Consider the reaction between Crystal Violet , which is a dye (much like the Bro
ID: 929472 • Letter: C
Question
Consider the reaction between Crystal Violet, which is a dye (much like the Bromophenol Blue), and OH- ions:
1) Write the rate law for this reaction. Use “k” for the rate constant. Use “m” and “n” to represent the order of reaction with respect to CV+and OH-, respectively:
2) In terms of m and n, what is the overall order of the reaction?
3) When there are multiple reactants in a reaction, it can be difficult to experimentally determine the reaction order of each reactant because both concentrations are changing as a function of time. To simplify the experiment, one reactant can be in extreme excess so that its concentration essentially remains constant during the reaction. Instead of using the rate constant, k, we use a “pseudo rate constant”, k’, which combines both k and the concentration of the reactant which remains constant. Thus, the rate law is written in terms of only one reactant concentration. If OH- is in extreme excess, write the simplified rate law, using the pseudo rate constant, k’, [CV+], and m:
4) Write an expression for k’ in terms of k, [OH-], and n:
N(CH3)2 N(CH3)2 OH N(CH3)2 + OH N(CH3)2 N(CH3)2Explanation / Answer
1) The rate law for this reaction can be written as
Rate = k * [CV+]m * [OH-]n
2) The overall order of the experiment is the summation of the orders of the reaction with respect to different reactants. Thus order of the reaction is (m+n).
3) When there is excess amount of OH- and its concentration can be presumed to be effectively constant through out the span of the reaction the reaction rate can be written as
Rate = k * [CV+]m * [OH-]n
= k' * [CV+]m
4) Thus the new rate constant is expressed as
k' = k * [OH-]n
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