- Le Chȃtelier’s principle: Changing concentration
- Le Chȃtelier’s principle: Changing volume
- Le Chȃtelier’s principle: Changing temperature
- Worked example: Calculating the equilibrium total pressure after a change in volume
- Worked example: Using Le Chȃtelier’s principle to predict shifts in equilibrium
- Le Châtelier's principle
Le Chȃtelier’s principle: Changing temperature
Le Chȃtelier’s principle can be used to predict the effect that a stress like changing temperature has on a system at equilibrium. If the temperature of the system is increased (at constant V), the system will shift in the direction that consumes the excess heat. If the temperature of the system is decreased, the reverse effect will be observed. Created by Jay.
Want to join the conversation?
- Is it always going to be blue over red(or pink)??
- If you mean how the equilibrium expression is defined, then it is always a ratio of the product (multiplication) of the products over the product of the reactants. It also includes exponents to the quantities of the ratio which come form the coefficients of the balanced chemical reaction. Here in the video all the coefficients were 1 so all the exponents were 1 as well, but most reactions involve the use exponents.
We also have to take into account physical states since only aqueous or gaseous chemical species are allowed in the equilibrium expression. We omit pure liquids and solids from equilibrium. But again we don't have to worry about that since all the chemical species are gases so they're both included.
Hope that helps.(5 votes)
- Why equilibrium constant (Kc) decreases if temperature increases?
When Qc>Kc, why there're more products then reactants? How we can know that?(1 vote)
- So the equilibrium constant of a reaction can either increase or decrease if the temperature increases. Whether it does one or the other depends on the enthalpy change; whether it is endothermic or exothermic. If a reaction is endothermic, where heat is absorbed as the reaction progresses from reactants to products, then increasing the temperature will increase the equilibrium constant. The opposite is true if a reaction is exothermic.
This has to do with how we define equilibrium constants and Le Chatelier's principal. Equilibrium constants are defined as the ratio of products to reactants; a fraction where the products are the numerator and the reactants are the denominator. So an increasing equilibrium constant would mean that the numerator, and concentration of products, are increasing too. And a decreasing equilibrium constant would mean that the denominator, and concentration of reactants, are creasing instead.
We also treat the enthalpy change as energy being added to either the reactants or products. An endothermic reaction views it as a reactant, while an exothermic reaction views it as a product. So increasing the temperature in means adding extra energy to that enthalpy energy. In effect it adds additional reactant or product similar to increasing the concentration of chemical species.
So if we increase the temperature of an endothermic reaction where the enthalpy is acting as a reactant then we are adding additional reactant and Le Chatelier's principal states that a reaction will attempt to resist that change. And in this case it will do this by increasing the concentration of the products thereby increasing the equilibrium constant. A similar logic is follow by an exothermic reaction where by increasing the temperature we add energy onto the enthalpy acting as product and cause an increase in the reactants thereby decreasing the equilibrium constant.
If Qc is larger than Kc, then thinking of how equilibrium is defined as a fraction, the numerator (the products) of QC will have a higher concentration compared to Kc.
Hope that helps.(4 votes)
- So when raising the temperature, the reaction will favor the direction where heat is absorbed and when decreasing the temperature, the reaction will favor the direction where heat is released?(1 vote)
- The affect temperature has on a reaction, and which side is favored, depends on its enthalpy, whether it is exothermic or endothermic.
If a reaction is exothermic, more energy is being lost by the system than is being input. So, we can think of the enthalpy in an exothermic reaction as a product. If we increase the temperature, then we have more product and so the reaction shifts towards the reactants. If we decrease the temperature, then have less product and so the reaction shifts towards the products.
If a reaction is endothermic, more energy is being put into the system than is being lost so we can think of enthalpy as a reactant now. If we increase the temperature, then we have more reactant and the rection shifts towards the products. If we decrease the temperature, then we have less reactant and so the reaction shifts towards the reactants.
Essentially enthalpy is viewed like a chemical species in a reaction and behaves similarly if we changed its concentration.
Hope that helps.(2 votes)
- Can we say if we increase T
P will increase if we keep V constant
So that will mean more reactants
decrease T, P will decrease with constant V
So that will mean more products(1 vote)
- I think it depends on which one has more moles - the reactant or the product. I don’t think you can make a general statement like that.(2 votes)
- Why is it that only temperature affects kc even though both pressure and concentration changes the position of equillibrium ?(1 vote)
- It's important here to notice the slight difference between the position of equilibrium and the equilibrium. The equilibrium constant is the invariable ratio of the products to the reactants at a certain temperature, while the position of equilibrium are the concentrations/pressures of the reactants and products. While the constant has a single value, the concentrations/pressures can take on many valid values so as long their ratio is still the same constant. For example 2/1 = 2, but also 4/2 = 2.
