i already posted this but i tried to solve it on my own can you check at the end of this paragraph?
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. the teacher said our k from last week was strong, and to use LnR instead.
so my ANSWER:
(-11.8)-(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
2 answers