Energy of activation and speed of reaction

[wall1two]

at room temperature two reactions with different activation energies have the same rate. When both at the same higher temperature which one will have a faster rate.







Answer is the one with the higher activation energy. I don't understand even when looking right at this equation

k= e^ -Ea/RT

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[qkchen]

i thought arrhenius equation was k = Ae^(-Ea / RT)

where did you see this question cuz i have a different answer in mind.

 

[recyrb]

if the question specified the use of enzymes, and the enzymes activation energys were in question, I would understand why the higher Ea would produce a faster reaction at higher temps

but, looking at that equation, i would not understand how that makes sense either, anyone?

 

[wall1two]

the question is in the ACS gen chem book. no explanation just the answer :/

i am trying to reason it out by thinking about 1,2-kinetic vs 1,4 thermodynamic addition of dienes. but then it is just the stability changing the equilibrium distribution not the rate.

with my luck with exam books it could be a misprint but the acs book has been awesome so far. the only way i see that answer being right is rate to completion, not just rate of forward and backward. i've googled that crap out of it and cant find a thing

 

[recyrb]

the question is in the ACS gen chem book. no explanation just the answer :/

i am trying to reason it out by thinking about 1,2-kinetic vs 1,4 thermodynamic addition of dienes. but then it is just the stability changing the equilibrium distribution not the rate.

with my luck with exam books it could be a misprint but the acs book has been awesome so far. the only way i see that answer being right is rate to completion, not just rate of forward and backward. i've googled that crap out of it and cant find a thing

googled the crap out of it, me as well... I hope to keep my fingers crossed and not worry about this concept. I only see how temperature increases rate and how increasing the activation engery will make it more difficult for the reaction to happen, thus slow it down. less particles moving at or above the Ea.
otherwise, maybe/ hopefully its a misprint.
seems like we are on the same pace, you are a little ahead of me. when do you take the exam?

 

[wall1two]

i'm set to take this bad boy the 25th. ive been studying 11-12 hours a day for 2 months now. and about 6-8 for 2-3 months before that. how about you

 

[AmpedUp]

at room temperature two reactions with different activation energies have the same rate. When both at the same higher temperature which one will have a faster rate.







Answer is the one with the higher activation energy. I don't understand even when looking right at this equation

k= e^ -Ea/RT

This one really sucks. If they could've thrown out the first part (in bold), it would be an easier question to answer. My guess is that the first part tells us that room temperature doesn't have an effect on activation energy - its thermodynamic effect. However, at a higher temperature, activation energy has a bigger effect on the enzymes - its kinetic effect. That's the only way I think you can explain the answer to this question; no other hunches come to mind.

Edit: wait. i don't think the answer makes sense. the lower the activation energy is at a higher temperature, the faster the reaction rate. this has to be a misprint! - i hope!

 

[MrFantastik]

Since you were gracious enough to answer my reduction potential question (even tho it was 3 days after i took my DAT😀) I'll take a crack at this one. Its gonna take me a while to explain but what the hell, I've been feeling extra helpful lately to give back for all the help i got for my DAT.

There are other things that affect the rate constant besides activation energy. Usually they arent mentioned too often but since the question mentions that at room temp the two reactions have different activation energies but the same rate we know that these other factors come into play.

The equation for the rate constant of a rxn (k) is:

k = z*f*p

z is the collision frequency or how often your reactants are going to collide

Now f is the fraction of collisions that cause the reaction
f = e^ -Ea/RT
And its what we're used to seeing when we think about rate constants

p is the orientation factor cuz if two reactants dont hit each other in the right orientation no reaction will happen.

When you put all of these together you get the Arrhenius equation

k = Ae^(-Ea / RT)

like qkchen mentioned

Usually we dont talk about the A too much but in this case its what makes the difference. Im not 100% sure about this but I think A = z*p
It accounts for the factors that affect the rate constant that dont depend on the activation energy.

