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Question 101 ptsWhich way will [he equilibrium shill; for tne exolhermic eaclion below; if addilional NHzlg) is added (and everylhing else is held constanl)?Nzlg) 3...

Question

Question 101 ptsWhich way will [he equilibrium shill; for tne exolhermic eaclion below; if addilional NHzlg) is added (and everylhing else is held constanl)?Nzlg) 3Hzlg)2NHzlg)To the right (more products)Ic will not shiftTo the left (more reactants)

Question 10 1 pts Which way will [he equilibrium shill; for tne exolhermic eaclion below; if addilional NHzlg) is added (and everylhing else is held constanl)? Nzlg) 3Hzlg) 2NHzlg) To the right (more products) Ic will not shift To the left (more reactants)



Answers

Starting with pure products, in which direction will an equilibrium shift if $\Delta G^{*}<0 ?$

In this problem, we want to determine in which direction and equilibrium reaction will shift in order to reestablish equilibrium. If we know that Q is less than K, we think about a generic equilibrium reaction of reacting a reverse oblique producing brought a B. We assume that this reaction is balanced with one mole of each A and B in that both A and B or either gases or equally species and solutions so that we can express both of them in terms of their concentrations. And we can write out the expression for the reactant reaction quotient. Q. Be equal to the concentration of product. Be divided by the concentration of reacting. A. This ratio is also equal to the equilibrium. Constant expression K Where again we have the concentration of product be invited by the concentration of reactive A at a given temperature, the value for the equilibrium constant K is constant. So that means for a given Were you given concentration of product beef, whatever, whatever the concentration of of reacting a is when we divide the concentration of B I. A. It has to always equal this constant value at equilibrium, and so in order for you to be less than cave which is the case in this problem, then that means that in order to decrease this ratio that either we decreased the concentration of product, be where we increased the concentration of reacting a and so in order to reestablish equilibrium, meaning in order for Q to to change the concentrations of being A respectively to maintain this ratio again so that when we divide them we get this equilibrium expression this that same constant for K that the system will respond to the disturbance in a way to counter act that. So if we decrease the concentration of B, which would lead to a decrease in Q, then we need to increase the concentration of B in order to get that came ratio back for equilibrium. In order to increase the concentration of product be, we have to four more product in favor the right side of this equilibrium reaction. Likewise, an increase in reacting a would be another way to decrease que so the system would respond by decreasing the concentration of react in a we decrease the concentration of reacting A. It must mean that were forming more product, be since the forward of the former direction of this reaction would lead to a decrease in the concentration of a so Either way you look at it when Q is less than K. In order to reestablish equilibrium, reaction will shift to the right to produce more product so that when you take the concentration of the products invited by the concentration of the reactant, you take that ratio, you can re establish that constant value for the equilibrium.

So this question gives us the reaction to be our and oh gas reversible e reacts to form to an old gas plus PR to guess and tells us that we are at equilibrium and then gives us several different situations and asks whether that will shift the reaction to the left to the right or have there be no change. And so this is essentially asking about the shot Elia's principle which says that at equilibrium when something has changed, the reaction will respond appropriately to return to equilibrium. So a says we are adding and oh, and so since we are adding and oh, we're adding something to the products so the reaction will shift to the left to the reactant ce Um, in order to compensate here, I'm gonna we write that a little a little cleaner. The reaction will shift to the left to compensate for letter B. We're adding ah b R and O to the reaction mixture. And so now, since we've added are reacting, we have to shift to the right in order to compensate Ah, in very similar way as in part A. When we added a product, we had to shift to the reactant CE. Now that we're adding a reactant, we have to shift to the products Let her see says we are losing Br two were removing BR to from the reaction mixture. So now that we're removing br two, we need thio compensate by moving back to the right to regenerate that you're too. And so the answers for A, B and C are right here, and the reasoning is that the shots liaise. Principle says that when a reaction is that equilibrium, if something has changed, it will compensate to go back to that equilibrium.

So let's say we have a reaction where A. Is in equilibrium with B. Let's assume for the moment that both of these are Aquarius. All right. So, if we wanted to uh come up with the law, mass action expression, it would be the concentration of B. Who were the concentration of A. We could find our equilibrium constant if we were at equilibrium, but say that we don't know whether we're at equilibrium or not. Well then we can't calculate K. We calculate the reaction quotient Q. But remember that the expression is the same mass action expression. And so if it turns out that K the numerical value of K is not equal to Q, then you're not at equilibrium. So you have to figure out which way does the reaction have to shift? Does have to shift to the right and make more B. Does it have to shift to the left and make more A. In order to achieve equilibrium? Well, let's say that we know that Q is less than K. Q is less than K. We're going to have to get you to become more like K. Q is going to have to become bigger to become the value of K. And so if Q needs to get bigger, look at the law of mass action expression, larger values of Q. R. When we have larger values of B. And smaller values of A. So if we make more B. B is going to go up, what also has to happen? If we make more B. We use more A. So the value of A goes down. And so if he goes up and it goes down, Q becomes larger. And so to make more B. And use more A. This reaction has to shift to the right and make more B. The opposite would be true of cure larger than Casey. We would have to move to the left. And if Q is equal to Casey, you are at equilibrium.

So were asked in this question to examine what happens to the equilibrium concentration of the reactant CE and products when the equilibrium position shifts to the right. Now, when the equilibrium position shifts to the right without saying is that for whatever reason, the forward reaction or the formation of products has effectively sped up. That means we're creating more products and we're reacting more react. It's now. Once the rate of the Ford reaction increases, eventually the concentration of the products is going to increase and that increased concentration of the products is going to cause. The reverse reaction to increase is well until the rates of the forward and reverse reactions are equal. However, the equilibrium concentration of the product is going to increase, and the equilibrium concentration of the reactant is going to decrease even though the rates of exchange between the two are equivalent at the new position of equilibrium. So in short, the concentration of the reactant CE will decrease once the equilibrium position has shifted to the right or once it's shifted towards the formation of products and the equilibrium concentration of the products will increase


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