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For A ⟶ products , time and concentration data were collected and plotted as shown. [ A ] (M) t (s) 0.900 00.0 0.474 30.0 0.321 60.0 0.243 90.0 A plot of the concentration of A versus time is concave up and decreases with time. A plot of the natural logarithm of the concentration of A versus time is concave up and decreases with time. A plot of one over the concentration of A versus time increases linearly with time. Determine the reaction order, the rate constant, and the units of the rate constant. order = 0 k = units=

User Aneeez
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Final answer:

The reaction described follows first-order kinetics, indicated by the linear relation in a plot of 1/[A] versus time. The rate constant, or k, can be found from the slope of this line, and its units are s^-1 for a first-order reaction.

Step-by-step explanation:

The reaction described is first-order, as indicated by the linear increase in a plot of one over the concentration of A versus time. For a first-order reaction, the integrated rate law is described by the equation [A] = [A]o e-kt, and taking the natural logarithm of each side gives a straight-line relationship between ln[A] and time. Since the plot of concentration versus time (concave upward) and the natural logarithm of concentration versus time also shows a decrease with time (concave upward), these are consistent with a first-order reaction.

The rate constant (k) can be determined from the slope of the plot of 1/[A] versus time, which is a straight line. The slope of this line is equal to k, and the rate constant can be found using any two points on this line.

As for units, in a first-order reaction, the rate constant k has units of s-1. This is because the rate law for a first-order reaction in terms of rate constant is rate = k[A], and since rate has units of M/s (molarity per second) and [A] has units of M (molarity), the units of k must be s-1 to balance the equation.

User Vivek Maru
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