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Topic: Effect of temperature on reaction rate  (Read 4603 times)

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Offline sodium.dioxid

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Effect of temperature on reaction rate
« on: May 08, 2012, 09:16:43 PM »
I read that at a certain temperature, only a fraction of the molecules have enough energy to react. Does this mean that (at that particular temperature), after the reactable fraction of molecules react, the remaining molecules won't react?

If yes, then why do certain problems ask to find the reaction constant k using the relationship rate=k[A][ B]? The reaction constant itself is changing as the reaction progresses.

This collision theory chapter is very complicated. Please answer all the questions and answer them directly. I greatly appreciate your help.
 
« Last Edit: May 09, 2012, 02:33:01 AM by Borek »

Offline Jorriss

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Re: Effect of temperature on reaction rate
« Reply #1 on: May 08, 2012, 11:19:11 PM »
I read that at a certain temperature, only a fraction of the molecules have enough energy to react. Does this mean that (at that particular temperature), after the reactable fraction of molecules react, the remaining molecules won't react?
It's not that you reach a certain fraction and the reaction just suddenly stops. It's that the reverse reaction also occurs and, at equilibrium, the forward and reverse reactions occur at the same rate.

If yes, then why do certain problems ask to find the reaction constant k using the relationship rate=k[A][ B]? The reaction constant itself is changing as the reaction progresses.
The reaction constant does not change as the reaction progresses if you are at a set temperature (and I imagine pressure, but that affect would likely be insignificant for nongas phase reactions). The rate constant is a function of temperature and only parametrically depends on collision frequency, activation energy, etc.  
« Last Edit: May 09, 2012, 02:32:32 AM by Borek »

Offline sodium.dioxid

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Re: Effect of temperature on reaction rate
« Reply #2 on: May 09, 2012, 12:08:42 AM »
It's not that you reach a certain fraction and the reaction just suddenly stops. It's that the reverse reaction also occurs and, at equilibrium, the forward and reverse reactions occur at the same rate.

Actually, I wasn't even referring to an equilibrium reaction. Maybe I didn't communicate the issue well enough. I'll try again:

Consider: A + B  :rarrow: C


at a particular temperature of A, the molecules don't all have the exact same kinetic energy - some have it higher and some have lower. The ones that are sufficiently kinetic meet the energy demands for breaking bonds (reacting); their KE (kinetic energy) is equal to or greater than the activation energy. Once the "reactable" bunch reacts, the reaction is complete. Unless:

1. heat is provided throughout the course of the reaction to get more of the A molecules into the reactable state

or

2. Molecule A, before letting it react with the B, was driven to a temperature that ensured all or most of the molecules were over the energy hump. Only then it was added to B to start the reaction (this way all of the A molecules had the necessary energy prior to introducing them to react with B)


The reaction constant does not change as the reaction progresses if you are at a set temperature (and I imagine pressure, but that affect would likely be insignificant for nongas phase reactions). The rate constant is a function of temperature and only parametrically depends on collision frequency, activation energy, etc.  

You got this right. The key phrase is "set temperature".

My point is this: if you set the temperature at an "edge" temperature (meaning a temperature at which only a fraction of the molecules posses the required reaction energy), then only half of those reactants will react. As the reactable reactants turn into products, the average reactant temperature goes down (because the only reactants remaining are the ones that don't have enough energy to react (as opposed to the start of the reaction when high energy molecules were factored into the average)).

« Last Edit: May 09, 2012, 12:19:59 AM by sodium.dioxid »

Offline Jorriss

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Re: Effect of temperature on reaction rate
« Reply #3 on: May 09, 2012, 12:24:09 AM »
Actually, I wasn't even referring to an equilibrium reaction. Maybe I didn't communicate the issue well enough. I'll try again:

Consider: A + B  :rarrow: C


at a particular temperature of A, the molecules don't all have the exact same kinetic energy - some have it higher and some have lower. The ones that are sufficiently kinetic meet the energy demands for breaking bonds (reacting); their KE (kinetic energy) is equal to or greater than the activation energy. Once the "reactable" bunch reacts, the reaction is complete. Unless:

1. heat is provided throughout the course of the reaction to get more of the A molecules into the reactable state

or

2. A, before letting it react with the B, was driven to a temperature that ensured all or most of the molecules were over the energy hump. Only then it was added to B to start the reaction (this way all of the A molecules have the necessary energy to react before introducing them to B )
Kinetic energy gets redistributed through the system. If you have a mole of particles, and half of them have an energy above X and half have an energy below X, the individual particles which possess kinetic energy above or below X will change. The energy gets moved around. There is no 'reactable' bunch in the way I believe you are using it.

Still, what I said stands. A reaction doesn't stop. Initially, a reactions forward rate is greater than its backwards rate. Eventually, the two rates are equal. It does not mean there are no more reactions. If you could shrink down to the size of a molecule, you'd see, effectively, EVERY molecule reacting all the time, just with no macroscopic change.


I could be misunderstanding your main point but your wording is very misleading.



Offline sodium.dioxid

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Re: Effect of temperature on reaction rate
« Reply #4 on: May 09, 2012, 12:42:49 AM »
The rate constant is a function of temperature and only parametrically depends on collision frequency, activation energy, etc.

I think this is where I am frustrated. I am trying to account for everything all at once. I am so frustrated that I am just throwing random questions out there without realizing it. I'll hit the book one more time and hopefully I can put it all together this time around. Thanks for your help.

Offline Jorriss

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Re: Effect of temperature on reaction rate
« Reply #5 on: May 09, 2012, 12:44:57 AM »
The rate constant is a function of temperature and only parametrically depends on collision frequency, activation energy, etc.

I think this is where I am frustrated. I am trying to account for everything all at once. I am so frustrated that I am just throwing random questions out there without realizing it. I'll hit the book one more time and hopefully I can put it all together this time around. Thanks for your help.
What's your background? Do you know any statistical mechanics?

Offline sodium.dioxid

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Re: Effect of temperature on reaction rate
« Reply #6 on: May 09, 2012, 12:54:33 AM »
What's your background? Do you know any statistical mechanics?

I am a student and I just finished my second year of college. I hadn't heard it before. I just looked it up and it looks like my kind of thing.

Offline fledarmus

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Re: Effect of temperature on reaction rate
« Reply #7 on: May 09, 2012, 08:39:39 AM »
Molecules have a distribution of energy, and the distribution is constantly being reestablished as the molecules bounce off of each other and off other molecules. As you remove the molecules with highest energy by reaction, the distribution is reestablished and the same proportion of molecules gets up into high enough energy to react.

This is why reaction rate depends on the concentration of the starting materials. While the proportion of molecules with high enough energy to react stays constant, the total amount of molecules, and therefore the total number with high enough energy to react, will decrease as the reaction proceeds.

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