If the reaction is repeated with $\pu{2 M}$ ethyl iodide the pyridine concentration decreases as shown below. Give the rate law of the reaction in terms of pyridine and ethyl iodide. Explain your reasoning.

$$ \begin{array}{l|rrrrr} \hline t/\pu{min} & 0 & 20 & 40 & 80 & 120\\ \hline [\ce{C5H5N}]/\pu{M} & 0.1 & 0.0746 & 0.0559 & 0.0312 & 0.0175\\ \hline \end{array} $$

I have found the rate constant and found that the orders of ethyl iodide and pyridine are both $1$ with respect to the reaction when their initial concentrations were both $\pu{0.1 M}.$ I would have thought this would be the case for any set of initial concentrations.

I would like to confirm this. In that case, I am thinking the only thing different would be the rate constant. But how would I go about finding that?

Could I plot the data, interpret the half life and go from there? Or is my approach totally wrong?

  • 3
    $\begingroup$ the order of a reaction is independent of starting concentration, if it were so then order would have kept changing at every moment sinc each moment can be considered as a starting point. $\endgroup$ Commented Apr 12, 2021 at 13:28
  • 3
    $\begingroup$ When the concentration of one reactant is much higher than the other, then it can appear to be zero order in the high concentration reactant because the percent change in that concentration is so small that it has a negligible effect on the overall rate. But it is still first-order in that reactant even if it's not observable under those conditions $\endgroup$
    – Andrew
    Commented Apr 12, 2021 at 13:33
  • $\begingroup$ This makes sense, thank you both. The rate constant would be a different value though wouldn’t it? $\endgroup$
    – user85551
    Commented Apr 12, 2021 at 13:42
  • $\begingroup$ Disregard that. I read it over and realise that wouldn’t make sense $\endgroup$
    – user85551
    Commented Apr 12, 2021 at 13:59

1 Answer 1


A reaction's rate does depend upon the consumption of its reactants, and the manner in which the reactants interact. Let's consider the following reaction: $$\ce{A + B -> final products}$$ The rate of consumption of reactants $$\dfrac{-d[A]}{dt}= k[A][B]$$ and $$\dfrac{-d[B]}{dt}= k[A][B]$$ It seems that this reaction is a second order reaction since the rate order appears to be:$$\text{Rate} = k[A][B] $$ As @Andrew astutely pointed out in the comments, a reaction can be of a lower order if one of the reactant's concentration is much higher than the other's.

Let's suppose, that one reactant is in great excess (say $10 \text{M of B vs } 0.01\text{M A}$) So what we're observing now is that the reactant B is barely consumed, yet the reaction terminates since reactant A, the limiting reagent is exhausted. In other words, the rate equation looks something like this: $$\text{Rate}=k'[A]$$, where $k^{'}=k[B]_o$ .Now, the reaction appears to be a first order reaction. This is known as a pseudo first-order reaction, since the rate of the reaction ultimately depends only on one reactant.

An example that I found here: $$ \rm{CH_3Br + OH^- \rightarrow CH_3OH + Br^-}$$

[...]"Imagine we had an initial concentration of CH3Br of 100 μM and and an initial concentration of OH- of 10 mM. Now even if all of the CH3Br has reacted the concentration of OH- will be essentially unchanged. Therefore during the course of the reaction, the concentration of OH- will be essentially constant. This makes the reaction "like a first order reaction", thus the name pseudo-first order."

Effectively, the rate now is: $$\ce{rate= k'[CH_3Br]}$$

This can also be mathematically represented in the integrated rate law: $$\dfrac{1}{\ce{[OH-]_0 - [CH3Br]_0}}\ln \dfrac{\ce{[OH-][CH3Br]_0}}{\ce{[CH3Br][OH-]_0}} = kt \quad \ce{where [OH-]\neq[CH3Br]}$$ Now, since $\ce{[OH-]_0 \gg [CH3Br]_0}$, we can write the equation as:

$$\dfrac{1}{\ce{[OH-]_0 - [CH3Br]_0}}\ln \dfrac{\ce{[OH-][CH3Br]_0}}{\ce{[CH3Br][OH-]_0}} \approx \dfrac{1}{\ce{[OH-]}}\ln \dfrac{\ce{[CH3Br]_0}}{\ce{[CH3Br]}}=kt $$

Remember, however that the rate and order of a reaction usually depends upon its mechanism.

For further reading, check here and here as well.


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