A reaction has the general form: aA rightarrow bB At a certain temperature and a

Question

A reaction has the general form: aA rightarrow bB At a certain temperature and at [ A ] versus time resulted in a straight line with a slope value of -2.97 Times 10-2 min-1 there is a lot going on this problem. Let's take it one step at a time. a. What is the order of the reaction? We can use the data given to find out. A reaction that has a linear plot for In [A] vs. time is a 1st order reaction The rate law for this reaction is: Rate = ln [ v ] + = -kt + ln [ x ]0 When a plot of In [A] vs. time is linear, the slope, according to the integrated rate law is equal to . Therefore, the rate constant for this reaction = For a first order reaction, the half life, t1/2 can be calculated using this elation: t1/2(0.693)/k The half life of this reaction is: How much time is required for the concentration to decrease to 2.50 Times 10-5 M? The last part of this question asks us to determine how much time must elapse to move from our initial concentration, [A]o, to a new concentration [A]. Use the integrated rate law and solve for t (Look at the equations on the back of the Unit 2 Schedule...)

Explanation / Answer

c. From the rate equation ln[A]t = -kt + ln[A]0 we have the slope of ln[A] vs t graph is -k.

from the given data -k = -2.97*10-2 M-1

                               k = 2.97*10-2 M-1

d. Half life of the reaction is calculated as

        t1/2 = 0.693 / k

                = 0.693 / 2.97*10-2 M-1

                = 23.33 M

e. The time rquired for the specific concentration is calculated as

              ln[A]t = -kt + ln[A]0

              kt =ln[A]0 - ln[A]t

               t = 1/k (ln[A]0 - ln[A]t)

                 = 1 / 2.97*10-2 M-1 * (ln0.02 - ln 0.0025)

                 = 70 min

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