Download Understanding Rate Laws: First and Second Order and more Study notes Law in PDF only on Docsity! 10/19/2020 1 Integrated Rate Law 2 types of rate laws The differential rate law (what we have already done, often called simply the rate law) shows how the rate of reaction depends on concentration. The integrated rate law shows how the concentrations of species in the reaction depend on time. Once you have either the differential rate law or the integrated rate law, you easily convert to the other. Whichever type is easiest to find usually dictates which type of rate law is determined experimentally. Rate Law A first-order reaction (n the exponent is 1) means that as the concentration doubles, the rate also doubles, etc. Differential Rate law Equation: Rate = k [A]1 The integrated rate laws is looking at how the concentration changes with time. Rate = k[A] Integrated rate law: ln[A]t - ln[A]o = –kt ln is the natural log, it is reversed by raising the expression to e [A]t = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A First-Order Rate Law Important things to note from the previous equation. The equation shows how the concentration of A depends on time. If the initial concentration of A and the rate constant k are known, the concentration of A at any time can be calculated. Plot of ln[N2O5] vs. Time First Order Reaction Thus, for a first-order reaction, plotting the natural logarithm of concentration vs. time always gives a straight line. This is used to test whether a reaction is first order or not. This integrated rate law for a first- order reaction also can be expressed in terms of a ratio of [A] and [A]o as follows: ln[A] = -kt +ln[A]o kt = ln[A]o - ln[A] kt = ln ([A]o/[A]) 10/19/2020 2 Decomposition of N2O5 The following data shows the concentration of over N2O5 time 2N2O5(g) 4NO2(g) + O2(g) [N2O5] (mol/L) Time (s) 0.1000 0.00 0.0707 50.00 0.0500 100.00 0.0250 200.00 0.0125 300.00 0.00625 400.00 [N2O5] over time Time (s) [N 2O 5] ( M ) 0 0.02 0.04 0.06 0.08 0.1 0.12 0 50 100 150 200 250 300 350 400 450 Problem Using this data, verify that the rate law is first order in [N2O5], and calculate the value of the rate constant. Answer You verify it is a first order rate law by graphing. If the graph of the ln[A] v t is a straight line then it is first order. slope The slope of this line is –k. Slope is rise over run or the change in y value, ln [N2O5], over the change in x value, time. You can choose any 2 points on a straight line since the slope will not change. I will arbitrarily chose the first and last points Slope = -5.075- (-2.303) = -.00693s-1 400-0s Slope = -k so k = 0.00693s-1 First Order Rate Laws Using the data in the previous example, calculate [N2O5] at 150 s after the start of the reaction. k = 0.00693s-1 [N2O5] (mol/L) Time (s) 0.1000 0.00 0.0707 50.00 0.0500 100.00 0.0250 200.00 0.0125 300.00 0.00625 400.00 Answer ln[N2O5] = -kt +ln[N2O5]o k = 0.00693 t = 150 s [N2O5]o = .100 M ln[N2O5]o= -2.303 ln[N2O5] = -(.00693)150+ -2.303 ln[N2O5] = -3.3425 [N2O5] = e-3.3425 [N2O5] = .035 M Half life of a first order reaction The time required for a reactant to reach half its original concentration is called the half-life of a reactant. Equation k = rate constant Half–life does not depend on the concentration of reactants. 1 2 0.693 = t k Problem A certain first-order reaction has a half-life of 20.0 minutes. a. Calculate the rate constant for this reaction. b. How much time is required for this reaction to be 75% complete?
