Are the orders of reactants with respect to a reaction different for different starting concentrations?

Question

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?

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.