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Topic: rates, temperature  (Read 9597 times)

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Offline 1stplace

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rates, temperature
« on: April 05, 2008, 08:36:58 PM »
Is it likely that the rate of an exothermic reaction will decrease as temperature increases?

I know that rates and enthalpies are different concepts but since the equilibrium will shift towards the reactants in an exothermic reaction, does the forward rate also decrease?

Thanks a lot

Offline Astrokel

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Re: rates, temperature
« Reply #1 on: April 06, 2008, 07:57:26 AM »
Is it likely that the rate of an exothermic reaction will decrease as temperature increases?

I know that rates and enthalpies are different concepts but since the equilibrium will shift towards the reactants in an exothermic reaction, does the forward rate also decrease?

Thanks a lot
Im confused too!

Im thinking abt..

When equilibrium shift towards the reactants in an exo reaction, the relative rate of backward reaction is higher than the rate of forward reaction. ----- (1)

According to Arrhenius equation, k = Ae^-(Ea of reactants to products/R(increases in temp))

Since products to reactants is an exo reaction, the increase in temp causes k to increase, therefore leading to rate of forward reaction increases. ---- (2)

So whats the relative result of (1) and (2)

or maybe i have some confused over fundamental concepts?

Hopefully you guys can guide me.  :)

Kelvin  ;D

No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline Yggdrasil

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Re: rates, temperature
« Reply #2 on: April 06, 2008, 02:03:30 PM »
Well, it depends on what you define as the forward rate of reaction.  When most people consider the  forward rate of reaction, they imagine a case where there are no products, so the rate of the reaction is not affected by the backward rate of reaction.  In this case, it is correct to apply the Arrhenius equation and conclude that rate increases with temperature.

Once you start looking at the overall rate of reaction (forward rate minus backward rate), it gets more complicated.

Offline Astrokel

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Re: rates, temperature
« Reply #3 on: April 07, 2008, 08:08:38 AM »
Well, it depends on what you define as the forward rate of reaction.  When most people consider the  forward rate of reaction, they imagine a case where there are no products, so the rate of the reaction is not affected by the backward rate of reaction.  In this case, it is correct to apply the Arrhenius equation and conclude that rate increases with temperature.

I don't quite get it,   A + B <--> no products??


Once you start looking at the overall rate of reaction (forward rate minus backward rate), it gets more complicated.

Do you have any link to further explain it?

Thanks much Ygg  :)

Kelvin  ;)
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline Valdorod

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Re: rates, temperature
« Reply #4 on: April 07, 2008, 12:33:25 PM »
Different but related concepts,

There are two  basic questions:

How fast do you want it?

How much do you want?

For exothermic reactions you have to give one up in order to get the other one.  As you increase the rate of the reaction by increasing the temperature, you are doing so at the expense of yield.  You are making the product faster but you are making less of it.

Say for example a theoretical reaction

A <---> B and it is exothermic

at 10 C you get 30% yield of B and it takes two days to get that 30%.

you increase the temperature to 20C to speed up the reaction so that it only takes 1 day, however, that increase in temperature will make it so that your yield decreases to 15%.

A commecial example is the industrial production of ammonia using the Harber Process.  The reaction is exothermic and it is ran at high pressures (usually between 500 atm and 1000 atm) and at high temperatures 500C.  This high temp is costly in terms of the yield of NH3, however, for this process it ends up being better to get NH3 faster than to get more of it at a slower speed.

Valdo

Offline Astrokel

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Re: rates, temperature
« Reply #5 on: April 08, 2008, 05:47:34 AM »
Different but related concepts,

There are two  basic questions:

How fast do you want it?

How much do you want?

For exothermic reactions you have to give one up in order to get the other one.  As you increase the rate of the reaction by increasing the temperature, you are doing so at the expense of yield.  You are making the product faster but you are making less of it.

Say for example a theoretical reaction

A <---> B and it is exothermic

at 10 C you get 30% yield of B and it takes two days to get that 30%.

you increase the temperature to 20C to speed up the reaction so that it only takes 1 day, however, that increase in temperature will make it so that your yield decreases to 15%.

A commecial example is the industrial production of ammonia using the Harber Process.  The reaction is exothermic and it is ran at high pressures (usually between 500 atm and 1000 atm) and at high temperatures 500C.  This high temp is costly in terms of the yield of NH3, however, for this process it ends up being better to get NH3 faster than to get more of it at a slower speed.

Valdo

Yes!, i do know about that concept, yet i appreciate your reply  :)

Kelvin ;)
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline Rabn

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Re: rates, temperature
« Reply #6 on: April 08, 2008, 09:29:43 AM »
Did you mean to ask: will the rate of an exothermic reaction slow down if you increase the temperature while the reaction is proceeding? i.e. add energy to push equilibrium toward reactants.  Just saying "increasing the temperature" implies a static operation.  I remember doing an experiment in gen chem, I believe it was with calcium hydroxide but I'm not sure, where at room temperature there was no precipitate but placing it into a boiling water bath made a white precipitate form. It did that because the reaction is exothermic, removing heat by placing the solution into a cold water bath caused the dissociation and adding heat caused the association. 

Offline Astrokel

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Re: rates, temperature
« Reply #7 on: April 09, 2008, 08:07:15 AM »
thanks for the replies,guys, i ve learnt pretty much from u all. cheerz !


Kelvin  :)
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

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