The reaction below is catalyzed by hydrogen ions. It is found that the rate of the reaction depends only on the concentration of sucrose, for a given concentration of hydrogen ions.
C12H22O11 + H20 > C6H12O6 + C6H12O6
At 25 degree Celsius, when the initial concentration of sucrose was 1.00moldm^-3, the following data were obtained:
http://i75.photobucket.com/albums/i308/Tolland/chem.jpg
How would halving the initial concentration of sucrose affect the half life?
2007-03-11 19:21:48 · 1 個解答 · 發問者 David 1 in 科學 ➔ 化學
Rate = k[A]^n
d[A] / dt = k[A]^n
Assume n = 1
d[A] / dt = k[A]
integrate gives ln([A]/[A]o) = kt
If the reaction is really first order, your data should fit this relatioship.
Reacted A
[A]=1-Reacted A
ln[A]
t
k X 103
0.195
0.805
-0.2169
60
-3.615
0.277
0.723
-0.3243
90
-3.6
0.373
0.627
-0.4668
130
-3.59
0.478
0.522
-0.65
180
-3.61
Actually, a graph should be plot and see if it is really a stright line. But I check this by calculating the value of k. It is observed that the value of k agree very good for all data.
The reaction should be first order. Since a first order reaction has half-life independent with concentration, the half-life time does not change with concentration.
2007-03-11 19:57:59 · answer #1 · answered by ? 7 · 0⤊ 0⤋