What is the order of the reaction #A -> B# if changing the concentration of #A# from #"0.05 M"# to #"0.1 M"# changes the rate from #2 xx 10^(-3) "M/s"# to #1.6 xx 10^(-2) "M/s"#?
1 Answer
The reaction is:
#"A" -> "B"#
The trick with this is that the order of the reactant
The order of a reactant is basically the contribution of its concentration to the change of reaction rate.
To put this in a larger context, the reaction rate
#\mathbf(r(t) = k["A"]^(m) = -(d[A])/(dt))#
where
The above equation is known as a rate law.
Note that you have two rates and two initial concentrations (you do not need equilibrium concentrations---in fact, those are not practical for kinetics studies, because the reaction is most active when it first starts).
These are for each trial run for a particular kinetics study, and can be used to determine the order of the reaction as follows.
Suppose we compared rate
#(r_2(t))/(r_1(t)) = (([A]_(i,2))/([A]_(i,1)))^m#
#(1.6xx10^(-2))/(2xx10^(-3)) = (("0.1 M")/("0.05 M"))^m#
#2^m = 8#
#color(blue)(m = 3)#
For this process, we thus describe it by saying it is a third-order unimolecular reaction.
What this order says is that if you double the concentration of reactant