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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

1 個解答

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

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