- Equilibrium questions
- Dynamic equilibrium
- Writing equilibrium constant and reaction quotient expressions
- Le Chȃtelier’s principle: Changing concentration
- Le Chȃtelier’s principle: Changing volume
- Le Chȃtelier’s principle: Changing temperature
- Changes in free energy and the reaction quotient
- Standard change in free energy and the equilibrium constant
- Galvanic cells and changes in free energy
Le Chȃtelier’s principle: Changing concentration
Le Chȃtelier’s principle can be used to predict the effect that a stress like changing concentration has on a reaction system at equilibrium. If the concentration of a reaction species is increased (at constant T and V), the equilibrium system will shift in the direction that reduces the concentration of that species. If the concentration of a reaction species is decreased, the reverse effect will be observed. Created by Jay.
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- Why is it that "If the concentration of a reaction species is increased, the equilibrium system will shift in the direction that reduces the concentration of that species"??
To clarify :
Why doesn't the equilibrium system shift toward the direction of the concentration?
- If that happened, the reaction would just go on and on until there’s only one compound left instead of two. Obviously, that does not happen in an equilibrium system.
- Let me see if my understanding is correct: Equilibrium is pretty much when the rate of forward reaction is equal and opposite to rate of the reverse reaction. There is no net change in product nor reactant. In other words, reactants and products coexist in the chemical system. In essence, it is pretty much just balance of chemical reactions in the system. Le Châtlier's Principle pretty much just says that if I disturb a system that has this equilibrium, the system will try to restore a 'balance' by shifting the reaction in the direction to minimize the stress you caused on the system. Take the Haber-Botsch Process to manufacture ammonia. N2+ 3H2=2NH3. If I increase the concentration of diatomic nitrogen gas, the system will most likely shift right because the best way to correct that stress you caused and more or less restore a sort of balance is to make more ammonia. Equilibrium also depends on things like temperature, pressure (which, according to Boyle's Law and KMT, is inversely proportional to volume), total pressure (even of non-reacting gas) and catalysts. For example, if you put a catalyst into the system, the equilibrium is not effected, but the reaction takes less time to happen because the Ea (the energy that is needed to be enough for the rxn) has been decreased, so you have increased the forward and reverse equally (both in magnitude and direction). Note, the equilibrium does not necessarily imply that the concentrations are the same, that may or may NOT be the same.(3 votes)
- What grade is this in?(1 vote)
- It's usually taught in the second semester of AP chem or gen chem if you're taking this in college.(3 votes)
- Why is it that we can calculate the order from a "balanced equation" (per Jay at1:16) here, but it's previously been stated that reaction order can only be determined experimentally and NOT from balanced equation? My best guess is that this equation an experimentally determined reaction mechanism and the previous item applies to balanced STOICHIOMETRIC equations?(1 vote)
- This is an equilibrium problem, not a kinetic problem. The rate law (kinetics) of a reaction is based on the coefficients of the slowest elementary reaction which we confirm through experimental data. The equilibrium expression uses the coefficients of the balanced overall chemical reaction.
Hope that helps.(2 votes)
- 2:17if we have the option of just being able to count the particles, then couldn't we say that Qc = Nb/Na?(0 votes)
- Well since equilibrium ratio/reaction quotients have products over reactants it would be Nb/Na. But I understand your main point. The issue with just using individual particles in a reaction is that here the number of particles has been heavily simplified so that we only have a few interacting with each other. In reality we have vastly more particles so expressing their actual amounts introduces large numbers and more cumbersome math.
But of course here in this problem we've been afforded the luxury of a particulate diagram so we are able to just count particles. However most equilibrium problems do not provide such a diagram so we have to rely instead solely on mathematical version of equilibrium where we use concentrations of the chemicals in molarity for the ratio.
