1
$\begingroup$

I'm an AP Chemistry student, and we're doing the Decomposition of $\ce{H2O2}$ lab, in which we decompose $\ce{H2O2}$ with $\ce{KI}$ as a catalyst. I've gotten this data so far:

+-------+-------------+----------------+-----------+--------------+-------------+--------------+--------------------+---------------------+-------------------+
| Trial | Volume H2O2 | Initial [H2O2] | Volume KI | Initial [KI] | Temperature | Rate (kPa/s) | Initial rate (M/s) | [H2O2] after mixing | [KI] after mixing |
+-------+-------------+----------------+-----------+--------------+-------------+--------------+--------------------+---------------------+-------------------+
|     1 |           4 | 0.882          |         1 | 0.5          | 21.2        | 0.4121       | 0.000168           | 0.706               | 0.10              |
|     2 |           4 | 0.882          |         1 | 0.25         | 21.1        | 0.2011       | 0.000082           | 0.706               | 0.05              |
|     3 |           4 | 0.441          |         1 | 0.5          | 21.3        | 0.1844       | 0.000075           | 0.353               | 0.10              |
|     4 |           4 | 0.882          |         1 | 0.5          | 31.2        | 0.8211       | 0.000334           | 0.706               | 0.10ds            |
+-------+-------------+----------------+-----------+--------------+-------------+--------------+--------------------+---------------------+-------------------+

I think that this is a first order reaction with regards to both, since when you halve the amount of each reactant, the rate of the reaction also halves.

Where I am running into trouble is calculating the rate constant (k) of the reaction. I don't really know where to start on this. I thought of dividing the initial rate by the $\ce{H2O2}$, but that wouldn't account for the amount of $\ce{KI}$. How would I go about calculating the rate constant of this reaction? I don't need an answer here, but pointing me in the right direction would be great.

Thanks!

$\endgroup$
0
$\begingroup$

As mentioned, you don't want a direct answer so I am just going to ask you some question to think about, that hopefully help in getting you to the answer. Please reply in the comments whenever something is unclear to you!

You're expecting first order in both reactants, so if you would write down the reaction rate equation, what does that look like?

With the equation you should have one unknown only: $k$ (or two that make up $k$: $k_0$ and one other, which?, if you are interested in the temperature effect as well).

Now calculate $k$ for each reaction separately. Are they the same? Do you expect them to be? If they're not, should you average them?

$\endgroup$
  • $\begingroup$ So the rate equation would be k[KI][H2O2], right? $\endgroup$ – John Pande Feb 7 '14 at 14:06
  • $\begingroup$ @JohnPande Yes, that is correct. r=k[KI][H2O2] to be precise $\endgroup$ – Michiel Feb 7 '14 at 14:23
  • $\begingroup$ So to find the rate of each reaction, would I use the equation k=r/([KI][H2O2])? $\endgroup$ – John Pande Feb 7 '14 at 14:25
  • $\begingroup$ Doing that, I got the rates of 0.001860, 0.001739, 0.001533, 0.003761 $\endgroup$ – John Pande Feb 7 '14 at 14:27
  • $\begingroup$ Rate constants you mean, but yes that seems right. Now, you also report the temperatures and should work those in to find $k_0$ (also often denoted as $A$) and another parameter. This wiki will be helpful: en.wikipedia.org/wiki/Reaction_rate $\endgroup$ – Michiel Feb 7 '14 at 15:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.