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myct/itemView?assignmentProblemID 78334951&offset; next tructor: Fiddler CRN 2... Help Close Cs) for First Order Reactions Resources previous l 11 of 25 l next. Hal-life equation for first-order reactions where t/2 is the half-life in seconds (s). and kis the rate constant in inverse seconds (s 1) Part A What is the half-tife of a first-order reaction with a rate constant of 8.00 10-4 s 1? Express your answer with the appropriate units. Value Units Submit Hints My Answers Give Up Review Part Part B to is constant of a first order reaction that takes 320 seconds for the reactant concentration to drop half of its initial value? Express your answer with the appropriate units.Explanation / Answer
Given t1/2 = 0.693/k where t1/2 = half life of a reaction in s and k is the rate constant in s-1.
Part A
Given k = 8.00*10-4 s-1; put in the equation for t1/2 and obtain
t1/2 = 0.693/k = 0.693/(8.00*10-4 s-1) = 866.25 s = (866.25 s)*(1 min/60 s) = 14.4375 min (ans).
Part B
The reaction takes 320 s to drop to one half of its initial concentration. Half life of a reaction is defined as the time taken by the reaction to drop to half of the initial concentration. Therefore, by definition, the half life of the given reaction is 320 s.
Plug the value of t1/2 = 320 s in the half life expression and obtain
t1/2 = 0.693/k
===> k = 0.693/t1/2 = 0.693/(320 s) = 2.1656*10-3 s-1 2.166*10-3 s-1 (ans).
Part C
The first order reaction has the rate constant 2.50*10-3 s-1. Therefore, k = 2.50*10-3 s-1. Plug in the equation and obtain
t1/2 = 0.693/k = 0.693/(2.50*10-3 s-1) = 277.2 s = (277.2 s)*(1 min/60 s) = 4.62 min
As per the definition of half life, the reaction will take 4.62 mins to reduce to half the initial concentration. The time taken by the reaction to drop to one fourth of the initial concentration (half of half concentration obtained after 4.62 min) = 2*4.62 min = 9.24 min.
The time taken by the reaction to drop to one eight of the initial concentration = 3*4.62 min = 13.86 min (ans).
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