AP Chem Work on warm up problem Important Dates: Iodine Clock Lab Write-Up due Tues 1/23 via Google Docs Unit 4 Test (Kinetics) ~ next Thurs 1/25 Today: integrated rate laws—determining reaction order from concentration vs. time data.

Variables we care about in integrated rate law equations [A]o = initial concentration of A [A]t = concentration of A at some time, t k = the rate constant t = time

Integrated Rate Laws: First Order Reactions ln[A]t – ln[A]o = –kt (this is the version on your formula sheet) Can be rearranged to look like: ln[A]t = –kt + ln[A]o (y = mx + b)

Examining First Order Reactions Calculate the missing values from the data table and plot the data for the two graphs as indicated. You can calculate each value individually or plug them into a “list” on your graphing calculator

For a first-order reaction, which of the following will generate a straight-line graph? [A] vs. time ln[A] vs. time 1/[A] vs. time

2.07−2.30 10−0 Calculating Slope =−𝟎.𝟎𝟐𝟑 𝒔𝒆𝒄 −𝟏 ln[A]t = –kt + ln[A]o (use any 2 data points from the straight-line graph): 2.07−2.30 10−0 =−𝟎.𝟎𝟐𝟑 𝒔𝒆𝒄 −𝟏 ln[A]t = –kt + ln[A]o (y = mx + b) The absolute value of the slope represents k, the rate constant.

Second Order Reactions Complete the questions for the second order reactions.

For a second-order reaction, which of the following will generate a straight-line plot? [A] vs. time ln[A] vs. time 1/[A] vs. time

Calculating Slope (use any two data points from the straight- line graph): 0.15−0.10 10−0 =𝟎.𝟎𝟎𝟓 𝑴 −𝟏 𝒔𝒆𝒄 −𝟏 . This represents the value of k, the rate law constant

Zero Order Reactions For a zero-order reaction, the rate of the reaction (i.e., the disappearance of A) does NOT depend on the concentration of the reactant A. The rate law for a zero-order reaction can be written as follows: rate = k The following integrated rate law can be obtained: [A]t – [A]o = –kt   For a zero-order reaction, which of the following will generate a straight-line plot? [A] vs. time ln[A] vs. time 1/[A] vs. time

Half Life (t1/2) the time required for the concentration of a reactant to reach half of its original value. A short half-life suggests a fast reaction rate; a long half-life suggests a slow rate. If a decomposition reaction yields data that shows that the half-life is constant over time, then the reaction must be a first-order reaction.

Summary

Example Questions Given the following reaction: 2 N2O5 --> 4 NO2 + O2, the concentration of N2O5 was monitored over time. *Note: sometimes you’ll only be given the concentration data. If so, you can create a table on your graphing calculator to determine the values for ln [x] and 1/[X] and then create separate graphs to determine the reaction order. Time (sec) [N2O5] (mol/L) ln [N2O5] 1/[N2O5] 0.100 -2.303   50 0.0707 -2.649 100 0.0500 -2.996 200 0.0250 -3.689 300 0.0125 -4.382 400 0.00625 -5.075

Determine the Reaction Order with respect to N2O5 Graph [N2O5] vs time. If linear  0 order Graph ln [N2O5] vs time. If linear  1st order Graph 1/[N2O5] vs time. If linear  2nd order

Graph of ln [N2O5] vs time (bottom graph) is linear, therefore the reaction is first order with respect to N2O5

First Order Integrated Rate Law Equation ln[N2O5]t – ln[N2O5]o = –kt OR ln[N2O5]t = –kt + ln[N2O5]o

Solve for k −5.075 − −2.303 400−0 slope= − 0.00693 k = 0.00693 s-1

If the [N2O5] is initially 0 If the [N2O5] is initially 0.224 M, what is its concentration after 450 seconds? ln[N2O5]t = –kt + ln[N2O5]o ln[N2O5]450 = –(0.00693)(450) + ln (0.224) ln[N2O5]450 = -4.614 [N2O5]450 = e-4.614 [N2O5]450 = 0.00991 M

How long will it take the N2O5 to decompose from 0.100 M to 0.0125 M? ln[N2O5]t – ln[N2O5]o = –kt ln[N2O5]t – ln[N2O5]o −𝑘 =𝑡 ln[0.0125] – ln[0.100] −0.00693 =𝑡 t = 300. seconds.

HW Answers 1a) 0.552 M 1b) 693 sec 2a) 0.00800 M 2b) 198 sec 3 100 sec 4a) Rate = k[H2O2] 4b) k= 1.77 x 10-5 sec-1 4c) 39,153 sec 5a) Rate = k[B]2 5b) 0.06 M-1sec-1 5c) 0.00204 M 6a) k = 0.11 sec-1 6b) 0.220 M 6c) 0.127 M 6d) 6.3 sec