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# 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.

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• Why equilibrium constant (Kc) decreases if temperature increases?

When Qc>Kc, why there're more products then reactants? How we can know that?
• 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.
• Is it always going to be blue over red(or pink)??

Thanks.

-THEWATCHER
• 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.
• 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?
• 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.
• In Kinetics, we saw that increasing temperature increases the rate of a reaction because it makes the molecules move faster and, therefore, collide with each other with greater frequency and force.

Now we see that the temperature will have a different effect depending on whether the reaction is exothermic or endothermic. I'm wondering why. Wouldn't collisions be more frequent in all cases? Or am I confusing phenomena?
• Kinetics are a different phenomenon than thermodynamics for a reaction.

Kinetics deals with the speed of a reaction and is related to the activation energy of a reaction. Where the activation energy is the minimum amount of energy required for molecules to have a successful collision/reaction. Increasing the temperature of a reaction does give more kinetic energy to the molecules which makes them move faster meaning the molecules collide with more force.

Really what is happening though is that there is a normal distribution of molecules with various energies, and temperature is simply the average of those energies. A fraction of this distribution has greater than the activation energy and can result in successful collisions. Increasing the temperature shifts the distribution so that a greater fraction of the molecules has this minimum activation energy. More potentially successful collisions mean the reaction rate (speed of reaction) increases.

Thermodynamics deals with the relative stability of the reactants to the products, or how energetically favorable either is. Essentially where will the reaction end up at, but now how quickly it will take to get there like in kinetics. Using Le Châtelier’s principle, heat can be seen as either a reactant or a product depending on if the reaction is endothermic or exothermic respectively. So adding or removing heat will shift the reaction left or right, or speed up either the forward or reverse reactions. Ultimately the reaction will reach a new equilibrium. But importantly, the principle doesn’t say how quickly it will take to get to this new equilibrium, just that it will occur.

Hope that helps.
• 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
• 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.
• All these videos on Le Chatelier's principle are raising a big question for me about infinite regress...

For the first example in this video:
1. You increase the temperature, which is like adding heat. This decreases the equilibrium constant to 0.5
2. B would combine with this heat and make more of A.
3. Now the Qc is 0.5, so it is in equilibrium.

But in stage 2, by consuming this heat, hasn't the temperature gone back down now?
Therefore, wouldn't the equilibrium constant rise back up towards 1?

And isn't this effect infinite? I.e. the equilibrium constant is constantly shifting, as the heat is absorbed or given off to rebalance the initial heat change.

So I guess my question can be summarised like this:
if a reaction gives off heat which changes the temperature, and changing the temperature affects the equilibrium constant, then how can equilibrium ever be achieved?

I'm visualising a race where every step you take moves the finish line...
• Why is it that only temperature affects kc even though both pressure and concentration changes the position of equillibrium ?
• 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)
• Why if you speed the forward and reverse reaction, you are no changing the rate of the overall reaction? Is not also faster?
(1 vote)
• There’s no single overall reaction rate. There’s two reaction rates for a reversible reaction; a forward and reverse reaction rate.
• So in the hypothetical reaction at the end of the video, if the goal was to have as much of the product as possible, couldn't you decrease the temperature?

Because this would raise the equilibrium constant to favour more product? (but also it would slow down the rate of reaction right?)
(1 vote)
• What determines the preferred way of writing a reaction?

E.g. is it just as valid to write A ↔ B with an exothermic ΔH and product-favoured Kc, as it is to write B ↔ A with an endothermic ΔH and reactant-favoured Kc?
(1 vote)
• There isn’t really a dominant way. Both ways are valid. It’s really down to which reaction, the forward or reverse reaction, you want to emphasize.
(1 vote)