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Reaction Rate Help.

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AnneBee:
Given the following reaction at a constant temperature:

CH3COCH3 + I2 --> CH3COCH2I + HI

1  0.100 mol/L          0.100 mol/L          1.16 x 10-7 mol/ls
2  0.0500mol/L         0.100 mol/L          5.79 x 10-8
3  0.0500 mol/L        0.500 mol/L          5.78 x 10-8

Determine the rate law for this reaction

HERE IS MY WORK:

Trial 1-->2 [CH3COCH3]x = rate
0.0500/0.100 = 5.79 x 10-8/1.16 x 10-7 mol/ls
0.5x = 2.9
This is incorrect why?

Trial 2 -->3   [I2]x = rate
[0.500/100]y=5.78 x 10-8/ 5.79 x 10-8
5y = 1
y = 0
This is also wrong why?

Any help would be greatly appreciated!!

MNIO:
where did you get
[CH3COCH3]x = rate
0.0500/0.100 = 5.79 x 10-8/1.16 x 10-7 mol/ls     what happened to the x
0.5x = 2.9                 x?, 2.9?
should be
([CH3COCH3#2] / [CH3COCH3#1])x = (rate#2 / rate#1)
(0.0500/0.100)x = (5.79 x 10-8 / 1.16 x 10-7)
0.5x = 0.5
x = 1

same thing for y, you wrote
[I2]x = rate
[0.500/100]y=5.78 x 10-8/ 5.79 x 10-8
5y = 1
y = 0
should be
([I2#3] / [I2#2])x = rate#3 / rate#2
([0.500/0.100])y = (5.78 x 10-8 / 5.79 x 10-8)
5y = 1
y = 0

*****************
finishing the problem
*****************
now we know
rate = k * [CH3CO2H]¹ * [I2]⁰
or simply
rate = k *  [CH3CO2H]

now we need to plug in any data point and solve for "k"
k = rate / [CH3CO2H] = (1.16e-7 M/s) / (0.100 M) = 1.16e-6 / sec

and now we can write the full rate equation
rate = (1.16x10^-6 / sec) * [CH3CO2H]