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3: Chemical Kinetics

Drill 1: 11/13 (A) 11/14 (B) Explain what is meant by rate.
Ratio of how much over time What are some units we can use to measure rate of a reaction in chemistry? g/ time, Δmoles/ Δtime, ΔM/ Δt What are some uses/ applications of understanding the rate of a reaction? Increase efficiency for mass production Study/ ID/ categorize the reaction Effective half-life of medicine How could color be used to study the rate of reaction? Time to turn clear Measure how pink over time Drill 1: 11/13 (A) 11/14 (B) Outcome: I can explain the relationship between concentration of a solution and its transmittance. Goal: CW 1, Skills lab Reactants (aq, pink)  Products (aq, clear)

CW 1: Spectroscopy to Find Concentration
Challenge How can light be used to study color and determine concentrations of chemical species in solutions?

Background Measuring how much of which wavelengths of light are absorbed by a substance and getting useful information about that substance from the results, is the scientific discipline of spectroscopy. In a visible spectrophotometer, students shine a beam of light into a solution containing the sample and detect how much of it comes out of the other side of the solution. By comparing the amount of light transmitted by the pure solvent to the amount transmitted when the sample is dissolved in it, we can calculate a quantity called the absorbance.

Background

Background Spectrophotometers can report measurements as percent transmittance (%T) or directly as absorbance. In this investigation, students will be guided to discover the relationship between transmittance and concentration and ultimately the relationship between transmittance, absorbance, and concentration of a solution. To determine these relationships, you will determine the %T for several concentrations of a solution of blue dye (FD&C blue #1). A calibration curve will be created by plotting this data. Using a linear fit, we can generate an equation that lets us find the concentration of the blue dye given its absorbance. You will then measure the absorbance of a sports drink to determine the concentration of the blue dye in the drink.

Background The wavelength of light absorbed by a substance is characteristic of its structure. The intensity of light absorbed depends on the amount of the substance in the solution. Generally, the more concentrated the solution, the more intense the color will be, and the greater then amount of intensity of the light it will absorb. %𝑇= 𝐹𝑖𝑛𝑎𝑙 𝐿𝑖𝑔ℎ𝑡 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐿𝑖𝑔ℎ𝑡 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 ×100% = 𝐼 𝐼 0 ×100%

You will be given a stock solution of Blue #1 dye with a known concentration, which you will dilute and calculate the concentration of. Complete the table below based on the concentration of the stock solution – ask Ms. L for this value. Concentration of Blue #1 Dye Stock Solution: _________ 7.0 𝜇𝑀 𝑀 1 =7.0 𝜇𝑀 𝑀 1 𝑉 1 = 𝑀 2 𝑉 2 𝑉 1 =8 𝑚𝐿 7.0 𝜇𝑀 8 𝑚𝐿 =( 𝑀 2 )(10 𝑚𝐿) 𝑀 2 =𝑥 7.0 𝜇𝑀 8 𝑚𝐿 (10 𝑚𝐿) = 𝑀 2 𝑉 2 =10 𝑚𝐿 𝑀 2 =5.6 𝜇𝑀

Sample Volume of Stock (mL) Volume of Water (mL) Concentration Final Concentration (diluted 10 fold) A 10 B 8 2 C 6 4 D E 3 7 F G 1 9 H 7.0 𝜇𝑀 0.70 𝜇𝑀 5.6 𝜇𝑀 0.56 𝜇𝑀 4.2 𝜇𝑀 0.42 𝜇𝑀 2.8 𝜇𝑀 0.28 𝜇𝑀 2.1 𝜇𝑀 0.21 𝜇𝑀 1.4 𝜇𝑀 0.14 𝜇𝑀 0.7 𝜇𝑀 0.07 𝜇𝑀 0.0 𝜇𝑀 0.0 𝜇𝑀

The spectrum below shows the absorbance of Blue #1 across all wavelengths of light. Based on the data given, what is the optimum wavelength to measure the absorbance of the dye? Absorbance measurements are most accurate and sensitive in the absorbance range 0.2 to 1.0. 630 nm: the peak of the spectrum with an absorbance of 0.85 What color of light corresponds to this wavelength? Orange Given the appearance of the dye in white light, do these observations make sense? Explain. The wavelengths (colors) of light absorbed (orange) by this dye are complementary to the transmitted color (blue).

As absorbance increases, the amount of light that reaches the detector decreases. Eventually, so little light is detected that we approach the limit of the precision of the spectrophotometer. For this reason, if the absorbance is above 1.5, it is recommended that you quantitatively dilute the solution and remeasure its absorbance.

You have a 1.0 M solution that measures an absorbance of 2.0. Explain how to dilute this solution by ten-fold, including calculations you may have to make. We want a solution that is 10% of the original 1.0 M, or 0.1 M Two choices for how to make the solution: Do M1V1 = M2V2, where M1 = 1.0 M V1 = volume of concentrated solution M2 = 0.1 M V2 = desired volume of diluted solution (say, 5 mL, enough to fill the cuvette) Use easy math: a solution that is 1 mL of 1.0 M and 9 mL water is 10% of 1.0 M, or 0.1 M. Assuming a linear relationship, what do you expect the new absorbance of the solution will be? 10% of the original value, or 0.2

Procedure Get ready Log on to the laptop. Open Logger Pro on the desktop and connect the SpectroVis. In the upper left-hand corner, click on the Set Up Sensors button, Select Transmittance as the Spectrometer Mode. Obtain prepared cuvettes containing samples A to H. To correctly use cuvettes: Wipe the outside of each cuvette with a lint-free tissue. Handle cuvettes only by the top edge of the ribbed sides. Dislodge any bubbles by gently tapping the cuvette on a hard surface. Always position the cuvette so the light passes through the clear sides

Procedure Calibrate with a blank Cuvette H only contains DI water. Place the cuvette into the SpectroVis according to the diagram. On Logger Pro, select Experiment, select Calibrate, select Spectrometer 1. Allow the lamp to warm up, then select Finish Calibration, then click OK. By doing this, you are telling the SpectroVis not to “see” the water – only the dye.

Procedure Set the wavelength Place Cuvette A into the SpectroVis. Click on the green Collect button on the top, right hand corner. You should see a complete spectrum showing %T vs. wavelength of light. Click on the red Stop button. Click on the Configure Spectrometer Data Collection button, Select % Transmittance vs Concentration as the Collection Mode. Under the list of wavelengths, click Clear Selection. Choose the wavelength nearest to 630 nm from the list. Click OK to continue. Select “Store last run” when prompted.

Procedure Collect absorbance data To take your first reading, leave Cuvette H in the SpectroVis. Start data collection by pressing the green Collect button. After the value displayed on the screen has stabilized, select Keep and enter the concentration in mol/L. Select OK. The % Transmittance and concentration values have now been saved. Repeat the necessary steps to test the remaining Blue #1 dilute solutions. After you have finished testing all the food dye solutions, stop data collection and save the data. In Logger Pro, choose Save As… from the File menu. Record the data in the data table.

Data Sample Dilution Ratio (mL stock/ mL water) Concentration (M) Measured % Transmittance (%T) Transmittance as a Decimal (T) A 10/0 0.70 1.0 0.010 B 8/2 0.56 1.2 0.012 C 6/4 0.42 2.1 0.021 D 4/6 0.28 6.3 0.063 E 3/7 0.21 12.6 0.126 F 2/8 0.14 24.0 0.240 G 1/9 0.07 47.1 0.471 H 0/10 0.00 99.9 0.999 Observations: More intense dyes have deeper color and transmit less light The Gatorade sample is about as intense as sample F or G, so we expect the concentration of dye to be close to those two cuvettes.

Skills Lab Analysis Open Excel and create the following columns:

Skills Lab Analysis Fill in your data for concentration and T (decimal form). Graph transmittance versus concentration. Select the concentration column, then select the transmittance column. Click on the Insert tab. In the Charts section on the navigation ribbon, click on the scatter plot symbol, Select a scatter plot with no lines between the dots. Note that concentration should be on the x-axis with T on the y- axis. If this is not the case, click the Switch Row/Column button to switch the axes. Once the graph appears, double click the graph and select the green plus sign, Check of the tick box for axes. Label your axes appropriately. Title the graph appropriately.

Skills Lab Analysis Describe the relationship between transmittance and concentration. Is the relationship linear? Explain. No, it appears to be logarithmic or exponential. The data curves, especially at higher concentrations. When presented with non-linear data, scientists may try different ways to graph the data such that the relationship is linear. Linear relationships are ideal because they make it easier to identify unknowns and predict outcomes of investigations. This is accomplished by applying various mathematical operations to the data and graphing the result against concentration.

Skills Lab Analysis Perform the operations shown in the table, (1/T, log T, and –log T). Create a graph for each operation, with concentration on the x-axis following the general series of steps in step 2. Add a linear trendline to each graph by right-clicking a data point and selecting Add trendline. Select Linear under Trendline Options and check off the Display Equation on chart and Display R-squared value on chart.

Skills Lab Analysis The closer the R2 value to one, the better the data is fitted by a linear equation. Which graph shows the most linear relationship? μM vs. –log(T) What is the equation and R2 value of this graph? Define the variables in the equation. 𝑦=4.0004𝑥 y = –log(T) x = concentration in μM

Skills Lab Analysis Based on your linear fit equation, if a sample of dye has a transmittance of 0.25, what is its concentration? 𝑦=4.0004𝑥 −𝑙𝑜𝑔(0.25)=4.0004𝑥 𝑥=0.14 𝜇𝑀 Note: transmittance data for the 0.70 μM and 0.56 μM have been excluded from the graph – they are too concentrated to accurately measure, which caused the tail end of the graph to not be linear.

