3: Chemical Kinetics www.leffellabs.com.

3: Chemical Kinetics

Drill 1: 11/18 (A) 11/19 (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 Drill 1: 11/18 (A) 11/19 (B) Outcome: I can use the rate law to describe a reaction and write the general form of a rate law. Goal: CW 1, CW 2

CW 1: 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 1: 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 1: 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 1: 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 1: Reaction Rate Compare the slope of each chemical as it changes during the reaction. Which one is negative? 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. Is the rate of consumption of NO2 is constant throughout the reaction? Explain. The slope is not a constant straight line, meaning the conversion of reactant into products is not occurring at a constant rate.

CW 1: 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 1: 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 1: 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, but negative. −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 2: 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

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. How to write the rate law… 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 1 and CW 2 (including the summary)
HW 1: Rates of Chemical Reaction Collected next class Unit 3 Progress Check (AP Classroom), 11/18 (A Day) and 11/19 (B Day) Percent by Mass of Copper in Brass, 11/22 (A Day) and 11/25 (B Day) Summary 1: 11/18 (A) 11/19 (B) Outcome: I can use the rate law to describe a reaction and write the general form of a rate law. Goal: CW 1, CW 2

Drill 3: 11/20 (A) 11/21 (B) A + 3B  2C + D
The reaction above is first order with respect to A and second order with respect to B. Write the rate law for the reaction. If [A] is doubled and [B] is halved, the rate of reaction would _____ by a factor of _____. Increase, 2 Decrease, 2 Increase, 4 Decrease, 4 Drill 3: 11/20 (A) 11/21 (B) Outcome: I can find rate law constants using the method of initial rates. Goal: CW 3 𝑅𝑎𝑡𝑒=𝑘[𝐴] [𝐵] 2

HW 2: Rates of Chemical Reaction

CW 3: 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 Now we can write the complete rate law! 𝑅𝑎𝑡𝑒=(2.7× 10 −4 )[ 𝑁𝐻 4 + ] 𝑁𝑂 2 − 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

Unit 3 Progress Check (AP Classroom), 11/18 (A Day) and 11/19 (B Day)
HW 2:Differential Rate Laws, Due: 11/22 (A) and 11/25 (B) (OWL) Percent by Mass of Copper in Brass, 11/22 (A) and 11/25 (B) Summary 2: 11/20 (A) 11/21 (B) Outcome: I can find rate law constants using the method of initial rates. Goal: CW 3

The reaction is overall second order, and first order with respect to CO:
CO + Cl2  COCl2 Write the rate law for the reaction. Describe what data you need to solve for the rate constant, k. Collect data: initial rate and concentrations of each reactant. Plug into the rate law, solve for k. Drill 4: 11/22 (A) 11/25 (B) Outcome: I can determine the order of a reaction using the integrated rate law. Goal: CW 4 Hand In: Percent by Mass of Copper in Brass Lab 𝑅𝑎𝑡𝑒=𝑘 𝐶𝑂 [ 𝐶𝑙 2 ]

CW 4: Integrated Rate Laws
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 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. 𝐿𝑛 𝐴 =−𝑘𝑡+𝐿𝑛 𝐴 0

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 1 and Question 2. Calculate the half-life for each reaction. Question 1: First order, k = 6.93x10–3 s–1 Question 2: 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 >1 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’ = (from the slope of Ln[NO] vs. t) O3: k’ = (from the slope of Ln[O3] vs. t)

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.

HW 3: Integrated Rate Laws, due 11/26 (A) and 12/2 (B).
Suggestion: Read Section 12.4 and complete notes (of your preference). Pay special attention to the worked math examples. Great summary on Pages 488 and 489. Unit 5 Progress Check (AP Classroom), due 12/11 (A) and 12/12 (B) Rate Law of a Crystal Violet Reaction Lab Report, due 12/17 (A) and 12/18 (B) Summary 4: 11/28 (A) 11/29 (B) Outcome: I can determine the order of a reaction using the integrated rate law. Goal: CW 4 Hand In: Percent by Mass of Copper in Brass Lab

Complete the drill for Day 5 in your unit packet.
Drill 5: 11/29 (A) 12/2 (B) Outcome: I can use spectroscopy to determine the rate law for the color-fading reaction of crystal violet with sodium hydroxide Goal: CW 5

CW 5: 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] # # This is going to require good Excel skills. Start early and ask questions often. 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 to measure the absorbance of CV? 590 nm

A calibration curve requires a set of known concentrations of CV solutions, 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 concentrations. A (CV Stock) B C D E F G (Blank) Conc. (μM) 25 2.5 5.0 7.5 10.0 12.5 0.0 Stock (mL) 1.0 2.0 3.0 4.0 Water (mL) 9.0 8.0 7.0 6.0

