Question

Notation question.

Answer 1

What's the question. 0-0

Answer 2

The Arrhenius equation:
\[k _{r}(T)=A \exp \left( \frac{ -E_{a} }{ RT } \right)\]
\[\ln(k _{r}(T))=\ln(A)-\frac{ E _{a} }{ RT }\]
\[\frac{ d \ln(k _{r}(T)) }{ dT ^{-1} }=\frac{ E _{a} }{ R }\]
Do we consider this a ok notation @amistre64 ?

Answer 3

define \(T^{-1}\), where does that apply in the stuff before it?

Answer 4

Sorry I forgot a minus sign btw.
Well the reason I want T^-1 is that I want the equation on linear form.

Answer 5

\[k _{r}(T)=A \exp \left( \frac{ -E_{a} }{ RT } \right)\]
\[ln[k _{r}(T)]=ln\left[A \exp \left( \frac{ -E_{a} }{ RT } \right)\right]\]
\[ln[k _{r}(T)]=ln(A)+ln\left[ \exp \left( \frac{ -E_{a} }{ RT } \right)\right]\]
\[ln[k _{r}(T)]=ln(A)+ \frac{ -E_{a} }{ RT }\]
is that (T) on the left side a multiplier? or is kr(T) the name of a function in T

Answer 6

No it is not a multiplier just a subscript that the variable Kr is depending on T.

Answer 7

if we let u = 1/t

\[ln[k _{r}(\frac1u)]=ln(A)+ \frac{ -E_{a} }{ R }u\]
\[\frac d{du}~\left[ln[k _{r}(\frac1u)]=ln(A)+ \frac{ -E_{a} }{ R }u\right]\]
\[\frac d{du}~\left[ln[k _{r}(\frac1u)]\right]=\frac d{du}\left[\frac{ -E_{a} }{ R }u\right]\]
\[\frac{[k _{r}(\frac1u)]'}{k _{r}(\frac1u)}=\frac{ -E_{a} }{ R }\]etc...

Answer 8

looks fine to me

Answer 9

since u = t^-1

\[\frac {d~\{ln[k _{r}(T)\}}{d(T^{-1})}~=-\frac{ E_{a} }{ R }\]may be more conventional

Answer 10

.... but then im rather unconventional at times so you might want to let the others weigh in

Answer 11

Okay, just don't understand why I got it send back but they just said it looked very crappy... I kinda make my own notation under the way and as I did not knew a sign to donate the slope I just choose to use the differentiated.

Answer 12

But thanks for checking, good to know if I get into future problems or discussions like this!! :)

Answer 13

lets try this for a test:\[y^2=t^{}\]\[\frac{d[y^2]}{dt}=1\]

let y^2 = f(t)
\[f=t\]\[\frac{d[f]}{dt}=1\]

i was wondering if just sucking up the function into the derivative looked ok, it seems to be fine to me that way as well

Answer 14

times up for me tho, so good luck :)

Answer 15

Danke vielmals :) I heavyly doubt that something is wrong but ones again thanks for clearing things out!