MCAT question a day 4/08/09

[vickpick]

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A particular reaction can be simplified to the following form: A -> B + Heat (i.e. an exothermic reaction). Assume that the reaction occurs in a closed environment. Say that the forward rate of the reaction is k1 and the reverse rate of the reaction is k-1. If temperature of the system was decreased, what would be the expected effect on k1 and k-1?

(a) k1 increases, k-1 decreases
(b) k1 decreases, k-1 increases
(c) Both k1 and k-1 decrease
(d) k1 and k-1 remain constant

I said (a) and that makes sense. The real answer is (c) and I disagree. Anyone else disagree, or has a better explanation.

their explanation: http://mcatquestionaday.com/evalc.php?arg1=316

The question did not say the T decreased to a point where activation won't be possible.

 

[VanBrown]

both rates are affected equally by the drop in temperature.

you probably got caught up on Le Chatlier's (sp?) and thought it meant heat was removed from the system... but that isn't the case...

 

[CielloStelatto]

I also got a bit caught up on this question. I am certain that Le Chatelier's principle does apply to the heat of a reaction. Maybe the true answer to this question would have to be a combination of both Le Chatelier's and reaction kinetics, wherein lowering the temperature decreases the rate of any reaction. We just don't know which one would have more of an effect.

My answer: not possible to determine without more information.

 

[determinedone]

Greetings all!

This is how I thought about it:

1. Closed Environment
2. Effects on Rate of Reaction, and not direction shifts/maintaining equilibrium (general effects on rate of reaction: Temp., concentration, catalysts, etc.)
3. System (and since, it's a closed environment then that means the surroundings has no bearing)

So, if Temperature decreases...we know because of KMT (Kintetic Molecular Theory) that the speed of the particles decreases...therefore, the rate (speed and rate are related) of the reaction must decrease for BOTH forward and reverse reactions (Le Chatelier's principle and since, it's in a closed environment the equilibrium can only be maintained by what's inside the system).

 

[alibai3ah]

I'm not 100% sure. But I believe k ALWAYS changes with tempeature.....and they have a linear relationship irrespective of direction

 

[InquisitiveDoc]

I co-sign on determinedone's response. I know the first thing that you want to do is go straight to Le Chatelier's Principle. However, I think the basic rule involving temperature is its direct relationship with kinetic energy and that is the foundation of the question. If temperature decreases, the rate of the molecules decrease and therefore the collision rate decreases as well. Also, heat wasn't removed from the equation. The temperature of the system was decreased and would therefore, affect the rate of collisions in both directions and cause them to decrease.

Hope that helps!!🙂

 

[Anthony Hartsoc]

First glance, was the obvious Le Chatelier's Principle. However, since it is a closed system nothing can be removed from the reaction therefore Kinetic Molecular Theory applies. Tricky Tricky...

 

[Der Der]

Agreed.

 

[chemtopper]

why you people are talking about Le Chatelier's Principle ???
Here K1 and K-1 are rates of the reactions .Rate of reaction decreases with decrease in temperature .Drop in temperature is for complete system so both the rates are affected in same manner.

 

[CielloStelatto]

ugh, what was I thinking? Ok. I know it has been answered above, but I just wanted to summarize (mostly for my own sake). Really, I have to clarify that the question is asking about the rate of the reaction - this has absolutely nothing to do with maintaining equilibrium, therefore le Chatelier's doesn't apply. Problem solved.

 

[sv3]

unless a question specifically states a reaction is at equilibrium, i resist the urge to use Lechatalier's principle. Not sure if this is a fool proof method but works so far for what i've come across. I always thought that kinetic molecular theory "ruled" until equilibrium was established - then it was all Lechatalier.

What's interesting to me now is what if this question stated that the reaction was in fact at equilibrium? Would both rates still slow down if temperature decreased? I think so. However, the forward reaction would be faster than the reverse (or the reverse that much slower) to restore equilibrium. Wow that's a bit confusing.........anyone confirm this or deny it?

 

[0Complications]

I bit on this one as well...

I think determinedone's explanation is the simplest and most direct.

I'm going to have to be careful with these in the future.