[THISMOMENTISFATE]

The study of kinetics

group 1:
Temp: 12.6 C
T(K): 12.6+273= 285.6
1/T(K): 3.50X10^-3 this one is the x value
Time: 816 sec
Rate ®: 7.66 x10^-6
LnR: -11.8 this is the y value



group 2:
Temp: 45C
T(K): 45+273= 318
1/T(K): 3.14X10^-3 this one is the x value
Time: 495 sec
Rate ®: 1.26 x10^-6
LnR: -11.3

group 3:
Temp: 12.6 C
T(K): 34.8+273= 285.6
1/T(K): 3.28X10^-3 this one is the x value
Time: 107 sec
Rate ®: 5.84 x10^-5
LnR: -9.75

now I need to find the

group 1:
Temp: 12.6 C
T(K): 12.6+273= 285.6
1/T(K): 3.50X10^-3 this one is the x value
Time: 816 sec
Rate ®: 7.66 x10^-6
LnR: -11.8

1. im supposed to find the
slope= -EA/R = triangle y/triangle x = (y2-y1) ________/(x2-x1)____________
how do I solve this? Also the teacher said our k from last week was strong, and to use LnR instead.


2. also how do I find the Ea?






3. can someone check to see if my graph looks right?
img207.imageshack.us/img207/6375/2ndgraph.jpg
We are supposed to find the y inter, and the R2.

any help would be appreciated.

[THISMOMENTISFATE]

update:
i tried to solve it on my own, can someone check?

the teacher said our k from last week was wrong, and to use LnR instead.
so my ANSWER:
(-11.-(9.75)
_____________________________
(3.50 X 10^-3)-(3.28 X 10^-3)

= -2.05/0.22 X 10^-3
= -9.32 X 10^3

2. also ANSWER for Ea:
9.32 x 10^3 x gas constant 8.3144125 j/mole

= 77.5 x 10^3 j/mole
=77.5 kj/mol