Skills Lab Analysis Save this file somewhere safe where you can all access it later. Write a detailed, step-by-step procedure for measuring the absorbance of a sample of a sports drink that contains Blue #1 dye. Include the reagents needed, the glassware and equipment that will be used, and the appropriate measurements and obser vations that must be made. Describe how to collect the data using LoggerPro (note – you are only measuring one sample, not creating a calibration curve as you did in the skills lab. Explain how you will use your linear fit to find the concentration of the dye in the sports drink.

Formal Experiment and Lab Write Up Pre-Laboratory Work Complete the following sections in your lab notebook before doing the experiment. Introduction: Explain theory behind the lab, as well as any equations, graphs, or mathematical equations that will be required. Make sure to discuss the relationship between transmittance and absorbance, and when/ why you may wish to dilute a sample. Material Safety and Data Sheets: None of the materials in this experiment are considered hazardous. Materials and Methods: Complete CW 1, including writing the procedure. List all chemicals and equipment (including type and size). Bring that procedure on a sheet of paper to class. You will meet with your lab groups to develop the final procedure and materials, which you will copy into your lab notebook.

Formal Experiment and Lab Write Up During Laboratory Work Complete the following sections in your lab notebook during the experiment. Materials and Methods: Finalize your procedure amongst your lab groups. Data Collection: Create a data table and record data in your lab notebook. Don’t forget to make detailed observations in addition to any measurements.

Formal Experiment and Lab Write Up Post-Laboratory Work Complete the following sections using a word processor after the experiment. Analysis and Conclusion: Download the Spectroscopy to Find Concentration Conclusion document from Complete this document by working with your group members. Submit one printed copy per group.

Formal Experiment and Lab Write Up Submission Check List Cover sheet: name, lab title, date, class, partners Staple carbon copy sheets from the lab notebook showing the pre-lab and during-lab work as outlined above to the cover sheet. But these into a stack, with each group member’s report on top of the previous. Print copy of post-lab work: one per group. But this behind you stack of individual components. Staple it all together.

Summary 1: 11/13 (A) 11/14 (B) Complete Pre-Lab:
Introduction MSDS Materials and Methods (procedure) on a sheet of loose-leaf paper. No “freebies.” Do it on time. Summary 1: 11/13 (A) 11/14 (B) Outcome: I can explain the relationship between concentration of a solution and its transmittance. Goal: CW 1, Skills lab

If the absorbance of a sample is found to be 1
If the absorbance of a sample is found to be 1.0, what % of light is transmitted through the sample? 𝐴𝑏𝑠= −𝑙𝑜𝑔 10 (𝑇) Why is this important experimentally? As absorbance increases, the amount of light reaching the sensor decreases ten- fold. Not enough light gets through to be measured accurately. Drill 2: 11/19 (A) 11/20 (B) Outcome: I can use the rate law to describe a reaction and write the general form of a rate law. Goal: CW 2, CW 3 Hand In: HW 1 10 −𝐴𝑏𝑠 = 10 ( 𝑙𝑜𝑔 10 𝑇) 𝑇= 10 −𝐴𝑏𝑠 = 10 −1.0 =0.1 %𝑇=100×0.1=10%

Spectroscopy Lab Assessment
15 minutes

CW 2: Reaction Rate Why Study Kinetics?
Applications of chemistry often focus on commercial use of chemical reactions. To economize a chemical reaction, we must understand the characteristics of the reaction, such as stoichiometry, energetics, and rate. The area of chemistry which studies reaction rates is called kinetics. One of the main goals of studying chemical kinetics is to understand the steps by which a reaction takes place. This series of steps for a given reaction is known as the reaction mechanism.

CW 2: Reaction Rate Consider the reaction of hydrogen and nitrogen to create ammonia for use in fertilizers. 3H2(g) + N2(g)  2NH3(g) This reaction occurs very slowly if you simply mix H2(g) and N2(g) at room temperature. Why is it important to study this reaction? A very high need for this product exists, we want to make it using the minimum amount of resources (efficiently). What kinds of things do we want to know about to make this reaction as efficient as possible? How long it takes, ideal concentrations, ideal temperature/ pressure for reaction, does it react at a constant rate, stoichiometry, reaction mechanism…

CW 2: Reaction Rate Rate of a Reaction
In the reaction below, the [NO2] (concentration of NO2) decreases with time as the [NO] (concentration of NO) and [O2] (concentration of O2) increase as they are formed. 2NO2(g)  2NO(g) + O2(g) Kinetics studies the speed (or rate) at which these changes occur. The rate is equal to the change in the concentration of a chemical over the time it took to occur. 𝑅𝑎𝑡𝑒= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑁 𝑂 2 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑡𝑖𝑚𝑒 𝑝𝑒𝑟𝑖𝑜𝑑 = 𝑁 𝑂 2 @ 𝑡𝑖𝑚𝑒 2 − 𝑁 𝑂 2 @ 𝑡𝑖𝑚𝑒 1 𝑡𝑖𝑚𝑒 2 − 𝑡𝑖𝑚𝑒 1 = ∆ 𝑁 𝑂 2 ∆ 𝑡

CW 2: Reaction Rate Time (s) [NO2] (mol/L) [NO] [O2] (mol/L) 0.0100 50
The following data were collected as the reaction occurred. Find the average rate at which the [NO2] changes over the first 50. seconds of the reaction. Time (s) [NO2] (mol/L) [NO] [O2] (mol/L) 0.0100 50 0.0079 0.0021 0.0011 100 0.0065 0.0035 0.0018 150 0.0055 0.0045 0.0023 200 0.0048 0.0052 0.0026 250 0.0043 0.0057 0.0029 300 0.0038 0.0062 0.0031 350 0.0034 0.0066 0.0033 400 0.0069 𝑅𝑎𝑡𝑒= ∆[ 𝑁𝑂 2 ] ∆𝑡 = 𝑚𝑜𝑙 𝐿 − 𝑚𝑜𝑙 𝐿 (50 𝑠 −0𝑠) =− 𝑚𝑜𝑙∙𝐿 𝑠

CW 2: Reaction Rate Compare the slope of each chemical as it changes during the reaction. Which one is negative? Why? How does this relate to the answer to Question 3? NO2 is a reactant, so its concentration decreases NO and O2 are products, so their concentration increases Question 3 gave a negative answer because we found the average rate a which NO2 is consumed.

CW 2: Reaction Rate Provide evidence to support or refute: The rate of consumption of NO2 is constant throughout the reaction. Refute: the slope is not a constant straight line, meaning the conversion of reactant into products is not occurring at a constant rate.

CW 2: Reaction Rate Instantaneous Rates
The instantaneous rate of a reaction is the value of the rate at a specific time during the reaction. It can be found by computing the slope of a line tangent to the curve at that point in time. 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑌 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑋 = ∆[ 𝑁𝑂 2 ] ∆ 𝑡 We have considered this reaction rate only in terms of the reactant. We can also write rates with respect to products.

CW 2: Reaction Rate = 0.0006 70 𝑠 =8.6× 10 −6 𝑚𝑜𝑙 𝐿∙𝑠
Find the instantaneous rate of production of NO and O2 at t = 250 seconds. 𝐼𝑛𝑠𝑡𝑅𝑎𝑡𝑒 𝑁𝑂 =𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = 𝑠 =8.6× 10 −6 𝑚𝑜𝑙 𝐿∙𝑠 𝐼𝑛𝑠𝑡𝑅𝑎𝑡𝑒 𝑂 2 =𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = 𝑠 =4.3× 10 −6 𝑚𝑜𝑙 𝐿∙𝑠

CW 2: Reaction Rate How do these rates compare? Explain using the balanced chemical equation. Given your answers to Question 6 and Question 7, what is the instantaneous rate of NO2 consumption at t = 250 seconds? The same as the rate of production of NO, and half of the rate of production of O2. 4.3× 10 −6 𝑚𝑜𝑙 𝐿∙𝑠 Rate of consumption of NO2 Rate of production of NO 2(Rate of production of O2) = −∆ 𝑁𝑂 2 ∆𝑡 = ∆ 𝑁𝑂 ∆𝑡 = ∆ 𝑂 2 ∆𝑡

CW 3: An Introduction to Rate Laws
What is a reversible reaction? A reaction in which products may react to re-form reactants. 2NO2(g) ⇌ 2NO(g) + O2(g) How does the reverse reaction impact the study of its rate? Initially, only the forward reaction occurs Eventually, enough products form that the reverse reaction occurs Now the change in [reactant] depends on the difference in the rates of the forward and reverse reactions (yikes!) To avoid this, we choose conditions where the reverse reaction is negligible and doesn’t contribute to the overall rate: We must study the reaction soon after reactants are mixed, before products have had time to build up to significant levels

How do you write the general rate law equation? 2NO2(g)  2NO(g) + O2(g) Order of the reactant Rate constant 𝑅𝑎𝑡𝑒=𝑘 𝑁𝑂 2 𝑛 Both the rate constant and the order of the reactant can only be determined using experimental data!

