Question

Can someone help me with this question

Answer 1

Hmmm I can't bring up any good regression technique that allows you to determine the constant from all the data.
So far what I can see that.
\(v=-k_{r} \times [A] \times [B]\)

Answer 2

???

Answer 3

@aaronq

Answer 4

there are 2 questions posted here, which one are we working on?

Answer 5

#1 annoys me.

Answer 6

Only solution I can see right now is multivariable regression analysis.... (way to over complicated I guess)

Answer 7

and that \[\Large \nabla v=-k\]
again not useful.

Answer 8

the seconnd one

Answer 9

latex isn't working for me :S

for the first one, if you figure out the rate law, i think you can just plug the values in for 2 of the trials and solve for k.

Answer 10

You get the same number in 1 and 3 but 2 is different.

Answer 11

i got the first one

Answer 12

its the 2ed one

Answer 13

for the second you can use the arrheius equation, the two point one

Answer 14

The Arrhenius equation
\[\Large k=A \times \exp(\frac{ -E _{a} }{ RT })\]
Set up the problem:
\[\Large k _{2}=5.5 k_{1}\]
Solve from there.

Answer 15

Also a way of doing it :P

Answer 16

haha that works too

Answer 17

got it

Answer 18

In other words what you need to solve is:
\[\Large A \times \exp(\frac{ -E _{a} }{ RT _{1} })=5.5 \times A \times \exp(\frac{ -E _{a} }{ RT _{2} })\]
Alright.

Answer 19

can u help with another question

Answer 20

Yeah?

Answer 21

Okay we use the Arrhenius equation again:
\[\Large k=A \times \exp(\frac{ -E _{a} }{ RT })\]
Rewrite it:
\[\Large \ln(A)=\ln(A \times \exp(\frac{ -E _{a} }{ RT }))\]
\[\Large \ln(k)=\ln(A) \times \frac{ -E _{a} }{ RT }\]
Move a bit around:
\[\Large \ln(k) = \frac{ E _{a} }{ R } \times T ^{-1} + \ln(A)\]
This is a straight line and is best for 2 data points.

Answer 22

So what you need to do is to ln() transform your rate constants and take the reciprocal temperature, plot it and find the slope. From the slope we then get:
\[\Large \frac{ d[\ln(k)] }{ d(T ^{-1}) }=\frac{ -E _{a} }{ R }\]

Answer 23

but where it says point one and point two....what do i put in

Answer 24

You put in the numbers:
\(\large 225^{-1}\) and \(\large 675^{-1}\) in the x.

Answer 25

And the numbers \(\large \ln(0.387)\) and \(\large \ln(0.833)\) in the y

Answer 26

You see the idea?

Answer 27

both y number came out negative

Answer 28

yea

Answer 29

It is alright.

Answer 30

ok for the rise i got -0.766

Answer 31

and run i got 2.96E10-3

Answer 32

Okay, calculate the slope then.

Answer 33

258.78

Answer 34

Perfect. now calculate the activation energy.

Answer 35

2151.49

Answer 36

Remember the minus :)

Answer 37

1 more question if u dont mind

Answer 38

Last one, then I got to be going

Answer 39

ok

Answer 40

Right.
\[\Large 1 cm ^{3}=0.001 l\]
and that
\[\Large N _{a}=6.02 \times 10^{23} \frac{ molecules }{ mol }\]
Think you can solve the first one from here?

Answer 41

i got 2.24E11

Answer 42

how would i turn it to torr

Answer 43

Emmmm a bit unsure

Answer 44

ok then thank you so much