Use the information given here to answer both questions (4 and 5).  For the fictitious reaction:  AB → A + B Some data was collected measuring the [AB] at certain time intervals.  This data was then graphed as described below. A graph of 1/[AB] versus time (s) gives a straight line with the equation: y = 0.0052x + 0.1135 and a R2 = 0.9995.  A graph of ln[AB] versus time (s) gives a straight line with the equation: y = -0.0068x + 1.725 and a R2 = 0.8925.   Assume the data doesn't fit a zero order reaction. What is the rate law for this reaction? Select one: a. Rate = k[AB]2 b. Rate = k[A][B] c. Rate = k[AB] d. Rate = k   Using the data from the previous question. What is the value of the rate constant for the given reaction? Select one: a. k = 0.0052 M-1s-1 b. k = 0.0157 Ms-1 c. k = 0.0068 s-1 d. There is no way to determine the value of the rate constant from the given information.

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Chapter12: Chemical Kinetics
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Use the information given here to answer both questions (4 and 5). 

For the fictitious reaction:  AB → A + B

Some data was collected measuring the [AB] at certain time intervals.  This data was then graphed as described below.

A graph of 1/[AB] versus time (s) gives a straight line with the equation: y = 0.0052x + 0.1135 and a R2 = 0.9995.  A graph of ln[AB] versus time (s) gives a straight line with the equation: y = -0.0068x + 1.725 and a R2 = 0.8925.   Assume the data doesn't fit a zero order reaction.

What is the rate law for this reaction?

Select one:
a. Rate = k[AB]2
b. Rate = k[A][B]
c. Rate = k[AB]
d. Rate = k
 

Using the data from the previous question.

What is the value of the rate constant for the given reaction?

Select one:
a. k = 0.0052 M-1s-1
b. k = 0.0157 Ms-1
c. k = 0.0068 s-1
d. There is no way to determine the value of the rate constant from the given information.
Expert Solution
Step 1

From the given information, it is clear that the graph of 1/[AB] versus time (s) gives a straight line with the equation: y = 0.0052x + 0.1135

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