Report

Using Rate Laws to Predict Concentrations at Time = t If our experimental study of a reaction shows that its rate is first order, we can calculate the concentrations of reactant and products at a time t. Process: A products rate = k[A] Using the form for the instantaneous rate − rearranging [] [] = − integrating from 0 to t rearranging [] − [] = − [] = [] − = [] Using Rate Laws to Predict Concentrations at Time = t For any first order reaction A products the concentration of A at time t is given by [] = − + [] where [A]o is the initial concentration and k is the rate constant. ln[A] is the natural logarithm of A and that is the power to which e (2.71828) must be raised to get A. Plots of First Order Reactions [] = − + [] y = mx + b A plot of [] versus time will be linear with slope = -k and y-intercept = [] . ln [A]0 slope = -k Order of a Reaction - Example 0 1.000 2500 0.947 5000 0.895 7500 0.848 10000 0.803 pressure of SO2Cl2 (atm) Analyze rate data for the reaction SO2Cl2(g) SO2(g) + Cl2(g) at 320°C to prove the reaction is first order and to find the rate constant. time (s) pressure of 1 SO2Cl2 (atm) 0.95 0.9 0.85 0.8 0 2000 4000 6000 8000 10000 time (s) A common mistake is to plot the data given without thought as to “why.” A plot of P vs. time is not linear. I used the “add trendline” feature of Excel. Plots of First Order Reactions If the reaction is 1st order, should a plot of P versus t be linear? (rate = k[A]) [] = − + [] [] = [] − A plot of [] versus time will be a decreasing exponential. Order of a Reaction - Example When we plotted P versus t, we did not get a straight line. So we calculate ln P and plot ln P vs t. time (s) pressure of SO2Cl2 (atm) ln P 0 1.000 0 2500 0.947 -0.0545 5000 0.895 -0.1109 7500 0.848 -0.1649 0.803 -0.2194 ln(pressure of SO2Cl2) 10000 0 -0.05 -0.1 -0.15 -0.2 -0.25 0 Why is it okay to use pressure instead of molarity when analyzing the data? 2000 4000 6000 8000 10000 time (s) A plot of ln P vs. time is linear. This proves that the reaction is first order in SO2Cl2. Order of a Reaction - Example Now we can determine the first order rate constant k: time (s) ln P k (s-1) 0 0 0 2500 -0.0545 2.18 x 10-5 5000 -0.1109 2.26 x 10-5 7500 -0.1649 2.16 x 10-5 10000 -0.2194 2.18 x 10-5 average k is k = 2.20 x 10-5 s-1 at T = 320°C. General equation for concentration in a first order reaction: [] = − + [] For this case: = − + since Po = 1, ln Po = 0 = − =− Plot of a Zero Order Reaction First, write the rate law: rate = k Integrating leads to: [] = − + [] y = mx + b A plot of [] versus time will be linear with slope = -k and y-intercept = [] . slope = -k Plot of a Second Order Reaction First, write the rate law: rate = k[A]2 Integrating leads to: [] = + y = A plot of [] [] mx + b versus time will be linear with slope = k and y-intercept = [] You mark the slope and y-intercept for this plot. . Determining the Order of a Reaction from Concentration vs Time Data On the Unit 3 test, I typically give concentration vs time data and ask a series of questions about the reaction. Give some thought as to how you would use conc vs time data to get: order of reaction, rate law, value of k, etc. If your thought is to plot the data, be sure you know HOW to plot data correctly, how to interpret the plot, and how to give written support to your conclusion. Half-Life of a Reaction The half-life of a reaction is a convenient way to describe the speed of a reaction. The half-life is the time it takes for the concentration of a reactant to drop to onehalf of its initial value. If a reaction has a short half-life, it is a fast reaction. The equation that gives the half-life depends on the rate law for the reaction. We will focus on the half-lives of first order reactions. Half-Life of a First Order Reaction The half-life (t½) is the time it takes for the concentration of a reactant A to drop to one-half of its initial value. For a first order reaction, ln[A]t = - kt + ln[A]o Rearranging [] [] = − [] At the end of one half-life, = − [] [] = − [] Half-Life of a First Order Reaction At the end of one half-life, [] [] = − ln ½ = - kt½ - ln 2 = - kt½ ln 2 = kt½ 0.693 = kt½ or . = . = t½ of a First Order Reaction - Example SO2Cl2(g) SO2(g) + Cl2(g) T = 320°C Determine the time required for the pressure of SO2Cl2 to drop from 1.000 atm to 0.500 atm. How long does it take for the pressure to drop from 0.050 atm to 0.025 atm? time (s) pressure of SO2Cl2 (atm) 0 1.000 2500 0.947 5000 0.895 7500 0.848 10000 0.803 We have already determined that the reaction is first order and that the average k is k = 2.20 x 10-5 s-1 at T = 320°C. t½ of a First Order Reaction - Example SO2Cl2(g) SO2(g) + Cl2(g) T = 320°C Determine the time required for the pressure of SO2Cl2 to drop from 1.000 atm to 0.500 atm. How long does it take for the pressure to drop from 0.050 atm to 0.025 atm? . . = . = = = . . − − The answer to BOTH questions is t½, because in both cases the final pressure is one-half the initial pressure. We did not have to convert pressure to molarity because halflives apply to the ratio of two pressures: P2 = M2RT = M2 P1 M1RT M1 Half-Life of a First Order Reaction The half-life applies to ANY drop in concentration (or pressure) of one-half. [] = [] − t½ of a First Order Reaction - Example How much time is required for a 5.75 mg sample of 51Cr to decay to 1.50 mg if it has a half-life of 27.8 days? Radioactive decay is a first order reaction process. Since the reaction is first order, we know [] [] = − Knowing the half-life gives us k (not t): . . = = = . − . . = −. − . t = 53.9 days