November 30, 2020, 12:34:41 AM
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Topic: Investigating the order of reaction between acetone and iodine in acid solution  (Read 2034 times)

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Offline 144730

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CH₃COCH₃ (aq) + I₂ (aq) ----> CH₃COCH₂I (aq) + H⁺ (aq) + I⁻ (aq)

Samples at 30, 60, 180, 270 and 360 seconds were neutralized and titrated with Na2S2O3.

The concentrations of I2 used were 0.01, 0.02, 0.03, 0.04 and 0.05 mol dm-3.
According to literature the reaction is zero order with respect to iodine.

Concentration of iodine solution (mol dm-3) -Initial rate of reaction (ml s-1)
0.010   - 0.000289
0.020 - 0.001000
0.030   - 0.001839
0.040   - 0.002984
0.050   - 0.004785

This produced an graph which was clearly first order for I2.

What could the reasons for this be?



 

Offline mjc123

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It's clearly not first order. Have you done a log-log plot?
Did you keep everything else constant?
What is your "initial rate" measurement? Loss of iodine? Are your units mol/L/s? If so, except in the first case, you would have used up all your iodine in less than 30 seconds. Something is obviously wrong with your data.

Offline 144730

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I plotted the initial rate of reaction against the concentration of the iodine solution.
I have not done a log-log plot, I am not sure how to.

Everything else was kept constant.

The initial rate measurement is the tangent to the line of bestfit for the volume (cm3) of Na2S2O3 needed to titrate the samples at 0.0001 seconds.

Offline mjc123

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It should be clear from your plot that the relationship is not linear.
For a log-log plot you simply plot ln(initial rate) vs. ln(concentration).
If y = axb, ln y = ln a + b*ln x, so the slope of the graph gives the exponent b.
But it's still wrong if it is meant to be zero order.
Are you drawing tangents by eye? That is notoriously difficult to do reliably.
What do you get if you estimate initial rate = {titre(30 s) - titre(0 s)}/30 ?

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