What is the rate law for each of these reactions? 4PH3(g)  P4(g) + 6H2(g) 2H2O2(aq)  2H2O(l) + O2(g) N2(g) + 3H2(g)  2NH3(g) 2NO(g) + Cl2(g)  2NOCl(g) I–(aq) + OCl–(aq)  OI–(aq) + Cl–(aq) 𝑅𝑎𝑡𝑒=𝑘 𝑃𝐻 3 𝑛 𝑅𝑎𝑡𝑒=𝑘 𝐻 2 𝑂 2 𝑛 𝑅𝑎𝑡𝑒=𝑘 𝑁 2 𝑛 𝐻 2 𝑚 𝑅𝑎𝑡𝑒=𝑘 𝑁𝑂 𝑛 𝐶𝑙 2 𝑚 𝑅𝑎𝑡𝑒=𝑘 𝐼 − 𝑛 𝑂𝐶𝑙 − 𝑚

What is the difference between the differential rate law and the integral rate law? Differential rate law (simply called the rate law): shows how the rate of a reaction depends on concentrations Integrated rate law: shows how the concentrations of chemical species depend on time If we know one, we can find the other (we usually experimentally measure which ever one is more convenient).

Why do we care to know the rate law for a given reaction? We can work backwards from the rate law to infer the steps of the reaction (most reactions are not completed in one step) If we know the rate determining step (slowest step in the series), we can speed up that step to increase reaction rate For example: A chemist making an insecticide must study the reactions involved in insect development to see what type of molecule might interrupt this series of reactions. A pharmaceutical chemist wants to increase the half life of a drug (how quickly it is used in the body) to make a longer lasting dose

Create a summary of your notes. There are 2 types of rate laws: The differential rate law… The integrated rate law… If you know one… We choose which one to find experimentally by… Our rate laws only involve reactants because they are completed under conditions where the reverse reaction… Knowing the rate law is important because…

Complete CW 2 and CW 3 (including the summary)
HW 2: Rates of Chemical Reaction Collected next class OTC Pain Reliever Lab, due 11/19 or 11/26 (A Day) and 11/20 or 11/27 (B Day) Spectroscopy to Find an Unknown Concentration, due 11/28 (A Day) and 11/29 (B Day) Summary 2: 11/19 (A) 11/20 (B) Outcome: I can use the rate law to describe a reaction and write the general form of a rate law. Goal: CW 2, CW 3 Hand In: HW 1

Write down your outcome in your unit packet
Take out a sheet of paper (which you will hand in) and put your name on it Answer the following (open notebook, not open person): 2H2O2(aq)  2H2O(l) + O2(g) Write the rate law for the reaction. Explain the difference between the differential and integrated rate laws. What data do you need to collect to find an instantaneous rate? Drill 3: 11/26 (A) 11/27 (B) Outcome: I can find rate law constants using the method of initial rates. Goal: CW 4 Hand In: OTC Pain Reliever Lab

Drill Answers 2H2O2(aq)  2H2O(l) + O2(g)
Write the rate law for the reaction. 𝑅𝑎𝑡𝑒=𝑘 [ 𝐻 2 𝑂 2 ] 𝑛 Explain the difference between the differential and integrated rate laws. Differential rate law (simply called the rate law): shows how the rate of a reaction depends on concentrations Integrated rate law: shows how the concentrations of chemical species depend on time What data do you need to collect to find an instantaneous rate? Concentration vs. time

HW 2: Rates of Chemical Reaction

CW 4: Determining the Order of the Rate Law
Reaction Order To understand how a chemical reaction occurs, we must determine the order of the rate law. The order of the rate law is the power to which each reactant concentration must be raised to in the rate law. 2N2O5(soln)  4NO2(soln) + O2(g) 𝑅𝑎𝑡𝑒= 𝑘 𝑁 2 𝑂 5 𝑛 Finding the value of n = “determining the form” or “determining the order”

Considering that one of the products is a gas, why are we able to ignore the reverse reaction in the equation above? The gas escapes the solution and therefore cannot react in the reverse reaction (remember, this was what we aim for) Using the data below, determine the instantaneous rate of the consumption of N2O5(g): @ 0.90 M @ 0.45 M How do these rates compare? When [N2O5] is halved, the rate is halved. The rate depends on just the [N2O5] 𝑅𝑎𝑡𝑒 𝑖𝑛𝑠𝑡 = 0.25 𝑚𝑜𝑙/𝐿 463 𝑠 =5.4× 10 −4 𝑚𝑜𝑙 𝐿∙𝑠 𝑅𝑎𝑡𝑒 𝑖𝑛𝑠𝑡 = 0.17 𝑚𝑜𝑙/𝐿 629 𝑠 =2.7× 10 −4 𝑚𝑜𝑙 𝐿∙𝑠

More specifically, the order of the rate law indicates to what extent the concentration of a species affects the rate of a reaction, as well as which species has the greatest effect. We can experimentally determine the order of a reaction with respect to each reaction component. The overall order of a rate law is the sum of the exponents of its concentration terms. Zeroth order: A zero-order reaction has a constant rate that is independent of the reactant’s concentrations. Many reactions that require catalysts fall in this category. First order: A first-order reaction has a rate that is proportional to the concentration of one reactant. Second order: A second-order reaction has a rate that is proportional to the product of the concentrations of two reactants, or to the square of the concentration of a single reactant. Higher order: Higher order reactions exist, but these chemical systems are very complicated, and it is unusual to see them in an introductory chemistry college course.

What is the reaction order with respect to [N2O5]? What is the overall order of this reaction? The reaction order with respect to N2O5 is first order: as the rate is proportional to the [N2O5] Only one reactant is involved, so the overall order of the reaction is first If there were other reactants, you would have to determine the order with respect to each and add them all up to get the overall order of the reaction.

Determining the Order Using the Method of Initial Rates The method of initial rates is a common method to determine the order of a rate law. It involves measuring the instantaneous rate (see CW 2) of a reaction just after the reaction begins, before products build up in significant concentrations and the reverse reaction complicates the reaction system.

NH4+(aq) + NO2–(aq)  N2(g) + 2H2O(l) Write the general form of the rate law for this reaction, (using variables as the exponents like in CW 3). Why are we able to ignore the products when writing this rate law? One of the products is a gas, which will escape the solution, thus preventing the reverse reaction from occurring. 𝑅𝑎𝑡𝑒=𝑘∙ 𝑁𝐻 𝑛 ∙ 𝑁𝑂 2 − 𝑚

NH4+(aq) + NO2–(aq)  N2(g) + 2H2O(l) The following data was collected for the reaction: Exp. Initial [NH4+] (mol/L) Initial [NO2–] Initial Rate (mol/L∙s) 1 0.100 0.0050 1.35x10–7 2 0.010 2.70x10–7 3 0.200 5.40x10–7

NH4+(aq) + NO2–(aq)  N2(g) + 2H2O(l) Determine the order of the reaction with respect to NO2–(aq) by comparing experiment 1 to experiment 2 𝐸𝑥𝑝. 2: 2.70× 10 −7 =𝑘∙ 𝑛 ∙ 𝑚 𝐸𝑥𝑝. 1: 1.35× 10 −7 =𝑘∙ 𝑛 ∙ 𝑚 2.70× 10 − × 10 −7 = 𝑘∙ 𝑛 ∙ 𝑚 𝑘∙ 𝑛 ∙ 𝑚 2.70× 10 − × 10 −7 = 𝑚 𝑚 2= 2 𝑚 𝑚=1

Which two experiments should you compare to determine the order of rate law with respect to NH4+(aq)? Experiments 2 and 3, where [NO2–] is constant and will cancel. Using the same method as in Question 7, determine the order of the reaction with respect to NH4+(aq). 𝐸𝑥𝑝. 3: 5.40× 10 −7 =𝑘∙ 𝑛 ∙ 𝑚 𝐸𝑥𝑝. 2: 2.70× 10 −7 =𝑘∙ 𝑛 ∙ 𝑚 5.40× 10 − × 10 −7 = 𝑘∙ 𝑛 ∙ 𝑚 𝑘∙ 𝑛 ∙ 𝑚 5.40× 10 − × 10 −7 = 𝑛 𝑛 2= 2 𝑛 𝑛=1

Plug the orders you found in Question 7 and 9 into the rate law from Question 5. Solve for the value of the rate constant (k). Choose any experiment to plug in, I choose Exp. 1: 𝑅𝑎𝑡𝑒=𝑘∙ 𝑁𝐻 ∙ 𝑁𝑂 2 − 1 𝑅𝑎𝑡𝑒=𝑘∙[ 𝑁𝐻 4 + ]∙ 𝑁𝑂 2 − 𝑅𝑎𝑡𝑒 𝑁𝐻 4 + ∙ [ 𝑁𝑂 2 − ] =𝑘 1.35× 10 − ∙ [0.0050] =𝑘 =𝑘

The reaction between bromate ions and bromide ions in an acidic aqueous solution is given by the equation: BrO3–(aq) + 5Br–(aq) + 6H+(aq)  3Br2(l) + 3H2O(l) Using these data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant. Exp. Initial [BrO3–] (mol/L) Initial [Br–] (mol/L) Initial [H+] Initial Rate (mol/L∙s) 1 0.10 8.0x10–4 2 0.20 1.6x10–3 3 3.2x10–3 4 𝑅𝑎𝑡𝑒=𝑘 𝐵𝑟𝑂 3 − 𝑛 𝐵𝑟 − 𝑚 𝐻 + 𝑝 𝑛=1 𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑜𝑟𝑑𝑒𝑟=4 𝑚=2 𝑘=8.0 𝐿 3 𝑚𝑜𝑙 3 ∙𝑠 𝑝=1

Summary 3: 11/26 (A) 11/27 (B) HW 3: Differential Rate Laws
Spectroscopy to Find an Unknown Concentration, due 11/28 (A Day) and 11/29 (B Day) Summary 3: 11/26 (A) 11/27 (B) Outcome: I can find rate law constants using the method of initial rates. Goal: CW 4 Hand In: OTC Pain Reliever Lab