So we have concentration or pressure/volume changes to a reaction it'll change the concentrations/pressures for each other chemical species as well (cause a change in the position of equilibrium), but it'll do so that the ratio (the equilibrium constant) is reestablished.
As for why temperature changes the equilibrium constant, this is because it is based on the reactions rates in the forward and reverse directions. Equilibrium is established when the forward and reverse reaction rates are equal. However reaction rates are sensitive to changes in temperature causing them to speed up (or slow down). One of the reasons why is that it changes the frequency of collisions necessary for a successful reaction between atoms/molecules. So if the reaction rates change because of temperature, then too does the equilibrium constant because it is based on the rates.
Hope that helps.(1 vote)
- [Instructor] Le Chateliers's Principle says if a stress is applied to a reaction at equilibrium, the net reaction goes in the direction that relieves the stress. One possible stress is to change the temperature of the reaction at equilibrium. As an example, let's look at the hypothetical reaction where gas A turns into gas B. Delta H for this reaction is less than zero, which tells us this is an exothermic reaction. And for an exothermic reaction, heat is given off, heat is released. Therefore we can go ahead and write plus heat on the product side. Let's say that our hypothetical reaction is at equilibrium, and then we change the temperature, so we're going to increase the temperature. According to Le Chatelier's Principal, the net reaction is gonna go in the direction that decreases the stress. And if we treat heat if we treat like a product and we've increased the temperature, it says if we've increased the amount of one of our products and therefore the net reaction is going to shift to the left to decrease a product. Let's use particulate diagrams and the reaction quotient Q to explain what's going on when we increase the temperature on our reaction at equilibrium. The first particular diagram shows the reaction at equilibrium, and let's prove that by calculating QC at this moment in time. We can get the expression for the reaction quotient, QC by looking at the balanced equation. So we have coefficients of one in front of A and in front of B, therefore QC is equal to the concentration of B to the first power divided by the concentration of A, also to the first power. To find the concentration of B, we know that B is represented by blue spheres, so there are 1, 2, 3 blue spheres, And if each sphere represents 0.1 moles of a substance three times 0.1 is equal to 0.3 moles of B and at the volumes equal to 1.0 liter, 0.3 divided by 1.0 liter is 0.3 Molar. So the concentration of B is equal to 0.3 Molar. There are also three particles of A therefore the concentration of A is also equal to 0.3 Molar. 0.3 divided by 0.3 is equal to one. So QC at this moment of time is equal to one. KC for this reaction is equal to one at 25 degrees Celsius. So QC is equal to KC. And when QC is equal to KC, the reaction is at equilibrium. So for this first particular diagram, the reaction's at equilibrium. Next, we introduced the stress to the reaction at equilibrium and the stress is an increase in the temperature. In general, for an exothermic reaction, increasing the temperature lowers the value for the equilibrium constant. So for this hypothetical reaction at 25 degrees Celsius, KC is equal to one, but since we've increased the temperature, the value for the equilibrium constant is going to decrease. So let's say it goes 2.5, if we increase the temperature to 30 degrees Celsius. So if we calculate QC for our second particular diagram, we still have three blues and three reds, and the volume is still the same, therefore QC is still equal to one, but the difference is KC has now changed, so QC is not equal to KC, so we are not at equilibrium. And in this case, QC is greater than KC 'cause QC is equal to one and KC is equal to 0.5. And when QC is greater than KC, there are too many products and not enough reactants. And therefore the net reaction goes to the left. When the net reaction goes to the left, we're going to have be turned into A. So we should see one blue sphere turn into one red sphere. So if we have three blues and three reds, and one blue turns into a red, that gives us only two blues and four reds now. So when we calculate QC for our third particular diagram, the concentration of B would be 0.2 Molar, and the concentration of A would be 0.4 Molar. So 0.2 divided by 0.4 is equal to 0.5. Well, KC is also equal to 0.5. Therefore QC is equal to KC, and the reaction is at equilibrium. And when a reaction is at equilibrium, the concentrations of reactants and products are constant. Let's go back to our hypothetical reaction at equilibrium, but this time we're going to decrease the temperature. If we treat heat like a product decrease in the temperature is like decreasing the amount of one of the products. Therefore the net reaction will go to the right to make more product. If we approach this problem by thinking about the reaction quotient Q for an exothermic reaction, a decrease in temperature in general causes an increase in the equilibrium constants. And if the equilibrium constant increases, then Q would be less than K. And when Q is less than K, the net reaction goes to the right. The net reaction would continue