urban_flavaz
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Report Thread starter 13 years ago
#1
i'm really stuck on this equation plz plz plzzzzzz help me its driving me crazy am totaly stuck :confused:

with the presence of hydrogen ions, propanone and iodine react according to the following equation.

I2 + CH3COCH3 = CH2ICOCH3 + HI

following results obtained

Exp. [CH3COCH3] [H+] [I2] Initial rate
1 1.25 0.1 0.01 3.5 x 10-4
2 1.25 0.2 0.01 7.0 x 10-4
3 1.5 0.2 0.01 8.4 x 10-4
4 1.5 0.1 0.02 4.2 x 10-4

i'm really sorry this isnt clear but i cant do a table arrgghh

find giving reason the order of reaction with respect to propanone, hydrogen ions and iodine and hence the overall order.

well i've found that the order of propanoone and hydrogen ions are both 1 but i dont get how to do iodine. plz ne help?

thanks
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alispam
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Don't see how you can do that as that info doesn't keep the other two concentrations the same as the I2 changes...
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mizfissy815
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Compare 1 & 2...than 2 & 3...and finally 3 & 4.

[CH3COCH3] = A
[H+] = B
[I2] = C

Compare 1 & 2-
A & C- remains the same
B- 2 x bigger
Rate- 2 x bigger

Order of reaction of B = 1st order

Compare 2 & 3-
B & C - remains the same
A- 1.2 x bigger
Rate- 1.2 x bigger

Order of reaction of A = 1st order

You already know that if you double
B the rate of reaction also doubles. Now when you compare 3 & 4 you’ll see that B is halved and the rate of reaction also decreased by half. This means that the concentration of iodine did not affect the rate of reaction.

Order of reaction of C= 0 …
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vinny221
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The reaction is zero order with respect to I2 so your rate equation is:

rate=k[H+][CH3C0CH3]

overall order 2

If you compare steps 3 and 4 youll see that halving the H+ conc halfs the rate and this must be true as the reaction is order 1 with respect to H+. The I2 has therefore got no effect on the rate.

This experiment hardly works anyways unless you add an acid catalyst.
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urban_flavaz
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Report Thread starter 13 years ago
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thank u so much, i understand nw ----> made my day!! lol
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