The reaction is overall second order, and first order with respect to CO:
CO + Cl2  COCl2 Write the rate law for the reaction. Describe how to solve for k. Collect data: initial rate and concentrations of each reactant. Plug into the rate law, solve for k. Drill 4: 11/28 (A) 11/29 (B) Outcome: I can determine the order of a reaction using the integrated rate law. Goal: CW 5 Hand In: Spectroscopy to Find an Unknown Concentration Lab 𝑅𝑎𝑡𝑒=𝑘 𝐶𝑂 [ 𝐶𝑙 2 ]

CW 5: Integrated Rate Laws
What is the difference between the (differential) rate law and the integrated rate law? Differential rate law: how the rate depends on concentrations Integrated rate law: how concentrations depend on time We will develop the integrated rate laws individually for first-order, second-order and zero-order reactions. For each case, we will consider a simple reaction involving only one reactant: aA  products

First-Order Rate Laws aA  products If this reaction is first-order (n = 1), we can write the rate law as: 𝑅𝑎𝑡𝑒=− ∆ 𝐴 ∆ 𝑡 =𝑘 𝐴 𝑛 =𝑘 𝐴 1 =𝑘[𝐴] Using calculus (integration), we can put this rate law into another form: 𝐿𝑛 𝐴 =−𝑘𝑡+𝐿𝑛 𝐴 0 Where: [A] = concentration of A k = rate constant t = time [A]0 = Initial concentration of A t = 0)

First-Order Rate Laws aA  products 𝐿𝑛 𝐴 =−𝑘𝑡+𝐿𝑛 𝐴 0 Some Important Observations: The integrated rate law shows how [A] depends on time. If [A]0 and k are known, we can find the [A] at any time during the reaction. The equation is linear, in the form y = mx + b: y = Ln[A] x = t m = –k b = Ln[A] We can test if a reaction is first-order by plotting Ln[A] versus t. If the plot is linear, the reaction is first-order.

The decomposition of N2O5 in the gas phase was studied at constant temperature. N2O5(g)  4NO2(g) + O2(g) Calculate the missing Ln[N2O5] values in the table above. Verify that the rate law is first order in [N2O5] by plotting Ln[N2O5] versus time. Time (s) [N2O5] (mol/L) Ln[N2O5] 0.1000 –2.302 50 0.0707 –2.649 100 0.0500 –2.996 200 0.0250 –3.689 300 0.0125 –4.382 400 –5.075

The decomposition of N2O5 in the gas phase was studied at constant temperature. N2O5(g)  4NO2(g) + O2(g) Calculate the value of the rate constant by finding the slope of the line. 𝑠𝑙𝑜𝑝𝑒= ∆𝑌 ∆𝑋 = ∆(𝐿𝑛 𝑁 2 𝑂 5 ) ∆𝑡 Time (s) [N2O5] (mol/L) Ln[N2O5] 0.1000 –2.302 50 0.0707 –2.649 100 0.0500 –2.996 200 0.0250 –3.689 300 0.0125 –4.382 400 –5.075 = −5.075−(−2.302) 400−0 = − 𝑘=6.93× 10 −3 𝐿 𝑚𝑜𝑙∙𝑠

The decomposition of N2O5 in the gas phase was studied at constant temperature. N2O5(g)  4NO2(g) + O2(g) Calculate the [N2O5] at 150 s after the start of the reaction. 𝐿𝑛 𝑁 2 𝑂 5 =−𝑘𝑡+𝐿𝑛 𝑁 2 𝑂 5 0 𝐿𝑛 𝑁 2 𝑂 5 =−( )(150)+𝐿𝑛(0.1000) 𝐿𝑛 𝑁 2 𝑂 5 =−3.342 𝑒 𝐿𝑛[ 𝑁 2 𝑂 5 ] = 𝑒 −3.342 [ 𝑁 2 𝑂 5 ]= 𝑒 −3.342 𝑁 2 𝑂 5 = 𝑚𝑜𝑙/𝐿

Second-Order Rate Laws aA  products If this reaction is second-order (n = 2), we can write the rate law as: 𝑅𝑎𝑡𝑒=− ∆ 𝐴 ∆ 𝑡 =𝑘 𝐴 𝑛 =𝑘 𝐴 2 Using calculus (integration), we can put this rate law into another form: 1 [𝐴] =𝑘𝑡+ 1 [𝐴] 0 Where: [A] = concentration of A k = rate constant t = time [A]0 = Initial concentration of A t = 0)

Second-Order Rate Laws aA  products 1 [𝐴] =𝑘𝑡+ 1 [𝐴] 0 Some Important Observations: A plot of 1/[A] versus t will produce a straight line of the form y = mx + b with a slope equal to k if the reaction is second-order. This can confirm reaction order. The integrated rate law shows how [A] depends on time and can be used to find [A] at any time during the reaction, provided [A]0 and k are known.

Butadiene reacts to form its dimer: 2C4H6(g)  C8H12(g) Calculate the missing Ln[C4H6] and 1/[C4H6] values in the table above. Time (s) [C4H6] (mol/L) Ln[C4H6] 1/[C4H6] –4.605 100 1000 –5.075 160 1800 –5.348 210 2800 –5.599 270 3600 –5.767 320 4400 –5.915 370 5200 –6.028 415 6200 –6.175 481

Butadiene reacts to form its dimer: 2C4H6(g)  C8H12(g) Determine if the reaction is first-order or second-order by completing the plots.

Butadiene reacts to form its dimer: 2C4H6(g)  C8H12(g) Calculate the value of the rate constant by finding the slope of the line. Time (s) 1/[ C4H6] 100 1000 160 1800 210 2800 270 3600 320 4400 370 5200 415 6200 481 𝑠𝑙𝑜𝑝𝑒= ∆𝑌 ∆𝑋 = ∆ 1 [ 𝐶 4 𝐻 6 ] ∆𝑡 = 481− −0 = 𝑘=6.14× 10 −2 𝐿 𝑚𝑜𝑙∙𝑠

Zero-Order Rate Laws aA  products If this reaction is zero-order (n = 0), we can write the rate law as: 𝑅𝑎𝑡𝑒=− ∆ 𝐴 ∆ 𝑡 =𝑘 𝐴 𝑛 =𝑘 𝐴 0 =𝑘 1 =𝑘 Using calculus (integration), we can put this rate law into another form: 𝐴 =−𝑘𝑡+ 𝐴 0 Where: [A] = concentration of A k = rate constant t = time [A]0 = Initial concentration of A t = 0)

Zero-Order Rate Laws aA  products 𝐴 =−𝑘𝑡+ 𝐴 0 Some Important Observations: A plot of [A] versus t will produce a straight line of the form y = mx + b with a slope equal to –k if the reaction is zero-order. This can confirm reaction order. The integrated rate law shows how [A] depends on time and can be used to find [A] at any time during the reaction, provided [A]0 and k are known. Zero-order reactions have a constant rate that is independent of the reactant concentrations. This most often occurs when a substance such as a metal surface or an enzyme is required for the reaction to occur.

The decomposition reaction below occurs on a hot platinum surface. Two experiments were carried out as shown: 2N2O(g)  2N2(g) + O2(g) Does the [N2O] effect the rate of the reaction? Explain. No – adding more doesn’t cause any more reaction. How do we know that this reaction is zero-order? Explain. Both experiments result in the same amount of products despite different concentrations.

Half-life and Reaction Order The half-life of a reactant (t1/2) is the time required for half of the reactant to be used up in the reaction. Because the integrated rate law shows how concentrations of reactants depend on time, we can use the integrated rate law to find the half-life of a reactant. When t = t1/2, one half-life has past, and the concentration of A is half of its original value: aA  products 𝐴 = [𝐴] @ 𝑡= 𝑡 1 2

We can plug in the equation above for [A] into the integrated rate law for each reaction order, allowing us to find the half-life of any reaction, if we know the correct reaction order. In fact, measuring half life of a reaction can be used to confirm reaction order. Order First Second Zero Rate Law 𝐿𝑛 𝐴 =−𝑘𝑡+𝐿𝑛 𝐴 0 1 [𝐴] =𝑘𝑡+ 1 [𝐴] 0 𝐴 =−𝑘𝑡+ 𝐴 0 Half-life Equation 𝑡 1/2 = 𝑘 𝑡 1/2 = 1 𝑘[𝐴] 0 𝑡 1/2 = 𝐴 0 2𝑘

A first order reaction is found to have a half-life of minutes. Calculate the rate constant for this reaction. How much time is required for the reaction to be 75% complete? 75% = 2 half lives 2×(20.0 min⁡)=40.0 𝑚𝑖𝑛 𝑡 1/2 = 𝑘 20.0= 𝑘 𝑘= 𝑚𝑖𝑛 −1

Return to your calculations from Question 2 and Question 3. Calculate the half-life for each reaction. Question 2: First order, k = 6.93x10–3 s–1 Question 3: Second order, k = 6.14x10–2 L/mol∙s 𝑡 1/2 = 𝑘 𝑡 1/2 = × 10 −3 𝑡 1/2 =100 𝑠 𝑡 1/2 = 1 𝑘 𝐴 0 𝑡 1/2 = 1 (6.14× 10 −2 )( ) 𝑡 1/2 =1630 𝑠

Integrated Rate Laws for Reactions with More Than One Reactant The kinetics of more complex reactions (with more than one reactant) is studied by observing the concentration of one reactant at a time. This is accomplished by having all the reactants in a large excess except for one. If the concentration of one reactant is much smaller than the concentrations of others, then the amounts of those reactants present in large numbers will not change significantly and can be considered as constant.

