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. By lowering the energy of the transition state, which is the rate-limiting step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly. Catalysis : A catalyst speeds up a reaction by lowering the activation energy required for the reaction to proceed.
To reiterate, catalysts do not affect the equilibrium state of a reaction. In the presence of a catalyst, the same amounts of reactants and products will be present at equilibrium as there would be in the uncatalyzed reaction. To state this in chemical terms, catalysts affect the kinetics, but not the thermodynamics, of a reaction. In this case the model has been set so the activation energy is high. Try running the reaction with and without a catalyst to see the effect catalysts have on chemical reactions.
Run the model to observe what happens without a catalyst. Pause the model. How does changing pressure and volume affect equilibrium systems? If you increase the pressure of a system at equilibrium typically by reducing the volume of the container , the stress will best be reduced by reaction that favors the side with the fewest moles of gas, since fewer moles will occupy the smallest volume.
Conversely, if you decrease the pressure by increasing the volume of the container , equilibrium will shift to favor the side with the most moles of gas, since more moles will occupy a greater volume.
If both sides of the equation have the same number of moles of gas, then there will be no change in the position of equilibrium.
Next, we think about the stress that was applied to the reaction at equilibrium. We decreased the volume. And if we look at the volumes here, we're going from 1. So we're decreasing the volume by a factor of two, which would cause an increase in the pressure by a factor of two. And changing the volume would change the concentration. So instead of 0. So the concentration has doubled. So if we calculate Q for our second particular diagram, we plug in the concentration of C, which is 0.
So Q is not equal to K, therefore the reaction is not at equilibrium for our second particular diagram. Let me write in here, not at equilibrium. In this case, Qc is greater than Kc, which tells us we have too many products and not enough reactants. Therefore the net reaction goes to the left to get rid of some of the products and to increase the amount of reactants.
The net reaction keeps going to the left until we reach equilibrium again. So if we calculate the concentration of C in the third particular diagram, here there are only two particles. So that'd be 0. And since Qc is equal to 0. So Qc is equal to Kc and we're at equilibrium. So equilibrium has been re-established. Since the reaction is at equilibrium in a third particular diagram, the net reaction stops going to the left and the concentration of C remains constant.
Let's apply what we've learned to another reaction, the synthesis of ammonia from nitrogen gas and hydrogen gas. If we have a mixture of these gases at equilibrium, and we introduce a stress, the system like a decrease in volume, the decrease in the volume of the container would cause an increase in the pressure.
And according to Le Chatelier's principle, the net reaction is gonna go in the direction that relieves the stress that was placed on the system. So if the stress is increased pressure, the net reaction says, I wanna move in the direction that decreases that pressure.
So the net reaction moves to the right, because there are four moles of gas on the left and only two moles of gas on the right. And by moving to the right, that goes from four moles of gas to two moles of gas, which decreases the amount of gas and causes a decrease in the pressure.
If we had a mixture of these gases at equilibrium, and we increased the volume and increase in the volume would cause a decrease in the pressure. So the stress this time, is decreased pressure. To relieve the stress, the net reaction, wants to move in the direction that increases the pressure. Therefore, the net reaction is going to move to the left because if it moves to the left, we're going from two moles of gas in the right to four moles of gas on the left.
So that's an increase in the moles of gas and increase in the amount of gas, causes an increase in the pressure. Now let's see what happens, when we have equal amounts of moles of gas on both sides of the equation.
For example, for the hypothetical reaction, where gas A turns into gas B, there's one mole of gas on the reactant side, and there's one mole of gas on the product side, and let's use particular diagrams and reaction quotients to understand what's going on here.
So for our first particular diagram, here's the Qc expression. It's equal to the concentration of B to the first power, divided by the concentration of A to the first power and the concentration of B since B is blue, there's two blue spheres in this first particular diagram.
So two times 0.
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