Q p results in the following:.

To reach equilibrium, the reaction must proceed to the right as written: the partial pressures of the products will increase, and the partial pressures of the reactants will decrease. Q p will thereby increase until it equals K p , and the system will once again be at equilibrium. Changes in the partial pressures of the various substances in the reaction mixture Equation Some of the added hydrogen is consumed by reacting with nitrogen to produce more ammonia, allowing the system to reach a new equilibrium composition.

We can force a reaction to go essentially to completion, regardless of the magnitude of K , by continually removing one of the products from the reaction mixture. Consider, for example, the methanation reaction, in which hydrogen reacts with carbon monoxide to form methane and water:. This reaction is used for the industrial production of methane, whereas the reverse reaction is used for the production of H 2 Example The expression for Q has the following form:.

Continuing to remove water from the system forces the reaction to the right as the system attempts to equilibrate, thus enriching the reaction mixture in methane. This technique, referred to as driving a reaction to completion , can be used to force a reaction to completion even if K is relatively small. For example, esters are usually synthesized by removing water.

The products of the condensation reaction are shown here. In Chapter 19 "Electrochemistry" , we will describe the thermodynamic basis for the change in the equilibrium position caused by changes in the concentrations of reaction components.

For each equilibrium system, predict the effect of the indicated stress on the specified quantity. Given: balanced chemical equations and changes. Asked for: effects of indicated stresses. Use Q and K to predict the effect of the stress on each reaction.

For each equilibrium system, predict the effect that the indicated stress will have on the specified quantity. Because liquids are relatively incompressible, changing the pressure above a liquid solution has little effect on the concentrations of dissolved substances. Consequently, changes in external pressure have very little effect on equilibrium systems that contain only solids or liquids. In contrast, because gases are highly compressible, their concentrations vary dramatically with pressure.

Hence the concentration of any gaseous reactant or product is directly proportional to the applied pressure P and inversely proportional to the total volume V. Consequently, the equilibrium compositions of systems that contain gaseous substances are quite sensitive to changes in pressure, volume, and temperature.

These principles can be illustrated using the reversible dissociation of gaseous N 2 O 4 to gaseous NO 2 Equation The syringe shown in Figure Decreasing the volume also constitutes a stress, however, as we can see by examining the effect of a change in volume on Q. If the volume is decreased by half, the concentrations of the substances in the mixture are doubled, so the new reaction quotient is as follows:. Because Q is now greater than K , the system is no longer at equilibrium.

The stress can be relieved if the reaction proceeds to the left, consuming 2 mol of NO 2 for every 1 mol of N 2 O 4 produced. This will decrease the concentration of NO 2 and increase the concentration of N 2 O 4 , causing Q to decrease until it once again equals K. Thus, as shown in part c in Figure Hence the color becomes more intense.

Increasing the pressure of a system or decreasing the volume favors the side of the reaction that has fewer gaseous molecules and vice versa. In general, if a balanced chemical equation contains different numbers of gaseous reactant and product molecules, the equilibrium will be sensitive to changes in volume or pressure.

Increasing the pressure on a system or decreasing the volume will favor the side of the reaction that has fewer gaseous molecules and vice versa. For each equilibrium system, write the reaction quotient for the system if the pressure is decreased by a factor of 2 i.

Given: balanced chemical equations. Asked for: direction of reaction if pressure is halved. Two moles of gaseous products are formed from 4 mol of gaseous reactants.