For example, to determine the order of the reaction in A, the reaction would be carried out such that the [B] and [C] are much higher than [A]. Under these conditions, as A reacts and consumes B and C, the [A] will change greatly while the [B] and [C] will remain close to constant. The change in [A] with time can then be used to determine the pseudo order of the reaction in that component. aA + bB + cC  products 𝑅𝑎𝑡𝑒=𝑘∙ [𝐴] 𝑚 ∙[𝐵] 𝑛 ∙[𝐶] 𝑝

Such a rate law would be known as a pseudo order rate law, as it was obtained by simplifying the rate law by holding [B] and [C] constant, for example: If m = 1, we would say the reaction is pseudo first order in A. This is repeated for each reactant until the complete rate law is found: 𝑅𝑎𝑡𝑒= 𝑘 ′ ∙ [𝐴] 𝑚 Where: 𝑘 ′ = 𝑘∙ 𝐵 𝑛 ∙ 𝐶 𝑝 Exp. Initial [A] (mol/L) Initial [B] Initial [C] Reaction Order 1 1x10–3 1.00 Determined experimentally in A 2 Determined experimentally in B 3 Determined experimentally in C

The reaction NO(g) + O3(g)  NO2(g) + O2(g) was studied in two rate experiments: What is the order with respect to each reactant? NO: 1st order O3: 1st order What is the value of the pseudo rate constant from each set of experiments? NO: k’ =0.0018 O3: k’ =

The reaction NO(g) + O3(g)  NO2(g) + O2(g) was studied in two rate experiments: Use the pseudo rate constant and [O3] from Experiment 1 to solve for the overall rate constant (or the pseudo rate constant and [NO] from Experiment 2). Write the overall rate law for this reaction. 𝑅𝑎𝑡𝑒=𝑘 𝑁𝑂 [ 𝑂 3 ] 𝑘′=𝑘[ 𝑂 3 ] 𝑅𝑎𝑡𝑒=𝑘′[𝑁𝑂] 0.0018=𝑘[1.0× ] 𝑘=1.8× 10 − (𝑐𝑚 3 ) 2 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 2 ∙𝑚𝑠 𝑘′=𝑘[ 𝑂 3 ] 𝑅𝑎𝑡𝑒= 1.8× 10 −17 (𝑐𝑚 3 ) 2 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 2 ∙𝑚𝑠 𝑁𝑂 [ 𝑂 3 ]

Summarize the kinetics for each reaction type.

Summary 4: 11/28 (A) 11/29 (B) HW 4: Integrated Rate Laws
Suggestion: Read Section 12.4 and complete Cornell notes. Pay special attention to the worked math examples. Great summary on Pages 488 and 489. Summary 4: 11/28 (A) 11/29 (B) Outcome: I can determine the order of a reaction using the integrated rate law. Goal: CW 5 Hand In: Spectroscopy to Find an Unknown Concentration Lab

Drill 5: 11/30 (A) 12/3 (B) Write down your outcome
Complete the Integrated Rate Law Drill from by the door. Remind Ms. L to show you the changes to question 2 in CW 6 Drill 5: 11/30 (A) 12/3 (B) Outcome: I can use spectroscopy to determine the rate law for the color-fading reaction of crystal violet with sodium hydroxide Goal: CW 6

CW 6: Rate Law of a Crystal Violet Reaction
Challenge The purpose of this laboratory activity is to determine the rate law for the reaction of crystal violet (CV) and sodium hydroxide (NaOH). In Part 1 of the investigation, you will generate a Beer’s law calibration curve for CV using similar techniques found in the Spectroscopy to Find Concentration Lab (CW 1). In Part 2 of the investigation, you will perform a reaction of CV with NaOH while monitoring in real time the concentration of CV remaining.

Part 1: Create a calibration curve of Abs vs. [CV] Abs [CV] # # Slope = ab A 𝐴 𝑐 =𝑎𝑏 c 𝐴=𝑎𝑏𝑐 Part 2: Find the rate law for the reaction CV+ + OH–  CVOH t Abs # # [CV] # Ln[CV] 1/[CV] # # Make these graphs to find order, slope gives k’, write the rate law. Use Beer’s Law to find the [CV] from the Abs

What is the optimum wavelength and concentration range to measure the absorbance of CV? Absorbance measurements are most accurate and sensitive in the absorbance range 0.2 to 1.0. 590 nm In the range of 0 μM to 100 μM

A calibration curve requires the preparation of a set of known concentrations of CV solutions, which are usually prepared by diluting a stock solution. Complete the table by finding the volume of 25 μM CV solution that would be required to prepare 10. mL of the desired concentration. Sample CV Concentration (μM) Volume of Stock CV (mL) Volume of Water Added (mL) A 2.5 1.0 9.0 B 5.0 2.0 8.0 C 7.5 3.0 7.0 D 10.0 4.0 6.0 E 12.5 F (blank) 0.0 10.0

A calibration curve requires the preparation of a set of known concentrations of CV solutions, which are usually prepared by diluting a stock solution. Show work for the sample B calculations below: M1: 25 μM V1: x M2: 5.0 μM V2: 10.0 mL Describe how to prepare the solutions above. Accurately measure and combine the volume of stock with the volume of water. 𝑀 1 𝑉 1 = 𝑀 2 𝑉 2 𝑉 1 = 𝑀 2 𝑉 2 𝑀 1 = (5.0 𝜇𝑀)(10.0 𝑚𝐿) (25 𝜇𝑀) =2.0 𝑚𝐿 𝑠𝑡𝑜𝑐𝑘

During the reaction of CV with NaOH, do you expect the measured absorbance reading to change? Will it increase, decrease, or remain the same as the reaction proceeds? Explain your reasoning. As CV reacts with NaOH, it forms a colorless product: As the CV fades, less light is absorbed by the dye Absorbance will decrease while transmittance will increase CV+ + OH–  CVOH violet colorless

Review the Integrated Rate Laws for Reactions with More Than One Reactant section from CW 6. How does using a ratio of 1000:1 of NaOH:CV allow you to simplify the rate law? The NaOH is in such great excess that we can consider its concentration as a constant We can wrap up the rate constant and [OH] as one constant How is the rate constant k found from the pseudo rate constant k’? Plug in values for k’ and [OH] wℎ𝑒𝑟𝑒: 𝑘 ′ =𝑘 [𝑂𝐻] 𝑅𝑎𝑡𝑒=𝑘 𝐶𝑉 [𝑂𝐻] 𝑅𝑎𝑡𝑒=𝑘′ 𝐶𝑉

In order to determine the order in CV, What kinds of graphs would you need to construct? [CV] vs. t Ln[CV] vs. t 1/[CV] vs. t What kind of data would you need to collect? Absorbance vs. time Use the calibration curve to find ab in Beer’s Law 𝑐= 𝐴 𝑎𝑏 Use Beer’s Law to find concentration Which ever one is linear tells us the order. Slope of calibration curve ([CV] vs. t)

In order to determine the pseudo-rate constant (k’), What kinds of graphs would you need to construct? Once we know the order of the reaction, we would use the integrated rate law to plot either [CV], Ln[CV], 1/[CV] vs. t Slope = k’ What kind of data would you need to collect? The slope of the linear graph when [OH] is in large excess

The data and graph below show how the concentration of a reactant changes over time for three different reactions: one that is first-order, one that is second- order, and one that is zero-order.

The data and graph below show how the concentration of a reactant changes over time for three different reactions: one that is first-order, one that is second- order, and one that is zero-order. For early parts of the three different reactions in Figure 3, all three curves seem relatively linear with different slopes. But as the reactions progress through time, at roughly what concentration level would you say some graphs start to look nonlinear? The reaction that is 0th order (blue) produces a straight line for a graph of concentration versus time. The other graphs (1st order, red) (2nd order, yellow), appear linear until the concentration dips from its starting value of 10 M to 1 M

The data and graph below show how the concentration of a reactant changes over time for three different reactions: one that is first-order, one that is second- order, and one that is zero-order. Given that you don’t yet know the order of the reaction of CV with NaOH, how might this help you to decide when to stop collecting data? Hint: Think in terms of percent completion instead of concentration. One must collect data for a sufficient amount of time to be able to see deviations from linearity if they are there. Collect data until the value of the absorbance of CV in its reaction with NaOH has dropped to roughly 10% of its initial value.

Open Excel and create the following columns: Fill in your data for Absorbance, including replacing the blank with the wavelength you set the spectrometer to measure.

Graph Absorbance versus concentration. Select the concentration column, then select the Absorbance column. Click on the Insert tab. In the Charts section on the navigation ribbon, click on the scatter plot symbol, Select a scatter plot with no lines between the dots. Note that concentration should be on the x-axis with Abs on the y- axis. If this is not the case, click the Switch Row/Column button to switch the axes. Once the graph appears, double click the graph and select the green plus sign, . Check of the tick box for axes. Label your axes appropriately. Title the graph appropriately.

