Why Does the Instantaneous Rate of Reaction Decrease Over Time?

[nesan]

We were doing a small assignment on rates of reaction and I plotted the data I received from the lab we did today. It looks like a perfect graph of log(x)

I have no idea why the instantaneous rate of change decreases over time. We haven't done this (we were only asked to graph, we're doing the slope stuff tomorrow) yet but I was just interested as to why that is. Thank you. :)

My guess is as time passes, there are less reactants?

 

[The Gringo]

If the reaction isn't 0 order, then it will generally decrease as time goes on if the conditions are held constant. You're correct, it's because there's less reactants. By collision theory, the reactants must collide with the correct alignment and with E > E_a. As you decrease the number of reactants, the number of effective collisions per unit time decreases.

Furthermore, suppose you have a reaction, A -> B + C.
d[A]/dt = -k[A]^n where n is order w.r.t A

Since [A] is decreasing, the magnitude of d[A]/dt decreases and so does the rate of the reaction.

 

[nesan]

If the reaction isn't 0 order, then it will generally decrease as time goes on if the conditions are held constant. You're correct, it's because there's less reactants. By collision theory, the reactants must collide with the correct alignment and with E > E_a. As you decrease the number of reactants, the number of effective collisions per unit time decreases.

Furthermore, suppose you have a reaction, A -> B + C.
d[A]/dt = -k[A]^n where n is order w.r.t A

Since [A] is decreasing, the magnitude of d[A]/dt decreases and so does the rate of the reaction.

Thank you. :)

This is actually very interesting stuff. Can't wait for class tomorrow. xD

Thanks again. :)