Things that affect the rate law and the rate constant, anyone?

[EECStoMed]

Ok. So I'm a bit confused on what affects the rate of a reaction.
I know that the rate of the reaction is dependent on 3 things,
1. the energy of the collision
2. the frequency of the collision
3. orientation of the collision

so from these 3 things I can assume that temperature, catalyst, pressure and concentration will affect the rates of a reaction among other more ambiguous things.

But the question I have is, what will change the rate constant, the little k in front of the [] in the rate law WITHOUT affecting the order of the reaction. I've always assumed that temperature, pressure, and catalyst would but at this point I'm no longer sure. Do those things change the rate constant without affecting the order of the reaction or do they Affect them? Can anyone please clarify? thanks.

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[axp107]

when a rate changes according to all those factors you stated above, the k is changing.. not the order of the reactions. So when they say a rate increases, they are saying K increases due to those factors.. correct me if I'm wrong

 

[EECStoMed]

yea, that's what I always thought. But the thing is, a catalyst can not only lower the activation energy, it can also induce a change in the mechanism of a reaction which would induce a change in the order of the reactions. So if it's this way, then a catalyst will NOT only change the k but also the order. Do you agree?

when a rate changes according to all those factors you stated above, the k is changing.. not the order of the reactions. So when they say a rate increases, they are saying K increases due to those factors.. correct me if I'm wrong

 

[EECStoMed]

when a rate changes according to all those factors you stated above, the k is changing.. not the order of the reactions. So when they say a rate increases, they are saying K increases due to those factors.. correct me if I'm wrong

OOh, also, [] does not change the k correct? because if it did, then we wouldn't be able to just pick any [] in the experimental data to get the rate constant k.

 

[meowkat444]

yes, that's right about the concentration not affecting k. k is specific to a certain temp, though.

 

[elektroshok]

yea, that's what I always thought. But the thing is, a catalyst can not only lower the activation energy, it can also induce a change in the mechanism of a reaction which would induce a change in the order of the reactions. So if it's this way, then a catalyst will NOT only change the k but also the order. Do you agree?

Correct me if im wrong, also, but i think that the K is derived for the specific experiment that you are trying to complete. If the catalyst is not consumed, i do not think it would change the order, just K.

A + B -> C might be slower than A + B + Enz -> C + Enz,

but this wouldnt change the order, and if it did it would be a totally different rxn.

 

[EECStoMed]

Correct me if im wrong, also, but i think that the K is derived for the specific experiment that you are trying to complete. If the catalyst is not consumed, i do not think it would change the order, just K.

A + B -> C might be slower than A + B + Enz -> C + Enz,

but this wouldnt change the order, and if it did it would be a totally different rxn.

Although it is not consumed (well of course because it's a catalyst) this does not mean that it will not use a different mechanism. When I mean mechanism I am referring to things that have intermediates such as the following

uncatalyzed: A + B -> C
catalyzed overall: A + B -> C, but has intermediates such as a D that is both made then consumed in the middle. This would in fact change the order wouldn't it?

 

[elektroshok]

Although it is not consumed (well of course because it's a catalyst) this does not mean that it will not use a different mechanism. When I mean mechanism I am referring to things that have intermediates such as the following

uncatalyzed: A + B -> C
catalyzed overall: A + B -> C, but has intermediates such as a D that is both made then consumed in the middle. This would in fact change the order wouldn't it?

In the case of changing mechanisms from say sn1 to sn2, then yes, it will change.

I dont think you would be able to say that reaction X will change orders if catalyzed w/o having the table of concentrations and rates.

 

[axp107]

Oh wait...

check this out.

From EK.. P. 187.. Chemistry.. answer to # 33

Increasing the amount of catalyst never increases the rate constant.

Hmm... so what does it mean in terms of the Rate formula.. not really sure

 

[EECStoMed]

Oh wait...

check this out.

From EK.. P. 187.. Chemistry.. answer to # 33

Increasing the amount of catalyst never increases the rate constant.

Hmm... so what does it mean in terms of the Rate formula.. not really sure

HOLY ****. I just read that.

 

[axp107]

lol I did well on the questions right after that chapter/section...

I got them all right except the one with the time and the dots in the rectangle lol

 

[EECStoMed]

But I think that's wrong, because since a catalyst lowers the Ea, according to the Arrhenius eqn for rate constants, it would increase it as well.

Oh wait...

check this out.

From EK.. P. 187.. Chemistry.. answer to # 33

Increasing the amount of catalyst never increases the rate constant.

Hmm... so what does it mean in terms of the Rate formula.. not really sure

 

[EECStoMed]

DUDE, ME TOO. Only missed the one with the dots because I thought it was a 2nd order reaction.

lol I did well on the questions right after that chapter/section...

I got them all right except the one with the time and the dots in the rectangle lol

 

[axp107]

But I think that's wrong, because since a catalyst lowers the Ea, according to the Arrhenius eqn for rate constants, it would increase it as well.

maybe it means increasing the amnt of catalyst won't do anything..

but adding a catalyst might? I dunno lol.. stilly studying .. 2 months left haha

 

[EECStoMed]

maybe it means increasing the amnt of catalyst won't do anything..

but adding a catalyst might? I dunno lol.. stilly studying .. 2 months left haha

1.4 months left. I'm taking it on the 24th of July. You?

 

[axp107]

Aug 20th... delaying as much as possible.. thats the last date before school starts

 

[Dr. Pepper]

maybe it means increasing the amnt of catalyst won't do anything..

but adding a catalyst might? I dunno lol.. stilly studying .. 2 months left haha

The presence of a catalyst will lower the activation energy, thus increasing k, thus increasing the reaction rate.

If we follow the basic equation: reaction rate = k[A]^2 for example and we use the arrhenius equation, then here's what we can conclude:

The rate constant k is affected by:
a. the PRESENCE of a catalyst (regardless of the concentration of a catalyst, the activation energy will decrease when a catalyst is added)
b. a change in temperature
c. a change in A (probability factor I think)....I'm kind of guessing on this one. A is different for each compound and it has to do with orientation. Therefore, highly theoretically, if you had an isomer of a certain compound, A could be different (esp. true if you switched between ring and straight-chain form). Once again, this is just a guess.

The rate is affected by:
a b and c
d. a change in concentration
e. a change in pressure

Correct me if I'm wrong. That's all I have for now.
-Dr. P.

 

[dayumgina]

The rate constant is affected by everything the rate of the reaction is affected by, minus the concentration of the reactants.

 

[yakuza]

The presence of a catalyst will lower the activation energy, thus increasing k, thus increasing the reaction rate.

If we follow the basic equation: reaction rate = k[A]^2 for example and we use the arrhenius equation, then here's what we can conclude:

The rate constant k is affected by:
a. the PRESENCE of a catalyst (regardless of the concentration of a catalyst, the activation energy will decrease when a catalyst is added)
b. a change in temperature
c. a change in A (probability factor I think)....I'm kind of guessing on this one. A is different for each compound and it has to do with orientation. Therefore, highly theoretically, if you had an isomer of a certain compound, A could be different (esp. true if you switched between ring and straight-chain form). Once again, this is just a guess.

The rate is affected by:
a b and c
d. a change in concentration
e. a change in pressure

Correct me if I'm wrong. That's all I have for now.
-Dr. P.

Sorry old thread, but how does pressure change rate?

 

[mmc48]

By increasing the pressure, you increase the number of overall collisions within a gas. This means, statistically, you should have more overall successful collisions (with correct orientation, energy, etc). Thus the rate constant should increase.