Based on your graph, what is the value of ab and how can this be used to find the concentration of CV? (HINT: check the end of the Background section.) Beer’s Law: A = abc A = absorbance a = molar extinction coefficient (L/mol∙cm) b = distance travelled by light (path length in cm) c = concentration Plot the absorbance of CV at several known concentrations The slope of the line is ab Measure the absorbance during the reaction, then plug in data and solve 𝑐= 𝐴 𝑎𝑏

Based on your answers to the questions in CW 6, design an experiment to determine the order of the reaction with respect to CV and the pseudo-rate constant (k’) for the reaction of CV with NaOH. Make sure to create a data table. The Collection Mode will be Absorbance vs Time, which can be selected by clicking on the Configure Spectrometer Data Collection button, . For the reaction, combine 5.00 mL of 0.25 μM CV and 5.00 mL M NaOH. The blank will therefore be 5.00 mL of distilled water and 5.00 mL M NaOH. You will not want to mix the reactants until you are ready to collect data (LoggerPro set up to collect absorbance at one wavelength, already calibrated with the blank). You will need to copy and paste the final LoggerPro data into an Excel spreadsheet and save and share this data with your group members to complete the data analysis

Pre-Laboratory Work Introduction: Explain theory behind the lab, as well as any equations, graphs, or mathematical equations that will be required. Make sure to discuss: Applications (why do we care?) Discussion of Beer’s law: How to find ab using a plot of absorbance vs. concentration How this allows us to find concentration given A over the course of the reaction (over time) Discussion of integrated rate laws: Why absorbance vs time data needs to be collected to study the rate law How to use absorbance vs time data to determine the order in CV (what graphs should you make?) How the order in CV can be used to determine the pseudo rate constant (k’) with respect to CV and the rate constant (k) for the reaction. Material Safety and Data Sheets: Look up the MSDS for the chemicals listed below and list major hazards (typically in the Hazards Identification section). Sodium hydroxide Crystal violet in aqueous solution Materials and Methods: Complete CW 6, including writing the procedure. List all chemicals and equipment (including type and size). Bring that procedure on a sheet of paper to class. You will meet with your lab groups to develop the final procedure and materials, which you will copy into your lab notebook.

During Laboratory Work Materials and Methods: Finalize your procedure amongst your lab groups. Data Collection: Create a data table and record data in your lab notebook. Don’t forget to make detailed observations in addition to any measurements.

Post-Laboratory Work Analysis: Use your group’s data to perform all required calculations using a spreadsheet program such as Excel or Goggle Sheets. Copy and paste into the lab report such that it is readable when the page is printed. Complete the following data analysis: Find the concentration of CV over the course of the reaction using your calibration curve data and the measured absorbance. Determine the order of the reaction in CV by creating 1st, 2nd, and 0th order plots of concentration vs. time. Copy and paste these into your post lab as well, ensuring appropriate axis labels and graph titles. Determine the value of the pseudo rate constant (k’) with respect to CV. Sample calculations: Show a sample calculation for finding the concentration of CV for a single time. For the other parts of the Analysis, explain how you arrived at your answers.

Post-Laboratory Work Conclusion: Include the following in narrative form: Discussion of what you did, including the purpose/ goal of the experiment. Describe any changes to the procedure from what you wrote in your lab notebook. Describe major findings and make claims with supporting evidence from the lab. Report the order of the rate law in CV and the pseudo rate law constant (k’). Describe what data you would need to collect in order to find the overall rate law constant (k). Comment on error, both inherit error (due to precision of equipment) and experimental/ human error. Remember that “the lab would have been better if we did it correctly” or “human errors were made” are not discussion of error. Errors should be clearly described along with how they impacted the calculations in the data analysis. Describe refinements or future experiments. What can we study next and why? Once you have completed your conclusion, print and highlight the answers to the above questions.

Submission Check List Put your documents in order and staple: Cover sheet: name, lab title, date, class, partners Carbon copy sheets from the lab notebook of the pre-lab and during-lab work. Printed post-lab work with highlighting

Summary 5: 11/30 (A) 12/3 (B) HW 4: Integrated Rate Laws Prelab work
Outcome: I can use spectroscopy to determine the rate law for the color-fading reaction of crystal violet with sodium hydroxide Goal: CW 6

The reaction is zero order in A with a rate constant of 5
The reaction is zero order in A with a rate constant of 5.0x10–2 mol/L·s at 25°C. An experiment was run where [A]0 = 1.0x10–3 M. A  B + C Write the integrated rate law. Calculate the half life of the reaction. Drill 6: 12/4 (A) 12/5 (B) Outcome: I can find the rate determining step of a reaction. Goal: CW 7 𝐴 =− 5.0× 10 −2 𝑡+1.0× 10 −3 𝑡 1/2 = [𝐴] 0 2𝑘 = 1.0× 10 −3 2(5.0× 10 −2 ) =0.01 𝑠𝑒𝑐

CW 7: Reaction Mechanisms
The series of steps by which a reaction occurs is known as the reaction mechanism. Consider the reaction between NO2 and CO: The balanced equation does not tell us how the reactants become products. This reaction is thought to have the steps: Where k1 and k2 are the rate constants of the individual reactions. In this reaction, NO3(g) is an intermediate, a species formed and consumed during the reaction. It does not appear in the balanced chemical equation. NO2(g) + CO(g)  NO(g) + CO2(g) Rate = k[NO2]2 k1 1 NO2(g) + NO2(g)  NO3(g) + NO(g) 2 NO3(g) + CO(g)  NO2(g) + CO2(g) k2

Molecularity Each proposed step in the reaction mechanism is called an elementary step, a reaction whose rate law can be written from its molecularity. Molecularity is defined as the number of species which must collide to produce the reaction indicated by that step. Unimolecular: only one molecule is involved in the elementary step. Bimolecular: two molecules are involved in the elementary step. Termolecular: three molecules are involved in the elementary step.

Determine the molecularity and rate law for the following elementary steps. Elementary Step Molecularity Rate Law A  products Unimolecular Rate = k[A] A + A  products (2A  products) Bimolecular Rate = k[A]2 A + B  products Rate = k[A][B] A + A + B  products (2A + B  products) Termolecular Rate = k[A]2[B] A + B + C  products Rate = k[A][B][C]

Molecularity For a proposed mechanism to be acceptable, it must meet two requirements. The sum of the elementary steps must give the overall balanced equation for the reaction. The mechanism must agree with the experimentally obtained rate law.

Check that the proposed mechanism below simplifies to the balanced equation. Balanced Equation: NO2(g) + CO(g)  NO(g) + CO2(g) (1) NO2(g) + NO2(g)  NO3(g) + NO(g) (2) NO3(g) + CO(g)  NO2(g) + CO2(g)

Rate Determining Step To test if the mechanism agrees with the experimentally determined rate law, we must identify the rate determining step, or the slowest elementary reaction. Multistep reactions often have one step that is slower than all the other steps. Reactants can become products only as fast as they can get through this slowest step.

For each step of the reaction mechanism, determine the molecularity of the step and the rate law. The rate of the reaction was experimentally determined to be: k = [NO2]2. Based on this data, which step from Question 3 is the rate determining (slowest) step? Step 1: Its rate law matches the experimentally determined rate law. # Reaction Molecularity Rate Law (1) NO2(g) + NO2(g)  NO3(g) + NO(g) Bi Rate = k[NO2]2 (2) NO3(g) + CO(g)  NO2(g) + CO2(g) Rate = k[NO3][CO]

Rate Determining Step How does a chemist determine the correct mechanism for a given reaction? First, the rate law is determined by experiment. Then, using the two requirements previously discussed, the chemist constructs several possible mechanisms. Further experiments are then conducted to eliminate mechanisms that are the least likely. A mechanism can never be proven absolutely. We can only say that a mechanism which satisfies both requirements is possibly correct. Deducing the mechanisms for chemical reactions is difficult and requires skill and experience – we are only touching on this topic.

The balanced equation for the reaction of NO2(g) and F2(g) is shown below, along with the rate law. A suggested mechanism for this reaction is: Is this an acceptable mechanism for this reaction? That is, does it meet both requirements? 1: Elementary steps sum to the overall balanced reaction 2: The experimentally determined rate law matches the proposed slow step of this reaction 2NO2(g) + F2(g)  2NO2F(g) Rate = k[NO2][F2] 1 NO2 + F2  NO2F + F 2 F + NO2  NO2F Slow Fast

The decomposition of H2O2 was studied, producing the following data. Time (s) [H2O2] (mol/L) 𝑅𝑎𝑡𝑒= ∆[ 𝐻 2 𝑂 2 ] ∆𝑡 1.00 120 0.91 300 0.78 600 0.59 1200 0.37 1800 0.22 2400 0.13 3000 0.082 3600 0.050

Use this data to determine the rate law, integrated rate law, and the value of the rate constant. Plot [H2O2], Ln[H2O2], 1/[H2O2] versus t, determine which is linear First order! 𝑘=− 𝐿𝑛 −(𝐿𝑛 1.00 ) −0 k = –slope 𝑘=8.3× 10 −4 𝑠 −1 𝑅𝑎𝑡𝑒=(8.3× 10 −4 𝑠 −1 )[ 𝐻 2 𝑂 2 ] 𝐿𝑛 𝐻 2 𝑂 2 =− 8.3× 10 −4 𝑠 −1 𝑡 +𝐿𝑛(1.00)

A possible mechanism for the decomposition of H2O2 is shown below. Specify which step is the rate determining step. (1) H2O2  2OH Rate = k[H2O2] (2) H2O2 + OH  H2O + HO2 Rate = k[H2O2][OH] (3) HO2 + OH  H2O + O2 Rate = k[HO2][OH] Step 1 is the rate determining step as it matches the rate law.

