EK In Lecture Chapter 4 Question 84

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StretchDoe

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So I did this as a simple Le Chatelier's principle problem and thought the reaction would shift to the reactant side aka the reverse reaction would increase. However, the answer is that both reactions increase. They saw temperature increases both reactions.

My question is how would I know when to look at a problem like this as a Le Chatelier's problem and when to look at the way EK looked at it.

I know I'm not verbalizing this the best way, so hopefully I make sense.

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So I did this as a simple Le Chatelier's principle problem and thought the reaction would shift to the reactant side aka the reverse reaction would increase. However, the answer is that both reactions increase. They saw temperature increases both reactions.

My question is how would I know when to look at a problem like this as a Le Chatelier's problem and when to look at the way EK looked at it.

I know I'm not verbalizing this the best way, so hopefully I make sense.

A Le Chatelier question will ask where the reaction will end up in equilibrium. That's thermodynamics.

EK was asking a kinetics question (rates), without regards to where the reaction will eventually end up. You could see a similar question on the biochemistry side, if the reaction involved an enzyme instead of heat. Enzyme or heat speeds up the rate of reaction in both directions (i.e. lowers the hump between left and right sides).


As an aside, the MCAT will have "trick" questions in this style, to see who gets excited and clicks on what they think the question is vs. who actually reads what the question is asking.
 
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A Le Chatelier question will ask where the reaction will end up in equilibrium. That's thermodynamics.

EK was asking a kinetics question (rates), without regards to where the reaction will eventually end up. You could see a similar question on the biochemistry side, if the reaction involved an enzyme instead of heat. Enzyme or heat speeds up the rate of reaction in both directions (i.e. lowers the hump between left and right sides).


As an aside, the MCAT will have "trick" questions in this style, to see who gets excited and clicks on what they think the question is vs. who actually reads what the question is asking.

Got it. I get those enzyme questions right, knowing that enzymes increase the forward and reverse rate of reaction. This just got me since it was worded differently. Need to me more careful.

Could you help me with this problem which is similar

 
So I did this as a simple Le Chatelier's principle problem and thought the reaction would shift to the reactant side aka the reverse reaction would increase. However, the answer is that both reactions increase. They saw temperature increases both reactions.

My question is how would I know when to look at a problem like this as a Le Chatelier's problem and when to look at the way EK looked at it.

I know I'm not verbalizing this the best way, so hopefully I make sense.

A simple way to put it, in addition to the answer given above, is that Le Chatelier's principle involves where equilibrium lies whereas rates involve how quickly equilibrium is reached, wherever that equilibrium may be. You're right - since this is an exothermic reaction, increasing the temperature will shift equilibrium towards the left. But that equilibrium will also be reached quicker - hence the increased rates.

Could you help me with this problem which is similar

Well, if you're just starting with all reactants, there will be a huge impetus for the system towards the equilibrium mixture of reactants and products. Since no product is present, the forward rate is extremely large. Since there's no product, there is no reverse rate. But once the system starts generating products, there will be a reverse rate. If there is only a little product, the system will still have a large forward rate and a smaller reverse rate because that's how it will reach the equilibrium distribution. In this way, the forward rate keeps decreasing and the reverse rate keeps increasing until equilibrium is reached.
 
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A simple way to put it, in addition to the answer given above, is that Le Chatelier's principle involves where equilibrium lies whereas rates involve how quickly equilibrium is reached, wherever that equilibrium may be. You're right - since this is an exothermic reaction, increasing the temperature will shift equilibrium towards the left. But that equilibrium will also be reached quicker - hence the increased rates.



Well, if you're just starting with all reactants, there will be a huge impetus for the system towards the equilibrium mixture of reactants and products. Since no product is present, the forward rate is extremely large. Since there's no product, there is no reverse rate. But once the system starts generating products, there will be a reverse rate. If there is only a little product, the system will still have a large forward rate and a smaller reverse rate because that's how it will reach the equilibrium distribution. In this way, the forward rate keeps decreasing and the reverse rate keeps increasing until equilibrium is reached.

So initially the forward rate will be high because only reactants are present. That's the answer I chose D, which was wrong. But you're saying gradually that forward reaction decreases and the reverse reaction increases to reach equilibrium?
 
So initially the forward rate will be high because only reactants are present. That's the answer I chose D, which was wrong. But you're saying gradually that forward reaction decreases and the reverse reaction increases to reach equilibrium?

Yes, because that's the only way you reach equilibrium. At equilibrium, the forward and reverse rates are equal. If the forward rate increases and the reverse rate decreases, those rates will never converge. That's like forward rate = 10 and reverse rate = 2 initially and then forward rate = 11, reverse rate = 1, etc. Those rates won't converge. The first series of charts on this page might help you visualize it: http://www.800mainstreet.com/7/0007-008-le_chatelier.html
 
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