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 4. 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 Write the rate law and pseudo rate law for this reaction. How is the rate constant k found from the pseudo rate constant k’? Plug in values for k’ and [OH]n wℎ𝑒𝑟𝑒: 𝑘 ′ =𝑘 𝑂𝐻 𝑛 𝑅𝑎𝑡𝑒=𝑘 𝐶𝑉 𝑛 𝑂𝐻 𝑚 𝑅𝑎𝑡𝑒=𝑘′ 𝐶𝑉 𝑛

To determine the order in CV, what kinds of graphs would you need to construct? (HINT: CW 3, Question 8) [CV] vs. t Ln[CV] vs. t 1/[CV] vs. t To determine the pseudo-rate constant (k’), what kinds of graphs would you need to construct? (HINT: CW 4, Question 7) Conduct the experiment where [NaOH] is in large excess to make it a constant. k’ is the slope of the linear plot under these experimental conditions. Which ever one is linear tells us the order.

Based on your answers from Question 5 and 6 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 Slope of calibration curve ([CV] vs. t)

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

Procedural Notes Change collection time to 300 seconds Fillable cuvette volume is 3 mL. Use pipette to measure: 1.50 mL 25 μM CV 1.50 mL M NaOH (note that this concentration is halved) Set up spectrometer: measure ABS vs. t at one wavelength, calibrate/ blank Add 1.50 mL CV first, then measure out 1.50 mL NaOH Add 1.50 mL NaOH to cuvette and at the same time press “collect” Disregard first couple of readings that happen during mixing – ABS jumps around Copy and paste data into Excel. Share with group members.

OWL HW 4: Reaction Mechanisms, due 12/9 (A) and 12/10 (B)
AP Classroom Unit 5 Progress Check, due 12/11 (A) and 12/12 (B) Rate Law of a Crystal Violet Reaction Lab Report, 12/17 (A) and 12/18 (B) Unit 2 Test Redos: by 12/11 Summary 5: 11/29 (A) 12/2 (B) Outcome: I can use spectroscopy to determine the rate law for the color-fading reaction of crystal violet with sodium hydroxide Goal: CW 5

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: 12/5 (A) 12/6 (B) Outcome: I can find the rate determining step of a reaction. Goal: CW 6 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

Rate Law of CV Post Lab Assessment

CW 6: 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 H2O2  H2O + O2

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)

The mechanism of ozone reacting to produce oxygen is given below, along with the observed rate law. 2O3(g)  3O2(g) 𝑅𝑎𝑡𝑒=𝑘 [ 𝑂 3 ] 2 [ 𝑂 2 ] Proposed Mechanism: O3 ⇌ O2 + O O + O3  2O2 (slow) k1 k2 k3

The mechanism of ozone reacting to produce oxygen is given below, along with the observed rate law. Write the rate law for the slow step. Does this match the observed rate law? Notice that an intermediate is found in the rate law. To deal with this, we must consider the fast equilibrium step which generates the intermediate. Write the rate law for the forward and reverse reactions. During equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. Set the rate law for the forward reaction equal to the rate law of the reverse reaction and solve for [O]. Return to the rate law you wrote for the slow step, substituting in your new expression for [O]. Combine rate constants together into one constant and simplify the expression.

The mechanism of ozone reacting to produce oxygen is given below, along with the observed rate law. The rate law does not match: 𝑅𝑎𝑡𝑒= 𝑘 3 𝑂 [ 𝑂 3 ] Forward: 𝑅𝑎𝑡𝑒= 𝐾 1 𝑂 3 Reverse: 𝑅𝑎𝑡𝑒= 𝐾 2 𝑂 2 [𝑂] Set equal: 𝐾 1 𝑂 3 = 𝐾 2 𝑂 2 [𝑂] Solve for [O]: 𝑂 = 𝑘 1 𝑂 3 𝑘 2 𝑂 2 Substituting in [O]: Simplify: 𝑅𝑎𝑡𝑒= 𝑘 3 𝑘 1 𝑂 3 𝑘 2 𝑂 2 [ 𝑂 3 ] 𝑅𝑎𝑡𝑒=𝑘 𝑂 𝑂 2

OWL HW 4: Reaction Mechanisms, due 12/9 (A) and 12/10 (B)
AP Classroom Unit 5 Progress Check, due 12/11 (A) and 12/12 (B) Rate Law of a Crystal Violet Reaction Lab Report, 12/17 (A) and 12/18 (B) Unit 2 Test Redos: by 12/11 Summary 6: 12/5 (A) 12/6 (B) Outcome: I can find the rate determining step of a reaction. Goal: CW 6

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