Question

Chap 5
14)

Answer 1

http://prntscr.com/4ezbbl

Answer 2

@midhun.madhu1987

Answer 3

How do I determine the scale .. to draw the graph

Answer 4

Sorry... :(

Answer 5

@ganeshie8

Answer 6

@sweetburger

Answer 7

@Callisto

Answer 8

How to determine the scale .. to draw the graph

Answer 9

exatly

Answer 10

@chmvijay

Answer 11

boop

Answer 12

try to convert in one unit either 10^-5 or 10^-4 :)

Answer 13

yaa

Answer 14

Btw, I need to draw temperature against the rate constant rite

Answer 15

0.0383 x 10^-4
17.1 x 10^-4
69.4 x 10^-4
2.57 x 10^-4
8.78 x 10^-4

Answer 16

@JoannaBlackwelder

Answer 17

I am working on it, but here is some good info.
http://www.chemguide.co.uk/physical/basicrates/arrhenius.html

Answer 18

We can calculate the Activation Energy by graphing lnk versus 1/T

When the lnk (rate constant) is plotted versus the inverse of the temperature (kelvin), the slope is a straight line. The value of the slope (m) is equal to -Ea/R where R is a constant equal to 8.314 J/mol-K.

Answer 19

http://www.chem.fsu.edu/chemlab/chm1046course/activation.html

Answer 20

The info I gave you is about halfway down. :)

Answer 21

ok

Answer 22

How to determine the scale

Answer 23

@JoannaBlackwelder

Answer 24

@JoannaBlackwelder

Answer 25

What are your ln(k) and 1/T values?

Answer 26

0.0383 x 10^-4 17.1 x 10^-4 69.4 x 106-4 2.57 x 106-4 8.78 x 10^-4

Answer 27

That's for ln K

Answer 28

So, lnk looks like it is from 0 to 1x10^-5 ish

Answer 29

3.5336 x 10^-3
3.4130 x ''
3.3003 x ''
3.1949 x ''
3.0960 x 10 ^-3

Answer 30

That's for 1/T

Answer 31

You just need to span your data points. So, 3x10^-3 to 3.6x10^-3 should do it.

Answer 32

Okay

Answer 33

What bout ln K

Answer 34

So, lnk looks like it is from 0 to 1x10^-5 ish

Answer 35

Alrite

Answer 36

Thank you @JoannaBlackwelder and also @chmvijay

Answer 37

No worries :)