2
$\begingroup$

For the hydrolysis of $trans-\ce{[Co(en)2Cl2]Cl}$ (and cis) we measured reaction rates via spectrophotometry. Plots of change in absorbance $(\ln(A-A_\infty))$ were used to determine the rate law and the rate for the reaction. Using my data, I created the first-order plot seen below:

enter image description here

The question is for the report was:

Justify assuming first-order kinetics

My explanation was that first-order reactions can be expressed as $$[\ce{A}]=[\ce{A}]_0\operatorname{e}^{-kT}$$ and because the concentration decreases exponentially over time, we can assume its first order. Other rate laws I looked up didn’t show an exponential function, so this seemed like sound reasoning, but I was told I was wrong.

What would be the justification for thinking this is first-order?

Improve this question
$\endgroup$

4 Answers 4

Reset to default
5
$\begingroup$

As you have explained, the reaction is a hydrolysis:

$$\ce{[Co(en)2Cl2]+ + H2O -> [Co(en)2(H2O)Cl]^2+ + Cl-}$$

Therefore, you may expect a second-order rate equation:

$$r = -\frac{\mathrm{d}c}{\mathrm{d}t} = k_1 \cdot \left[\ce{[Co(en)2Cl2]+}\right] \cdot \left[\ce{H2O}\right]$$

However, you are using a dilute aqueous solution. Compared to the concentration of the complex, the concentration of water is very large. Thus, the concentration of water remains approximately constant and it can be included in the rate constant, obtaining a pseudo-first-order rate equation:

$$r = -\frac{\mathrm{d}c}{\mathrm{d}t} = k_2 \cdot \left[\ce{[Co(en)2Cl2]+}\right]$$

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ I like this answer, but doesn't that imply then that the reaction rate is actually second order (which is actually not the case)? $\endgroup$
    – John Snow
    Commented Apr 9, 2015 at 20:10
  • 1
    $\begingroup$ @JohnSnow That is correct; the overall reaction equation implies a second-order rate equation (even if it effectively is pseudo-first-order). However, the detailed mechanism may be more complicated. If the first step (removal of one chlorido ligand) is very slow (i.e. rate-determining), the reaction actually has a first-order rate equation (not just pseudo-first-order). $\endgroup$
    – user7951
    Commented Apr 9, 2015 at 20:43
  • $\begingroup$ Finally talked to my professor and this was the correct answer! $\endgroup$
    – John Snow
    Commented May 7, 2015 at 7:31
5
$\begingroup$

What would be the justification for thinking this is first-order?

The KISS principle :)

Before you start to model complex kinetics, begin with a simple model. If it fits, consider your job done.

Good enough is good enough!

(Imagine that you perform therapeutic drug monitoring. It's not your job to elucidate the exact metabolism of a drug here, you just want to be sure that the patients receive the right dose.)

Improve this answer
$\endgroup$
4
$\begingroup$

This might be an oversimplification, but my reasoning for assuming that it's first order would be because the plot of the log of the change in absorbance versus time gave a straight line.

The log of the concentration vs time will normally indicate a first order equation (given a straight line) but using the Beer-Lambert law:

A = ecl (where A is absorbance, e is the extinction coefficient, c is the concentration and l is the path length)

and assuming that the path length (l) is 1, you can see that the absorbance is proportional to the concentration. So plotting the graph of absorbance is analogous to plotting the graph of concentration, in that you should get a similar straight line and the result for the reaction constant k.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I feel like thats similar to my answer of it decreasing exponentially? $\endgroup$
    – John Snow
    Commented Apr 9, 2015 at 17:35
  • 1
    $\begingroup$ Pretty much. I feel like you may have been told you were wrong because they wanted you to use the graph to justify your answer, which is what I tried to do above. I did a similar experiment in freshman chemistry and that's the sort of thing they were looking for me to say. I think it was poor form of them to tell you that you were wrong and not to give you proper feedback on your answer. $\endgroup$
    – Folie
    Commented Apr 10, 2015 at 1:02
  • $\begingroup$ To be fair, I didn't approach the professor about it. I think I'll talk to him next class and see what the appropriate answer would be. $\endgroup$
    – John Snow
    Commented Apr 10, 2015 at 1:15
3
$\begingroup$

You were told you are wrong, but you are right. This is clearly a first order reaction, just because of the reasoning you exposed. But I will try to present you a simple mind exercise to realize why in an heuristic way (your argument is the more accurate one in my opinion, but many chemist need an inaccurate representation to feel that the concept is understood).

I bet that it is just a misunderstand of reaction order concept. In the case of elemental reactions in gas phase, it will be (just for statistical reasons) the sum of reactive molecules involved. BUT, here we are talking about a reaction in aq. phase, where one of the reactives is not another solute. So, thinking this hydrolysis just as molecule $A$ plus only one water molecule is not accurate.

The water is making a field around each molecule of your reactive, so represent (in your head) it like your molecule surrounded a electric field is more accurate that only thinking in two molecules. Thinking so, the first order for this reaction arises naturally.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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