Summary 6: 12/4 (A) 12/5 (B) HW 5: Reaction Mechanisms (OWL)
A Day: 12/10 B Day: 12/11 Rate Law of a Crystal Violet Reaction lab will be next class Summary 6: 12/4 (A) 12/5 (B) Outcome: I can find the rate determining step of a reaction. Goal: CW 7

Agenda 12/6 (A) and 12/7 (B) CV Part 1: Calibration Curve
Determine ab (slope of Abs vs. c) Follow directions in CW 6 (measure at 590 nm) Include a data table of this and sketch of the plot, along with the equation of the linear fit and R2 value in your observations section CV Part 2: Reaction with NaOH Blank and warm up, set SpectroVis to measure at 590 nm Go to Experiment > Data Collection and change the duration to 900 sec and check off the continuous data collection box Measure out 2 mL NaOH into a cuvette, insert into SpectroVis Using the pipette in the bin, measure out 2 mL of 25 μM CV Partner A adds the CV solution to the cuvette, Partner B presses play so soon as Partner A starts adding the CV Record the first 2 seconds of data in your lab notebook under observations, noting that you will disregard the first two seconds of the data to account for the time to mix the reactants Copy and paste data into an Excel sheet, share data with partners (but analysis is individual) Post Lab Discussion Report due 12/20 and 12/21

Drill 7: 12/10 (A) 12/11 (B) Log on to a laptop with one partner
Write down your outcome Complete the reaction mechanism drill from by the door Drill 7: 12/10 (A) 12/11 (B) Outcome: I can relate the rate of a reaction to the energy changes of a reaction. Goal: CW 8

CW 8: Chemical Kinetics Learning Goals
I can describe how the reaction coordinate can be used to predict whether a reaction will proceed including how the potential energy of the system changes. I can describe what affects the potential energy of the particles and how that relates to the energy graph. I can describe how the reaction coordinate can be used to predict whether a reaction will proceed slowly, quickly or not at all. I can use the potential energy diagram to determine: The approximate activation energy for the forward and reverse reactions. The sign difference in energy between reactants and products. I can draw a potential energy diagram from the energies of reactants and products and activation energy. I can calculate the value of activation energy of a reaction using graphical and algebraic methods.

CW 8: Chemical Kinetics Important Vocabulary
Reaction coordinate: reflects the extent to which a reaction has progressed from reactants to products, starting with reactants near the y-axis (the energy coordinate) and progressing toward products. Potential energy: energy that comes from position and a force (in chemistry, this force is the Coulomb attraction and repulsion that atoms experience when they collide). When the charges have the same sign, they repel and will accelerate away from each other; the potential energy has a positive sign. When the charges have the opposite sign, they attract each other and have negative potential energy. If they get closer together, the potential energy will get more negative. If they are separated, d gets bigger and the potential energy approaches zero. 𝑃𝐸= 𝑘 𝑄 1 𝑄 2 𝑑 Where Q1 and Q2 are charges, k = 8.99x109 J∙m∙C2 (constant), and d = distance between the two charges

Coulomb attraction: The attraction between opposite charges due to size of the charges and distance separating the charges. Kinetic energy: energy that comes from motion. This definition should make sense: big things moving fast have the most energy, the most ability to shove other things or knock them over, etc. Activation energy (Ea): the minimum amount of energy two reactants must posses in order to react successfully. Activated complex: The temporary structure formed when reactant molecules collide in the correct orientation and with enough activation energy to react. This must be formed before a reaction can occur. 𝐾𝐸= 1 2 𝑚 𝑣 2 Where m = mass of particle and v = velocity

Forward reaction: Reactants  Products Reverse reaction: Products  Reactants Enthalpy (H): Sum of all the heat content of a system: potential energy, kinetic energy, energy due to work (pressure times volume). Enthalpy change (∆H): 𝐻 𝑓𝑖𝑛𝑎𝑙 − 𝐻 𝑖𝑛𝑖𝑡𝑖𝑎𝑙

CW 8: Chemical Kinetics

CW 8: Chemical Kinetics Talk with your partner about what you think is happening on a microscopic level when the iron (III) nitrate and sodium thiocyanide mix. Fe3+ + SCN–1 ⇌ FeSCN2+ Draw pictures that would help you describe the process. Make a list of what things that might make the color change happen faster and explain your reasoning. Cause more collisions (change concentration, maybe temperature) Make a list of what things might make more of the colored complex form and explain your reasoning. Shift equilibrium conditions yellow red Fe3+ SCN–

CW 8: Chemical Kinetics Run experiments using Single Collisions to determine on a simplest level what contributes to a successful reaction. Make sure that you use the Energy view and Separation view to help you explain how the energy changes in a reaction can help you make predictions. Explain the difference between total energy and potential energy. Describe how each can be changed. Try to get the sliding gray bar “stuck” on each side of the curve. How does the Separation view help you? What are two ways you can change the kinetic energy of the colliding particles? Make sketches of energy graphs to help describe how the energy diagram can be used to predict if the reaction will occur or not.

CW 8: Chemical Kinetics Run experiments using Many Collisions to determine what contributes to a successful reaction and what affects the speed of the reaction. Describe how this model relates to the single collision model. Make a table to demonstrate that you have thoroughly used all the simulation features.

CW 8: Chemical Kinetics Sketch the energy graph could look like for the forward reaction to be an exothermic reaction. What would the sign for ΔH be for the forward reaction? Reverse reaction? Select the Design Your Own Reaction to make your own exothermic reaction. Explain how the Activation energy for the forward and the one for the reverse reaction are similar and how they differ.

CW 8: Chemical Kinetics Sketch the energy graph could look like for the forward reaction to be an endothermic reaction. What would the sign for ΔH be for the forward reaction? Reverse reaction? Select the Design Your Own Reaction to make your own endothermic reaction. Explain how the Activation energy for the forward and the one for the reverse reaction are similar and how they differ.

CW 8: Chemical Kinetics Sketch the energy graphs for the following situations. The reactants have a lower potential energy than the products. The activation energy of the reverse reaction is greater than the forward reaction The products have a lower potential energy than the reactants. The forward reaction has a positive ΔH. The reverse reaction has a negative ΔH.

Predicting Activation Energy
CW 8: Chemical Kinetics Predicting Activation Energy Two requirements must be satisfied for reactants to collide and produce a reaction: The collision must involve enough energy to break chemical bonds (the collision energy must equal or exceed the activation energy). The molecular orientation of the reactants must allow formation of any new bonds necessary to produce products. In other words, most collisions do no result in a successful reaction, and the rate of reaction is much lower than the predicted rate of collisions.

CW 8: Chemical Kinetics Taking these factors into account, we can re-write the rate constant as the Arrhenius equation: 𝑘=𝐴 𝑒 − 𝐸 𝑎 /𝑅𝑇 Where: k = rate constant A represents the fraction of collisions with effective orientations Ea = activation energy R = universal gas law constant ( J/K∙mol) T = Kelvin temperature (K) 𝑒 − 𝐸 𝑎 /𝑅𝑇 represents the fraction of collisions with enough energy to produce a reaction.

CW 8: Chemical Kinetics Method 1 to Find Ea
We can take the natural logarithm of both sides of the Arrhenius equation to produce a linear equation of the type y = mx + b: This means we can find the activation energy for a reaction by experimentally measuring the rate constant k at several temperatures and then plot Ln(k) versus 1/T. The slope of the line along with the universal gas law constant can then be used to find Ea. Ln 𝑘 =− 𝐸 𝑎 𝑅 1 𝑇 +Ln(𝐴) Where: y = Ln(k) m = – Ea/R ( the slope) x = 1/T b = Ln(A) (the intercept)

CW 8: Chemical Kinetics The decomposition of N2O5 was studied at several temperatures, and the following values of k were obtained 2N2O5(g)  4NO2(g) + O2(g) Open a spread sheet (or calculator) and type in the data above. Find the Kelvin temperature, 1/T and Ln(k) for all data points. T (°C) T (K) 1/T (K–1) k (s–1) Ln(k) 20 293 3.41x10–3 2.0x10–5 –10.82 30 303 3.30x10–3 7.3x10–5 –9.53 40 313 3.19x10–3 2.7x10–4 –8.22 50 323 3.10x10–3 9.1x10–4 –7.00 60 333 3.00x10–3 2.9x10–3 –5.84

CW 8: Chemical Kinetics 𝑚= − 𝐸 𝑎 𝑅 Ln 𝑘 =− 𝐸 𝑎 𝑅 1 𝑇 +Ln(𝐴) 𝐸 𝑎 =−𝑚𝑅
The decomposition of N2O5 was studied at several temperatures, and the following values of k were obtained 2N2O5(g)  4NO2(g) + O2(g) Create a plot of 1/T versus Ln(k) and use a linear fit to find the slope. Use the slope and the universal gas law constant ( J/K∙mol) to find Ea. 𝑚= − 𝐸 𝑎 𝑅 Ln 𝑘 =− 𝐸 𝑎 𝑅 1 𝑇 +Ln(𝐴) 𝐸 𝑎 =−𝑚𝑅 𝑦 = 𝑚 𝑥 𝑏 𝐸 𝑎 =−(−12186)( 𝐽/𝑚𝑜𝑙∙𝐾) 𝐸 𝑎 =1.0× 𝐽/𝑚𝑜𝑙

CW 8: Chemical Kinetics Method 2 to Find Ea
Another method to find Ea is to measure the rate constant k at two different temperatures, then solving algebraically using the equation: 𝑙𝑛 𝑘 2 𝑘 1 = 𝐸 𝑎 𝑅 1 𝑇 1 − 1 𝑇 2 Where: k1 = rate constant at T1 k2 = rate constant at T2 T1 = Kelvin temperature 1 T2 = Kelvin temperature 2 Ea = activation energy R = universal gas law constant ( J/K∙mol)

