How can equilibrium change

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. Since Qc is equal to 0. 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. Conversely, lowering the temperature on an endothermic reaction will shift the equilibrium to the left, since lowering the temperature in this case is equivalent to removing a reactant. For an exothermic reaction, heat is a product.

Therefore, increasing the temperature will shift the equilibrium to the left, while decreasing the temperature will shift the equilibrium to the right. In which direction will the equilibrium shift if the temperature is raised on the following reaction? Our heat of reaction is positive, so this reaction is endothermic. Since this reaction is endothermic, heat is a reactant. It also demonstrates an easy and convenient method for making predictions about the effects of temperature, concentration, and pressure.

Catalysts speed up the rate of a reaction, but do not have an affect on the equilibrium position. Reactions can be sped up by the addition of a catalyst, including reversible reactions involving a final equilibrium state. Recall that for a reversible reaction, the equilibrium state is one in which the forward and reverse reaction rates are equal. In the presence of a catalyst, both the forward and reverse reaction rates will speed up equally, thereby allowing the system to reach equilibrium faster.

However, it is very important to keep in mind that the addition of a catalyst has no effect whatsoever on the final equilibrium position of the reaction. It simply gets it there faster. Recall that catalysts are compounds that accelerate the progress of a reaction without being consumed. Common examples of catalysts include acid catalysts and enzymes. Catalysts allow reactions to proceed faster through a lower-energy transition state. We can add some cheese to the mouse side.

How will this affect the relative ratios? This means if we add reactant, equilibrium goes right, away from the reactant. If we add product, equilibrium goes left, away from the product. If we remove product, equilibrium goes right, making product. If we remove reactant, equilibrium goes left, making reactant. It can do this by favouring the reaction which produces the fewer molecules. If there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium.

Let's look at the same equilibrium we've used before. This one would be affected by pressure because there are 3 molecules on the left but only 2 on the right. An increase in pressure would move the position of equilibrium to the right. Once again, it is easy to suppose that, because the position of equilibrium will move to the right if you increase the pressure, K p will increase as well. Not so! Note: If you aren't happy with this, read the beginning of the page about K p before you go on.

If you sort this out, most of the "P"s cancel out - but one is left at the bottom of the expression. Now, remember that K p has got to stay constant because the temperature is unchanged. How can that happen if you increase P? To compensate, you would have to increase the terms on the top, x C and x D , and decrease the terms on the bottom, x A and x B. Increasing the terms on the top means that you have increased the mole fractions of the molecules on the right-hand side.

Decreasing the terms on the bottom means that you have decreased the mole fractions of the molecules on the left. That is another way of saying that the position of equilibrium has moved to the right - exactly what Le Chatelier's Principle predicts.

The position of equilibrium moves so that the value of K p is kept constant. There isn't a single "P" left in the expression. Changing the pressure can't make any difference to the K p expression. The position of equilibrium doesn't need to move to keep K p constant. Equilibrium constants are changed if you change the temperature of the system.

K c or K p are constant at constant temperature, but they vary as the temperature changes. Note: You might possibly be wondering what the units of K p are.