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At a particular temperature and [A]0 = 2.00 x 10-2 Molar, concentration versus time data were collected for this reaction and a plot of ln[A]t versus time resulted in a straight line with a slope value of -2.97 x 10-2 min-1.

What is the half-life of this reaction?

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Final answer:

The half-life of the reaction is calculated using the formula t1/2 = 0.693/k, where the rate constant k is the magnitude of the slope of the ln[A] versus time plot for a first-order reaction. Given a slope of -2.97 x 10^-2 min^-1, the half-life is approximately 23.4 minutes.

Step-by-step explanation:

The half-life of a reaction, denoted as t1/2, is defined as the time required for the concentration of a reactant to decrease to half its initial value. In the context of a first-order reaction, where a plot of ln[A] versus time yields a straight line, the slope of this line is equivalent to the negative of the rate constant (k). The equation for the half-life of a first-order reaction is given by t1/2 = 0.693/k. With a given slope (rate constant) of -2.97 x 10-2 min-1, we can use this equation to calculate the half-life.

By substituting the given slope into the half-life equation, we find:

t1/2 = 0.693/(-2.97 x 10-2 min-1)

As the rate constant will be a positive value (since it is the magnitude of the negative slope), the half-life calculation would be:

t1/2 = 0.693/(2.97 x 10-2 min-1)

After performing the calculation, the half-life is determined to be approximately 23.4 minutes.

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