Hope that helps.(2 votes)
- [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 0.1 moles of a substance, and the volume of the container is one litter, since we have three particles, that'd be three times 0.1, which is 0.3 moles divided by a volume of one litter is 0.3 molar.. So the concentration of B is 0.3 molar. For A, we have one particles, that's 0.1 mole divided by one litter, which is 0.1 molar. So the concentration of A is 0.1 molar. And 0.3 divided by 0.1 is equal to three. 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. So I should have written a C in here. So when Qc is equal to Kc, the reaction is at equilibrium. So in this first particular diagram here where Qc is equal to Kc, the reactions are at equilibrium. Next, we're gonna introduce a stress to our reaction at equilibrium. We're going to increase the concentration of A. So here, we're gonna add four particles of A to the reaction mixture at equilibrium. The second particulate diagram shows what the reaction looks like right after we add those four red particles. So we started with one red particle and we added four. So now there's a total of five red particles. And we still have the same three blue particles that we had in the first particular diagram. Let's calculate Qc at this moment in time. So just after we introduced the stress. Since there are three blue particles and five red particles, Qc is equal to three divided by five, which is equal to 0.6. Since Qc is equal to 0.6, and Kc is equal to three, at this moment in time, Qc is less than Kc. So there are too many reactants and not enough products. Therefore, the net reaction is going to go to the right and we're going to decrease in the amount of A, and we're gonna increase in the amount of B. The third particular diagram shows what happens after the net reaction moves to the right. So we said, we're gonna decrease the amount of A and increase in the amount of B. We're going from three blues in the second particular diagram to six blues in the third. And we're going from five reds to only two reds. Therefore, three reds must have turned into blues to get the third particular diagram on the right. And if we calculate Qc for our third particular diagram, it'd be equal to six divided by two, which is equal to three. So at this moment in time, Qc is equal to Kc. They're both equal to three. So equilibrium has been reestablished in the third particular diagram. It isn't always necessary to calculate Q values when doing a Le Chatelier's changing concentration problem. However, for this hypothetical reaction, it's useful to calculate Q values to understand that we're starting at equilibrium and then a stress is introduced such as changing the concentration of a reaction or product. And that means the reaction is no longer at equilibrium. Le Chatelier's principle allows us to predict which direction the net reaction will go or we could also use Q to predict the direction of the net reaction. The net reaction will continue going in that new direction until Q is equal to K again and equilibrium has been reestablished, let's look at another reaction. This is the synthesis of ammonia from nitrogen gas and hydrogen gas. And let's see the reaction is at equilibrium. So let's also look at this on a graph of concentration versus time. At equilibrium, the concentrations of reactants and products are constant, which is why we see these straight lines here for the concentration of hydrogen, ammonia, and nitrogen. And let's introduce a stress to the system at equilibrium. So right now we are at equilibrium and all the concentrations are constant. And let's increase the concentration of hydrogen. So we can see that on our graph. So right here, there's a sudden increase in the concentration of hydrogen. Adding hydrogen means that Q is no longer equal to K and therefore the reaction is not at equilibrium. So let's go ahead and write over here, Now we're not at equilibrium. And Le Chatelier's principle allows us to predict which direction the net reaction will move. So since we added a stress, the stress being increased concentration of hydrogen, the net reaction is gonna move to the right to get rid of some of that hydrogen that was added. And when the reaction goes to the right, the amount of ammonia will increase. And that's what we can see right here on this red line here, the amount of ammonia is increasing. And the amount of ammonia increases because nitrogen and hydrogen are reacting to form ammonia. Therefore, the amount of nitrogen and hydrogen will decrease. Here we can see the amount of hydrogen is decreasing. And down here, we can see the amount of nitrogen is decreasing. The reaction will continue to go to the right until equilibrium is reestablished. And that happens at the second dotted line here. And we know that because we can see all of these concentrations are now constant. So the reaction has reached equilibrium. So far we've only talked about change in the concentration of a reactant so for example, if we increase the concentration of hydrogen, the net reaction goes to the right. We could also say shifts to the right. So for a reaction at equilibrium, if you increase the concentration of reactants, such as the concentration of hydrogen or the concentration of nitrogen, the reaction will shift to the right to decrease the amount of one of those reactants. And if our reaction is at equilibrium and we were to increase the amount of our product, increase the amount of ammonia, so this time the stress place is increased concentration of a product, Le Chatelier's principle says the net reaction is going to move in the direction that decreases the stress. So in this case, the net reaction would go to the left to decrease the amount of ammonia. And if our reaction is at equilibrium and we were to decrease the concentration of our product, the net reaction would shift to the right to make more of the product. Or if we decrease the concentration of one of our reactants, let's say of nitrogen, in this case the reaction will shift to the left to make more of our reactant.