CH4(g) + 2S2(g)  CS2(g) + 2H2S(g)
CW 8: Chemical Kinetics The reaction between methane and diatomic sulfur is: CH4(g) + 2S2(g)  CS2(g) + 2H2S(g) At 550 °C the rate constant for this reaction is 1.1 L/mol∙s, and at 625 °C the rate constant is 6.4 L/mol∙s. Using these values, calculate Ea for this reaction. 𝐿𝑛 𝑘 2 𝑘 1 = 𝐸 𝑎 𝑅 1 𝑇 1 − 1 𝑇 2 𝐿𝑛 = 𝐸 𝑎 𝐽/𝑚𝑜𝑙∙𝐾 𝐾 − 𝐾 𝐸 𝑎 = 𝐽 𝑚𝑜𝑙 ∙𝐾 𝐿𝑛 𝐾 − 𝐾 =1.4× 𝐽/𝑚𝑜𝑙

CW 8: Chemical Kinetics Create a chart to provide evidence that you have met each learning goal defined above. Hand in your results. A template is available on LEFFELlabs > Unit 3

Complete Kinetics Learning Goal Evidence Chart
Due: 12/12 (A) & 12/13 (B) Section 12.6 HW 6: Review for Unit test Due: 12/14 (A) & 12/17 (B) Rate Law of a Crystal Violet Reaction Due: 12/20 (A Day) and 12/21 (B Day) Summary 7: 12/10 (A) 12/11 (B) Outcome: I can relate the rate of a reaction to the energy changes of a reaction. Goal: CW 8

A first order reaction has the following rate constant data:
What is the value of Ea? Drill 8: 12/12 (A) 12/13 (B) T k 0 °C 4.6x10–2 s–1 20 °C 8.1x10–2 s–1 Outcome: I can explain how catalysts impact the rate of a reaction. Goal: CW 9 Hand In: Kinetics Learning Goal Evidence Chart 𝐿𝑛 𝑘 2 𝑘 1 = 𝐸 𝑎 𝑅 1 𝑇 1 − 1 𝑇 2 𝐿𝑛 8.1× 10 −2 4.6× 10 −2 = 𝐸 𝑎 𝐽/𝑚𝑜𝑙∙𝐾 𝐾 − 𝐾 𝐸 𝑎 = 𝐽 𝑚𝑜𝑙 ∙𝐾 𝐿𝑛 8.1× 10 −2 4.6× 10 − 𝐾 − 𝐾 =1.9× 𝐽/𝑚𝑜𝑙

Crystal Violet Rate Law Post Lab Discussion
Goal: Measure [CV] using spectroscopy to find the rate law Part 1: Graphed [CV] vs. Abs to get slope, which is ab in Beer’s Law. Part 2: Found the pseudo rate constant, which was used to find the overall rate constant. We also determined the order of the reaction. Findings: order in [CV], k’, k Order: Plot [CV] vs. t, Ln[CV] vs. t, 1/[CV] vs. t, look for linear Pseudo rate constant: slope of the linear plot Overall rate constant: solve by plugging in [NaOH] (constant) Errors: Too slow to get into cuvette, reaction already started, may miss initial rate Instrumental errors: not letting the lamp warm up/ blanking it (inaccurate Abs values) Errors in solution making, could be incorrect concentration, which affects the linear fit in part 1 Refinements/ future experiments: Improve mixing technique of reactants Study other reactions Calculate activation energy by finding the rate constant at another temperature

Goal: Measure rate of reaction between CV and NaOH Part 1: Measured Abs of several known [CV], graphed [CV] vs. Abs, slope is the term ab in Beer’s Law (A = abc) – allows us to find [CV] from Abs Part 2: Measured Abs over t, use Abs to solve for [CV]. Determine order of the reaction by graphing [CV] vs. t, Ln[CV] vs. t, 1/[CV] vs. t, look for the linear graph. Find the rate law by finding k from k’ (slope of the linear graph – NaOH was in large excess, so the slope is k’) Findings: Report rate law, order, values of k and k’. k’ is the slope Rate = k[CV]n[NaOH]m Rate=k’[CV]n k’ = k[NaOH]m Errors: Residue on cuvettes impacts measured Abs Challenges in mixing the reactants and measuring the initial rate Assumed [CV] was correct in Pt 1, may skew the linear fit (ab) Refinements/ future experiments: Improve reactant mixing technique so we don’t lose the first 2 seconds Improve consistency in cuvettes Study other more real world reactions

Goal: Determine the rate law (order and k) for CV and NaOH Part 1: Calibration curve: measured Abs at different [CV], graph [CV] vs. Abs, slope = ab (Beer’s Law: A=abc) Part 2: Reaction between NaOH and CV, measured Abs over time. Convert Abs to [CV] using Beer’s Law. Findings: order in CV, k’, k Order: Plot [CV], Ln[CV], and 1/[CV] vs. t, look for linear k’: slope of the linear plot (k’ = k [NaOH]) k: plug in k’ and [NaOH] (constant: M) and solve Errors: Problems with mixing reactants (some CV left behind in pipette), CV splatters out of cuvette, affects Abs, timing of mixing/ collecting data Errors in preparing Part 1 solutions, concentration could be wrong, skews the linear fit, wrong ab value Refinements/ future experiments: Improve mixing technique Calculate activation energy by finding the rate constant at another temperature, use this to speed up a desired reaction Study other reactions, or maybe a more real world reaction

Goal: Determine the rate law of the CV + NaOH reaction Part 1: Used spectrophotometer to measure the Abs of several [CV] solutions. Plotted [CV] vs. Abs and found the slope (ab term in Beer’s Law, A = abc) Part 2: Used spectroscopy to monitor [CV] over time as it reacted with NaOH, to determine the rate law Findings: order in CV, k’, k Order: Graph [CV] vs. t, Ln[CV] vs. t, and 1/[CV] vs. t, linear graph tells us the order k’ = k[NaOH]m. Obtain k’ from the linear graph k: Solve by plugging in [NaOH] (constant) Errors: CV on the side of the cuvette in part 2 could cause a “bump” in the Abs data (Abs too high, concentration too high) Inaccurate measuring of reactants may cause different concentrations Stopped data collection too soon Too slow to get into cuvette, mixing errors Errors in calibration curve solutions could skew ab and thus concentration Refinements/ future experiments: Improve mixing technique of reactants Test the reaction under different conditions to see if we get the same rate law, study other reactions (perhaps a more real world one)

Crystal Violet Rate Law Lab Assessment

CW 9: Kinetics and Catalysts (Section 12.6)
How does the collision model describe a chemical reaction? Molecules must collide with enough KE: KE is converted into PE as the molecules are distorted during a collision to break bonds and rearrange atoms into product molecules. Collision must have the correct orientation in 3D space. This means that most collisions do not produce reactions and the rate of collisions is much higher than the rate of the reaction.

What is the transition state? The correct arrangement of atoms that produces a reaction; labeled at the top of the potential energy “hill”.

What is activation energy and why is it required? Minimum amount of energy needed to produce a reaction. Comes from the KE of the molecules, which is translated into PE. Rate depends on the size of Ea (large Ea = slower, molecules less likely to collide with enough energy to occur). Only collisions that occur past Ea will result in reactions

How can we find the value of the activation energy? Use the Arrhenius Equation: 𝑘=𝐴 𝑒 − 𝐸 𝑎 𝑅𝑇 Graphical Method Ln 𝑘 =− 𝐸 𝑎 𝑅 1 𝑇 +Ln(𝐴) Find the rate constant (k) for the same reaction at several temperatures. Plot Ln(k) vs 1/T (Kelvin), find slope Ea = –(slope)( J/K∙mol) Algebraic Method L𝑛 𝑘 2 𝑘 1 = 𝐸 𝑎 𝑅 1 𝑇 1 − 1 𝑇 2 Find the rate constant at two different temperatures. Plug in K1, K2, T1, T2. Solve for Ea using J/K∙mol.

How are enzymes and catalysts related? Catalyst: a substance that speeds up a reaction without being consumed itself. Enzyme: biological catalyst for biological reactions (life). Provides a new reaction pathway with lower Ea, reaction happens faster.

What types of catalysts exist? Heterogenous: catalyst is in a different phase than the reacting molecules Often involves a gaseous reaction with a solid catalyst Homogenous: catalyst is in the same phase as the reacting molecules

What are the steps in heterogenous catalysis? Reactants are above the metal surface Hydrogen is adsorbed (not absorbed), forming H–metal bonds, breaking H–H bonds. Adsorbed reactants react, forming new H–C bonds The ethane cannot bind to the metal, and thus escapes

How is ozone decomposed by homogenous catalysis? NO(g) (produced by emissions) reacts with O2 to make NO2. In the presence of UV light, NO2 decomposes, forming unstable O. The unstable O radical reacts with O3 (ozone), producing O2. A similar set of reactions occurs with CFCs catalyzing the reaction:

How can acids and bases act as catalysts? Acids can supply H+, bases can supply OH– or remove H+.

Create a summary of your notes. Collisions and conditions to produce a reaction Activation energy: relate to the rate solving for activation energy (2 methods) How do catalysts impact the activation energy and kinetics (rate) Examples of catalysts Two types of catalysis

Summary 8: 12/12 (A) 12/13 (B) HW 6: Review for Unit test
Due: 12/14 (A) & 12/17 (B) Rate Law of a Crystal Violet Reaction Due: 12/20 (A Day) and 12/21 (B Day) Summary 8: 12/12 (A) 12/13 (B) Outcome: I can explain how catalysts impact the rate of a reaction. Goal: CW 9 Hand In: Kinetics Learning Goal Evidence Chart

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