to go to the right until Q is equal to K, and equilibrium has been re-established. Next let's look at an endothermic reaction where Delta H is greater than zero. When six water molecules complex to a Cobalt two plus ion the resulting complex ion is pink in color. And when for chloride anions complex to a cobalt two plus ion, the resulting complex ion is blue in color. When the pink ion reacts with four chloride anions, the blue ion is formed. And since this reaction is endothermic, we can put heat on the reactant side. We're gonna use these particular diagrams down here to help us understand what happens to an endothermic reaction at equilibrium when the temperature changes, however, these drawings aren't designed to be completely accurate for this particular reaction. They're just to help us understand what color we would see. For example, let's say that this middle particulate diagram represents the reaction at equilibrium. And if there are decent amounts of both the blue ion and the pink ion at equilibrium, the resulting equilibrium mixture, so this is an aqueous solution would appear to be purple or violet. If we were to increase the temperature for this endothermic reaction, we treat heat like a reactant. So increasing the temperature is like increasing the amount of a reactant. And therefore the net reaction will shift to the right to get rid of some of that reactant. Whether the net reaction goes to the right, we're gonna increase in the amount of blue ion, and we're going to decrease in the amount of pink ion. Therefore looking at this particular diagram on the right there are now more blue ions than there are pink ions compared to our equilibrium mixture in the middle. Therefore for this third particular diagram, the resulting aqueous solution is going to look blue. If we think about those using Q in general, for an endothermic reaction, an increase in temperature causes an increase in the equilibrium constant K. And if K increases, then the reaction quotient Q is less than K. And when Q is less than K, the net reaction goes to the right. Now let's go back to the middle particular diagram. And so there are reactions at equilibrium and this time we're going to decrease the temperature. If we treat heat as a reactant, and we decrease the temperature, it's like we're decreasing one of our reactants. Therefore the net reaction will shift to the left to make more of our reactant. And when that reaction goes to the left, we're gonna decrease in the amount of the blue ion, and we're going to increase in the amount of the pink ion. So when we compare the middle particulate diagram to the one on the left and the one that left, there's a lot more of the pink ion, then there is of the blue ion. Therefore the overall solution, the overall aqueous solution is going to appear pink. If we think about what's happening using Q for an endothermic reaction in general, when you decrease the temperature, you decrease the equilibrium constant. And if the equilibrium constant decreases now, Q would be greater than K, which means too many products and not enough reactance. Therefore the net reaction would go to the left. Let's go back to our exothermic reaction. At this time let's pretend like we're starting with only A, so we start with only A and we have none of B. And our goal is to make as much as we possibly can and to do it as fast as possible. One way to increase the rate of a reaction would be to increase the temperature. However, for an exothermic reaction, increasing the temperature decreases the equilibrium constant K. And if you decrease the equilibrium constant K, you would decrease the amount of B that you would have when you reach equilibrium. So we can't run our hypothetical reaction at too high of a temperature because that would decrease the equilibrium constant. So instead to speed up the rate of the reaction, we could add a catalyst, let's look at a graph of concentration of B versus time, and we're gonna start with this blue curve here, which represents the hypothetical reaction without a catalyst, so the uncatalyzed reaction. When time is equal to zero, the concentration of B is zero because we start with only A. And as A turns into B the concentration of B increases over time, and eventually the concentration of B becomes constant. And when the concentration of B becomes constant, the reaction reaches equilibrium. So this dotted line here represents the concentration of B at equilibrium. The yellow line represents the reaction with a catalyst added. So once again, we're starting with only A, so when time is equal to zero, the concentration of B is equal to zero. And as time increases A turns into B, so the concentration of B increases and eventually the concentration of B becomes constant. And the reaction reaches equilibrium. Notice that the reaction reached equilibrium much faster with the addition of the catalyst than it did without the catalyst. So the addition of a catalyst allows a reaction to reach equilibrium faster. However, notice that the final equilibrium concentration of B is the same for both the uncatalyzed reaction and the catalyzed reaction. Therefore, the addition of a catalyst does not change the composition of the equilibrium mixture. And that's because the catalyst speeds up both the forward reaction and the reverse reaction, but the rates are still equal. And since the rates are equal, there's no change in the composition of the equilibrium mixture.