At room temperature the reaction with the higher activation energy also has a higher A. This is how it can keep up with the reaction that has a lower activation energy. Now when you raise the temp youre increasing the rate of both of the reactions but you're helping out the reaction with the higher activation energy more because more energy is exactly what it needs to go faster. The reaction with the lower activation energy doesn't really need the energy as much.

You might be able to prove this mathematically by setting

Ae^(-Ea / RT) = (A+y)e^(-(Ea+w) / RT) at a certain temp T, when y and w are positve

and then showing that

Ae^(-Ea / R(T+x)) < (A+y)e^(-(Ea+w) / R(T+x)) when x, y and w are positive

But i have absolutely no desire to do that😉

I'd be pretty impressed if someone else does tho

 

[qkchen]

this is pretty interesting. hope this will not be the scope of the level of the gen chem conceptual questions on the DAT, cuz i think i'm gonna puke

if A were to be constant for both reactions (let's assume), then that would contradict the original answer provided, right? meaning, the reaction with lower Ea will actually proceed faster at higher temperature

Since you were gracious enough to answer my reduction potential question (even tho it was 3 days after i took my DAT😀) I'll take a crack at this one. Its gonna take me a while to explain but what the hell, I've been feeling extra helpful lately to give back for all the help i got for my DAT.

There are other things that affect the rate constant besides activation energy. Usually they arent mentioned too often but since the question mentions that at room temp the two reactions have different activation energies but the same rate we know that these other factors come into play.

The equation for the rate constant of a rxn (k) is:

k = z*f*p

z is the collision frequency or how often your reactants are going to collide

Now f is the fraction of collisions that cause the reaction
f = e^ -Ea/RT
And its what we're used to seeing when we think about rate constants

p is the orientation factor cuz if two reactants dont hit each other in the right orientation no reaction will happen.

When you put all of these together you get the Arrhenius equation

k = Ae^(-Ea / RT)

like qkchen mentioned

Usually we dont talk about the A too much but in this case its what makes the difference. Im not 100% sure about this but I think A = z*p
It accounts for the factors that affect the rate constant that dont depend on the activation energy.

At room temperature the reaction with the higher activation energy also has a higher A. This is how it can keep up with the reaction that has a lower activation energy. Now when you raise the temp youre increasing the rate of both of the reactions but you're helping out the reaction with the higher activation energy more because more energy is exactly what it needs to go faster. The reaction with the lower activation energy doesn't really need the energy as much.

You might be able to prove this mathematically by setting

Ae^(-Ea / RT) = (A+y)e^(-(Ea+w) / RT) at a certain temp T, when y and w are positve

and then showing that

Ae^(-Ea / R(T+x)) < (A+y)e^(-(Ea+w) / R(T+x)) when x, y and w are positive

But i have absolutely no desire to do that😉

I'd be pretty impressed if someone else does tho

 

[MrFantastik]

this is pretty interesting. hope this will not be the scope of the level of the gen chem conceptual questions on the DAT, cuz i think i'm gonna puke

if A were to be constant for both reactions (let's assume), then that would contradict the original answer provided, right? meaning, the reaction with lower Ea will actually proceed faster at higher temperature

Lol...yea i think this might be beyond the level of the DAT. But you should be aware that other things beside the activation energy and temperature can affect the rate constant of a reaction. If you think about it in orgo terms just think of the orientation factor (p) as whether or not the reactants are sterically hindered.

Also, if A was the same for both reactions than the reaction with the lower activation energy would be faster at ALL temperatures. They wouldnt be able to have the same rate at room temp like it says at the beginning of the question.

edit: Actually if A was the same for both reactions then i think at a high enough temperature the two reactions might eventually end up at almost the same rate. This is because as T --> infinity the value of Ea/RT AND (Ea + w)/RT would both get really really small and you would no longer be able to tell the difference between them.

 

[AmpedUp]

yuck!!!

