Catalyst........

Question

有一個反應用左一種catalyst..........

結果發現佢令反應時間縮短(即係快左)...
但個activation energy 高左

會唔會有咁ge事??


現在有下列data....

Temperature/ K : 323 , 318 , 313 , 308 , 303
Time / s : 2.2 , 5.4 , 8.3 , 11.2 , 15.9

點樣計個activation energy?

Answer 1

有一個反應用左一種catalyst..........
結果發現佢令反應時間縮短(即係快左)...但個activation energy 高左
會唔會有咁ge事??

不會的。

=====
現在有下列data....
Temperature/ K : 323 , 318 , 313 , 308 , 303
Time / s : 2.2 , 5.4 , 8.3 , 11.2 , 15.9
點樣計個activation energy?

Denote T as temperature in K, and t as time in s. Calculate corresponding values of ln(1/t) and (1/T) :
ln(1/t) : -0.79, -1.68, -2.12, -2.42, -2.77
(1/T) / K-1 : 0.003096, 0.003145, 0.003195, 0.003247, 0.003300
Plot a group of ln(1/t) (y-axis) against (1/T) (x-axis). A straight line is obtained.
Slope = -(Ea/R) where R = gas constant = 8.314 J mol-1 K-1
Then, activation energy (Ea) in J mol-1 = -(slope) x (8.413)

Reason :
One form of Arrhenius equation : ln k = -(Ea/R)(1/T) + ln A
where k = rate constant at temperature T, and A = Arrhenius constant
Also k is directly proportional to (1/t), and thus k = C(1/t) where C = proportional constant
Therefore, ln(1/t) = -(Ea/R)(1/T) + lnCA
It is the form of a linear equation in Mathematics : y = mx + c
where y = ln(1/t), x = (1/T) and slope = -(Ea/R)

Answer 2

有一個反應用左一種catalyst..........

結果發現佢令反應時間縮短(即係快左)...
但個activation energy 高左

會唔會有咁ge事??

係唔會的！catalyst係provide另一條way去進行個反應，如果係postive catalyst o既話，應該係低左的。