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.
0 answers