Can an excessive reactant amount result in a longer reaction time?
Let me first say that I don't know this specific reaction, so I am not sure whether my answer holds for it, but let me answer the general question: can a high concentration slow down the reaction rate?

The answer is, yes it can. The mechanism by which this happens is called inhibition and it is often (although not necessarily) connected to reactions on surfaces. I will explain inhibition in terms of a surface reaction here, but keep in mind that it can also happen in solution due to 'crowding', in which there is so much of 1 reactant that it basically 'cannot find' the other reactant.

For surface reactions the story is as follows. You will get a competition between adsorption on the surface and reaction on the surface. 

Qualitatively, if you have a large amount of one of the two reactants than many of the surface sites will have this reactant adsorbed to it meaning that some of the reactant on the surface has neighbors that it cannot react with (because it is the same molecule), this can slow down the reaction rate.

Quantitatively, the reaction order will be concentration dependent. A famous example of this is Langmuir-Hinshelwood kinetics
$$r =k C_s^2 \frac{k_1 k_2 C_A C_B}{(1+k_1 C_A+k_2 C_B)^2} $$ 
where $C_s$ is the concentration of available adsorption sites and $C_A$ and $C_B$ are the concentrations of your reactants. 

There are many possible ways in which this reaction rate equation can behave, depending on the values of the rate constants $k$, $k_1$ and $k_2$, but let me single out two cases that will explain a shift from increasing to decreasing reaction rate with increasing concentration:


a very high concentration of $C_A$ will result in $k_1C_A&gt;&gt;&gt;1, k_2C_B$. This allows us to neglect the $1$ and $k_2C_B$ terms in the denominator, turning the rate equation into this:
$$r =k C_s^2 \frac{k_2 C_B}{k_1 C_A} $$ 
which indeed decreases with increasing concentration of $C_A$. 
a very low concentration of $C_A$ will result in $k_1C_A&lt;&lt;&lt;1, k_2C_B$. Then the rate equation becomes:
$$r =k C_s^2 \frac{k_1 k_2 C_A C_B}{(1+k_2 C_B)^2} $$ 
which has an increasing rate with increasing concentration of $C_A$


So what you see here is that a change of $C_A$ from low to high concentration can change the reaction order in $C_A$ and therefore indeed result in an initial increase of the reaction rate with increasing $C_A$, but later a decrease if you increase it even further. When this crossover occurs depends on the values of the rate constants ($k$, $k_1$ and $k_2$) and the concentration of the other reactant.