Sapling #12

[IreneSeo3F]

The problem states:
A particular reactant decomposes with a half‑life of 109 s when its initial concentration is 0.288 M. The same reactant decomposes with a half‑life of 241 s when its initial concentration is 0.130 M.

Can someone explain why the reaction order is 2?

[Mikayla James 2A]

Hi! So initially, we can cancel out the reaction being first order because the half-life of first order reactions is not dependent on concentration (t1/2=0.698/k). For a zero-order reaction, the equation is t(1/2)=[A]0/2k, but it indicates that as [A]0 increases, half-life also increases, which contradicts the data given in the question, where as the concentration decreases, the half-life increases. So, the reaction has to be second-order because its half-life equation demonstrates an inverse relationship between half-life and concentration, which agrees with the data given. Hope this helps!

[Andreas Krumbein 1L]

Since the half life changes depending on the initial concentration, we know that the reaction order cannot be 1, since first order half life is independent of initial concentration. Looking at the half life equations for 2nd and 0th order reactions, the 2nd order half life reaction model of t_1/2=1/k[A₀] perfectly fits both sets of data, so the reaction order is 2nd order.

[Sharon Kim 2A]

The order of the reaction would be 2 because in an inverse relationship. The easiest way to tell if this is a second order is by plugging values and rearrange k to see which fits. We know that this half-life is dependent on [A] initial so that automatically cancels out the first order!