 

[wall1two]

yuck!!!

agreed. MrF. you just dominated that question, good looks again! and i just found this

http://www.tutorvista.com/content/chemistry/chemistry-iv/chemical-kinetics/energy-barrier.php

sorry i didn't answer that question in time, but it turns out it didn't make much of a difference eh, you rocked the test.

 

[qkchen]

Lol...yea i think this might be beyond the level of the DAT. But you should be aware that other things beside the activation energy and temperature can affect the rate constant of a reaction. If you think about it in orgo terms just think of the orientation factor (p) as whether or not the reactants are sterically hindered.

Also, if A was the same for both reactions than the reaction with the lower activation energy would be faster at ALL temperatures. They wouldnt be able to have the same rate at room temp like it says at the beginning of the question.

edit: Actually if A was the same for both reactions then i think at a high enough temperature the two reactions might eventually end up at almost the same rate. This is because as T --> infinity the value of Ea/RT AND (Ea + w)/RT would both get really really small and you would no longer be able to tell the difference between them.

i bow to your gen chem prowess, in fact, i bow to your DAT-dominating prowess in general

 

[recyrb]

at room temperature two reactions with different activation energies have the same rate. When both at the same higher temperature which one will have a faster rate.







Answer is the one with the higher activation energy. I don't understand even when looking right at this equation

k= e^ -Ea/RT

Hey wall to wall... I wish i could get the time for my last two months to be 8-10 hour days... I hope i get through everything i want. I have set aside 5 months of 4 hours a day, which i just rolled by 3 months to the day. 2 months left. I also work 40 hours a week so with the 30 hours of studying thats about all i got. good luck let me know how everything goes!

 

[SurferGirlinSB]

This question has to do with the Arrhenius equation (ignoring the "A"), k = e^ -Ea/RT. This equation is frequently provided on standardized exams.

Two reactions a --> products and b --> products.
We will assume identical concentrations for both a and b. The rate laws for each are: Rate = k[a](raised to the x) and Rate = k (raised to the y which may or may not be the same as x).

According to the Arrhenius equation the rate constant, k, is related to Ea and temperature. Since the temperatures are stated to be the same for the two reactions, if the Eas are different, the rate constants, k, must be different since R is a constant.

So how can the rates be the same if the rate constants are different? The order of the reactants must be different for the two reactions i.e. first order for "a" (x = 1) and second order for "b" (x = 2, or something like that). But this isn't part of the answer, just an explanation for the original premise.

From the Arrhenius equation, the larger the Ea, the smaller will be the k. This is a math thing based on the form of the equation. So the reaction with the larger Ea will have a smaller rate constant, k, and be slower (in general unless other things like order or concentration are introduced as changes).

The important thing is how the rate constant changes with temperature. This may be more conceptually obvious to someone that is mathematically inclined, but you can make up some numbers and check it out empirically. It turns out that a reaction with a larger Ea, when you change the temperature, will have a larger effect on the rate constant, k. Recall that the rate constant changes with temperature.

That is, the reaction with the larger Ea will have a bigger change in the rate constant (making it faster) than the reaction with the smaller Ea (which will also be faster, but not as much).

Using the Arrhenius equation, you can see this by using three values of temperature (273, 283 and 373) and two values for Ea (100 J and 200 J). Then solve for k and plot lnk versus 1/T according most general chemistry textbooks. The slope is the (negative) value for the energy of activation (divided by R, a constant) or -Ea/R. For the reaction with an Ea of 100J, a slope of -11.91 is obtained. For the reaction with an Ea of 200J, a slope of -24.16 is obtained. Bigger slope equals bigger change (effect).

Clearly this problem wasn't meant to be solved by plotting values. So I think it is the ability to see how the value of k is affected by a change in the denominator (temperature) of the exponent in the Arrhenius equation that, again, for someone mathematically versed, could be seen conceptually without using numbers and plotting them. Or simplistically a bigger energy of activation has a bigger slope (with the variables ln k and T) and thus will change more with temperature.