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The data collected from using the method of initial rates is tabulated above.

Suggest a rate law and a mechanism consistent with the data. State any assumptions.

My Attempt

Obviously B and D has no impact on the rat of the reaction. From the first table it appears that the reaction is first order in A, however this doesn't make sense when you examine the last two rows of the second table. Since the rate law is first order in A and is independent of B, decreasing the concentration of A by 4 and increasing the concentration of B by 4 should result in a fourfold decrease in the rate. However the rate of the reaction decrease by 2.5 times. So then my logic is somehow wrong.

Furthermore I am even more confused about the order of C. I know that concentration of C is inversely proportional to the rate of the reaction. However the reaction still occurs without the presence of C (evident by the first table). Doesn't this imply that the rate law is independent of the concentration of C?

Any help is much appreciated.

Edit 1

From the comments below, I have realised that the rate order of B is in fact not 0. However now I am even more confused. How can it be that the rate is not dependent on B when there is no C, but it is dependent on B when there is C present? What would its rate law be like?

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  • $\begingroup$ It looks like B does affect the rate. Compare cases 2 and 6 in the lower table. $\endgroup$ – Chet Miller Feb 21 '16 at 12:54
  • $\begingroup$ @ChesterMiller thanks for that, I didn't see that. However this just confuses me even more. I don't see how it can be possible that has B has no impact on the rate, but when in the presence of C, it does. I have no idea what the order of B is suppose to be now. $\endgroup$ – Nanoputian Feb 21 '16 at 20:18
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Empirical rate law is:

$rate\,law=\frac{\left [ A \right ]*\left [ B \right ]}{\left [ B \right ]+\left [ C \right ]}$

In the first block of data you can notice that C and D are not necessarly required for reaction to proceed because otherwise the rate would be zero. You can also deduce that their concentration must appear as a sum in the rate law for the same reason. The sum will appear in denominator because C slows down the overall reaction.

Next step is to guess elementary reactions. From the law's denominator sum you can conclude that intermediate will competitvley react with B and C.

$A\overset{k1}{\rightarrow}I\, (Probably\, some\, kind\, of\, exitaction)$

$I+C\overset{k2}{\rightarrow}D$

$I+B\overset{k3}{\rightarrow}Product$

Then you derive the overall rate via steady state approximation (assumption).

$rate\, law=\frac{k_{1}k_{2}AB}{k_{3}B+k_{2}C}$

${k_{1}}={k_{2}}={k_{3}}$

Hope it helps.

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I think you should consider some sequence of reactions involving mass action kinetics. The rate they give in the table must be the rate of disappearance of A (I guess). So the rate might be represented by something like $R=C_1A+C_2B+C_3C+C_4A^2+C_5B^2+C_6C^2+C_7AB+C_8BC+C_9AC$. Not all the coefficients would be positive because of reverse reactions. There is not enough data in the table to determine all the coefficients even for this set up to 2nd order. So you might have to play with the coefficients to determine the simplest version which matches the table I hope this helps and I hope it works.

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  • $\begingroup$ Thanks for you thoughts. But I think that there must be a simpler way of doing it since we have never been taught mass action kinetics and it is expected that we should be able to do this question within 7-8 minutes. $\endgroup$ – Nanoputian Feb 22 '16 at 5:23

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