Question

The deadline for this question has passed, so I feel as if it is safe to post here:

Answer 1

The rate constants for a reaction are found to be

k(110 C) = 7.0 x 10 -4 s-1

k(165 C) = 41.9 x 10 -4 s-1

Determine the rate constant in units of s-1 at 269 C

For starters, I have /just/ started 11th grade so I know very little about calculus, I am familiar with logarithms though.

I know I have to answer this question using the Rate of Reaction Equation , but I don't know what the symbols in the equation mean, or how to apply it to this question.

Any hints?

Thanks in advance! c:

Answer 2

@Rushwr

Hope you don't mind the ping! ;u;

Answer 3

what you need to do is use the arrehnius parameter:

\[k=Ae ^{-\frac{ E _{a} }{ RT }}\]

Answer 4

perhaps take the logarithm on both sides to get

\[\ln(k)=\ln(A)-\left( \frac{ E }{ R } \right)\left( \frac{ 1 }{ T } \right)\]

Answer 5

we know R right, 8.314, etc...universal gas constant

Answer 6

we also know two rate constants at its specific temperature.

Therefore, we can make a system of two linear equations

Answer 7

to solve for the pre-exponential factor and the activation energy

Answer 8

then we just simply use the general rate equation and just substitute the known pre-exponential factor and the activation energy values as well as the temperature at which we are asked to find the rate constant.

Answer 9

Yeah, Boltzman Constant I believe?

I see where you're going with this. So this is what it should look like with certain values plugged in before I solve the linear equations?

\[\ln(7*10^{-4}) = \ln(A) - \frac{ E }{ 8.314} * \frac{ 1 }{ 110}\]

\[\ln(41.9*10^{-4}) = \ln(A) - \frac{ E }{ 8.314} * \frac{ 1 }{ 165}\]

Answer 10

make sure your temperature is in kelvin

Answer 11

\[\ln(7*10^{-4}) = \ln(A) - \frac{ E }{ 3185.5091} \]

(I converted C to K)

\[\ln(41*10^{-4}) = \ln(A) - \frac{ E }{ 3642.7791} \]

So subtracting the first from the second:

\[\ln\frac{ 41*10^{-4} }{ 7*10^{-4} } = \frac{ E }{ 3642.78 } - \frac{ E }{ 3185.5091 }\]

Answer 12

Ah sorry, was a bit behind you, let me catch up.

Answer 13

oops, my equations are slightly incorrect. the lHS of each equation should have ln(..)

Answer 14

i'll re type

Answer 15

My \[\frac{ E }{ 3642.78}\] looks different to yours? perhaps I did something wrong? (I checked my first E as well, it's the same)

Answer 16

yeah my bad

Answer 17

i'll fix it again

Answer 18

I feel your pain haha.

Answer 19

hahaha

Answer 20

\[\ln(7\times10^{-4})=lnA-3.14\times10^{-4}E\]
\[\ln(41.9\times10^{-4})=lnA-2.75\times10^{-4}E\]

Answer 21

thats better

Answer 22

so i'm going to do it quick,

but what you should get is

ln(A)=7.1423
E=45881

Answer 23

hence we know have a general solution for such rate constant at any temperature.

\[\ln(k)=7.1423-\left( \frac{ 45881.43}{ 8.314 } \right)\left( \frac{ 1 }{ T } \right)\]

Simply just sub T=269 into the equation to get a value for ln(k)

\[\ln(k)=7.1423-\left( \frac{ 45881.43}{ 8.314 } \right)\left( \frac{ 1 }{ 269+273 } \right)\]


Then evaluating
\[\ln(k)=-3.03957...\]

Taking exponential to isolate k, rate constant at 269 degrees celcius is:

\[k=0.048 {\space} s ^{-1}\]

Answer 24

Okay

After that:

\[\ln \frac{ 41*10^{-4} }{ 7*10^{-4} } =0.39*10 ^{-4} E\]

And then you take the value of E, stick it in there and find A, Gotcha.

Then you plug them into the first equation you gave me, the Arrhenius parameter? Gotcha, it's pretty clear now. I was staring at the equation for general molecules, super confusing.

Answer 25

Thanks a lot! Your detailed explanations were just what I was looking for. c:

Answer 26

remember, R is not the baltzman constant. Thats a different variable. Here, R is the universal gas constant

Answer 27

Yes, my bad. But essentially it's the same value of Boltzman constant but with different units? Right? (hence a different number)

Answer 28

remember when you deal with rate constants always think KINETICS. the Arrehnius equation is so fundamental for reaction kinetics

Answer 29

yeah pretty much

Answer 30

i completed a 3rd year uni kinetic and reactor design course, so anything reaction rates, i'll boss

Answer 31

Alright, I understand. I wasn't aware of that equation before.

Just want to confirm, E was activation energy and A was the pre-exponential constant?

Impressive! If I have trouble with this stuff in future I hope you won't mind if I give you a ping haha.

Answer 32

yep that is correct!

so you never came across that equation? hmm

Answer 33

Well, I probably /did/ but in such a way that it wasn't highlighted. I'm doing the EDX MIT Solid Chem course, and I'm struggling because I don't even know some of the math I should.

Answer 34

Either way I know know, so if I some something in a similar vein I'll kick its a**

Answer 35

*know now

Answer 36

haha awesome! good luck with it! i'm off to bed. catchya

Answer 37

Good night! Thanks again!