13.4 The Integrated Rate Law: The Dependence of Concentration on Time

The Integrated Rate Law Integrated rate law for a chemical reaction is a relationship between the concentration of the reactant(s) and time. The following equations are on your reference table! The first order integrated rate law is: [A]t ln[A]t = -kt + ln[A]0 OR ln ------ = -kt [A]0 Where [A]t is the concentration at any time t, k is the rate constant, and [A]0 is the initial concentration of A. The rate law above has the form of an equation of a straight line with decreasing slope: y = mx + b. This is because a plot for a first order reaction is a straight line, and even though the slope is negative the rate constant is always positive.

Let’s Try a Practice Problem! Cyclopropane rearranges to form propene in the gas phase. The reaction is first order in cyclopropene and has a measured rate constant of 3.36X10-5s-1 at 720 K. If the initial cyclopropane concentration is 0.0445 M, what will the cyclopropane concentration be after 235.0 min? ln[A]t = -kt + ln[A]0 ln[cyclopropane]t = -(3.36X10-5s-1)((235 min X (60 s / 1 min)) + ln(0.0445 M) ln[cyclopropane]t = -3.586, so [cyclopropane]t = e-3.586 = 0.0277 M

Second Order Integrated Rate Law 1 1 ---- = kt + ---- [A]t [A]0 This equation is also in the form of a straight line, but has an increasing slope because the plot is of the inverse of concentration of the reactant with respect to time.

Zero-Order Integrated Rate Law [A]t = -kt + [A]0 Again here, the equation is in the form of a straight line. These straight lines that can be produced by these integrated rate law equations can be used to aid in interpreting data, and extrapolating points on the graph.

Half-Life of a Reaction The half-life (t½) of a reaction – the time required for the concentration of a reactant to fall to one-half of its initial value. First order reaction half-life: (here t½ is independent of the initial concentration. This causes the half-life to be constant, making the concept of half-life particularly useful for first-order reactions). 0.693 t½ = -------- This formula is k on your reference table From the College Board:

Let’s Try a Practice Problem! Molecular iodine dissociates at 625 K with a first-order rate constant of 0.271 s-1. What is the half-life of this reaction? 0.693 t½ = ---------- k k = -------------- = 2.56 s 0.271s-1

Let’s Try Another! A first-order reaction has a half-life of 26.4 s. How long does it take for the concentration of the reactant in a reaction to fall to one-eighth of its initial value? If you wanted to use an equation instead of a table here, you would need to use: Time = half-life X log2(1/y) where y= fraction of material remaining… so the table may seem easier. Half-life Time Amount Initial value 1 26.4 s ½ initial value 2 52.8 s ¼ initial value 3 79.2 s 1/8 initial value

Summarizing Basic Kinetic Relationships The reaction order and the rate law must be determined experimentally. The rate law relates the rate of the reaction to the concentration of the reactants. The integrated rate law (which is mathematically derived from the rate law) relates the concentration of the reactant(s) to time. The half-life is the time it takes for the concentration of the reactant to fall to one-half of its initial value. The half-life of a first order reaction is independent of the initial concentration. (We will only worry about half-lives of first order reactions in this course)

13.4 pgs. 640-641 #’s 45, 48, 53 & 56 Read 13.5 pgs. 615-622