[Shadow]

Why is the concentration drop for a first order reaction bigger than for a second order reaction?

[Babcock_Hall]

The general forum rule is that you must show an attempt before anyone can help you.  However, in this case, you might start by rephrasing this question.  What do you mean by concentration drop?  Over what time period?

[Shadow]

Over the whole course of  a reaction.

[Babcock_Hall]

Over the whole course of a reaction, the reactant drops from its initial concentration to is equilibrium concentration, regardless of the order of the reaction.  I have to suspect that I don't yet understand your question.

[Shadow]

See the diagram just above Case2 http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/Second-Order_Reactions.

[Corribus]

Compare the two reactions:

A P
A + A P

In the second order reaction, a collision is required between two A molecules in order for one product conversion to take place. As A is converted into P, there is less A in the reaction vessel, so the probability of a collision becomes lower and lower. The reaction rate is very slow as the concentration of A becomes vanishingly small. 

In the first order reaction, no collision is required. The probability per unit time of a successful conversion event still decreases as time elapses, because there is still less A in the reaction vessel at later times, but this drop in the reaction probability is not as severe as it is where a collision is required.

The amount of A left in the reaction vessel per unit time is inversely dependent on the reaction probability per unit time. This is why the concentration of A drops faster in the first order case than it does in the second order case.

[Shadow]

Thank you very much sir.