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Rate constant question

Hello, I was wondering if someone can help me understand why in some experiments (e.g. disappearing cross test) ln k is proportional to 1/time. When you put it into the format lmk = -Ea/R(1/T) + ln A is it like saying lnk is proportional to the rate, which is proportional to 1/time..? I understand that rate constant relates the rate of reaction and the larger the value of k, the faster the reaction so this is my understanding so far, but I'm wondering if there is a more complete explanation... hope this makes sense!
Original post by lavely
Hello, I was wondering if someone can help me understand why in some experiments (e.g. disappearing cross test) ln k is proportional to 1/time. When you put it into the format lmk = -Ea/R(1/T) + ln A is it like saying lnk is proportional to the rate, which is proportional to 1/time..? I understand that rate constant relates the rate of reaction and the larger the value of k, the faster the reaction so this is my understanding so far, but I'm wondering if there is a more complete explanation... hope this makes sense!

T in the Arrhenius equation (k = Ae^-Ea/RT or lnk = -Ea/R (1/T) + lnA) is the temperature in kelvins.

t denotes time.

I think there has been a bit of a misunderstanding here of the notation
(edited 10 months ago)
Reply 2
Original post by TypicalNerd
T in the Arrhenius equation (k = Ae^-Ea/RT or lnk = -Ea/R (1/T) + lnA) is the temperature in kelvins.

t denotes time.

I think there has been a bit of a misunderstanding here of the notation


this is what aqa have written themselves... that k is proportional to 1/t..
Original post by lavely
this is what aqa have written themselves... that k is proportional to 1/t..

k is proportional to 1/t, but in your first post, you said “ln k is proportional to 1/time. When you put it into the format lmk = -Ea/R(1/T) + ln A is it like saying lnk is proportional to the rate, which is proportional to 1/time..?”

I interpreted that to mean you had assumed T in the Arrhenius equation represents the time when it instead represents the temperature in kelvins.

ln(k) is not proportional to the rate of the reaction, as you has said in your first post. k, however, is proportional to the rate of reaction as rate equations can be expressed in the form:

rate = k x

Since the rate is defined the change in concentration per unit time, then it follows that rate α 1/t. As such, 1/t α k, or 1/t = ak, where a is a constant.

If you then apply this relationship and substitute it into the logarithmic Arrhenius equation, then yes, you will get a linear relationship between ln(1/t) and 1/T, but if you instead substitute it into the exponential form, you will find instead that 1/t is directly proportional to e^-Ea/RT.
Reply 4
thank you so much!! I was confused but now I understand it. Just to clarify, could you express in terms of: rate α 1/t α k so they are proportional to each other?
Original post by TypicalNerd
k is proportional to 1/t, but in your first post, you said “ln k is proportional to 1/time. When you put it into the format lmk = -Ea/R(1/T) + ln A is it like saying lnk is proportional to the rate, which is proportional to 1/time..?”

I interpreted that to mean you had assumed T in the Arrhenius equation represents the time when it instead represents the temperature in kelvins.

ln(k) is not proportional to the rate of the reaction, as you has said in your first post. k, however, is proportional to the rate of reaction as rate equations can be expressed in the form:

rate = k x

Since the rate is defined the change in concentration per unit time, then it follows that rate α 1/t. As such, 1/t α k, or 1/t = ak, where a is a constant.

If you then apply this relationship and substitute it into the logarithmic Arrhenius equation, then yes, you will get a linear relationship between ln(1/t) and 1/T, but if you instead substitute it into the exponential form, you will find instead that 1/t is directly proportional to e^-Ea/RT.
Original post by lavely
thank you so much!! I was confused but now I understand it. Just to clarify, could you express in terms of: rate α 1/t α k so they are proportional to each other?

Yes, that is correct

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