The reaction #2"Ce"^(4+)(aq) + "Tl"^(+)(aq) -> 2"Ce"^(3+)(aq) + "Tl"^(3+)(aq)# is slow; write the rate law?
1 Answer
Oct 5, 2017
This asks you to know the definition of an elementary step --- it is simply saying that the reaction stoichiometry exactly describes the molecularity of the reaction, i.e. the number of particles involved in each collision.
And the molecularity in such an elementary step then gives the orders of each reactant in the reaction.
Therefore, for
#color(red)(2)"Ce"^(4+)(aq) + "Tl"^(+)(aq) -> 2"Ce"^(3+)(aq) + "Tl"^(3+)(aq)# ,
we assume the rate law, being for a very slow reaction, is:
#color(blue)(r(t) = k["Ce"^(4+)]^color(red)(2)["Tl"^(+)])#
