Chem 104 1st EditionExam #3 Study Guide Lectures: 22-28Lecture 22I. Rate Law:a. Within the study of kinetics there is a mathematical representation of the rate of a reaction and the concentrati0ons of the reactants. This is the rate law. The rate law must be determined experimentally. This is because the rate of reaction is directly proportional to the concentration of each reactant raised to a power. This is represented through a sample equation:aA+bB=cC+dDThe lowercase letters represent coefficients of an equation and the uppercase letters represented atoms in an equation. The rate law by definition is represented below with k being the rate constant:B ¿mA ¿n∗¿Rate Law=k ¿a. The remartson rate law is the basic rules of rate law:1. m and n are the order of the reaction with respect to A and BExample: CH4+2O2→CO2+2H2OO2¿3C H4¿2¿rate=2.10∗1 0−4¿The reaction is second order because of the two exponents. Respect to CH4 & third order with respect to O2. Overall its 5.2. Rate law is only written for the concentration of a reactant. So no [C]n * [D]m. This means it does not solve for the products3. m and n are different from the stoichiometry coefficient4. m and n are obtained from experimental data.5. K is called the rate constantWe will study A→P near the end of the chapterb. Reaction orderi. Order is the exponent per reactant with respect to the reactantii. The reaction order is the sum of the exponents.Example: NO with ozone rate = k[NO][O3]. Rate = 6.60*10-5 M/s when NO=1.00*10-6 M and O3=3.00*10-6 M. Calculate the rate constantRate = k[NO][O3]These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.6.60∗10−5=k (1.00∗1 0−6)(3.00∗1 0−6)6.60∗10−5(1.00∗1 0−6) (3.00∗10−6)=kk =2.20∗1 0−7c. Finding the rate lawi. The initial rate method tells us that the rate must be determined experimentally. The rate shows how the rate of a reaction depends on theconcentration of the reactants. Therefore, changing the initial concentration of reactant will change the initial rate of reaction.Example:d. Zero orderi. Zero order is when the rate of the reaction is always the same. If you decided to double [A] the rate wouldn’t be affected. This is when n=0 (hence zero order). When solving you don’t even have to place the initial concentration since it won’t change.e. How to prove mathematicallyi. (This is copied directly from his slides)when comparing experiment 1 to 2, “[A]” changes but “[B]” does not. To determine the rate law, divide experiment 2 by experiment 1:0.020 ¿n¿0.010 ¿n¿k ¿k ¿rate2rate1= ¿Experiment 2Experiment 10.0200.010¿n=1¿2n=1n=0Lecture 23I. The goals of rate law is to measure:1. Time2. Concentration of reactants at a given time3. Control kineticsb. This way an experiment can be repeated, controlled, and discussed with accuracyII. Rate lawa. The rate by rate law is a represented through the equation:B ¿mA ¿n∗¿rate law=k ¿Remember that n and m must come from experimental data.b. The initial rate method is represented by:A ¿nrate=k ¿To solve for the n between two experiments, divide them by each otherIII. When there are multiple reactants with an effect, this results in changing the overall rate of the reaction.Example: Rate = k[NO2]n[CO]m. determine the rate law and k. (this chart was taken directly from the slides)Solve for n then m..10¿m¿.10¿m¿.10 ¿n∗¿k ¿.20 ¿n∗¿k ¿rate2rate1= ¿.20.10¿n=.0021.0082¿2n=4=n=2.20 ¿n¿.20 ¿n¿.10 ¿m∗¿k ¿.20 ¿m∗¿k ¿rate3rate2=¿.20.10¿m=.0083.0082¿m=0Lecture 24I. Rate lawa. The graphical methods are practically similar to any rate law representation. The rate must be determined experimentally. However, a graph of concentration of reactant vs time can be used to determine the effect of concentration on the rateof reaction. However, this method involves calculus. The area under the curve can be measured as such:rate(by definition)=−∆[A]∆ tA ¿nrate(by rate law)=k ¿II. The integrated rate law is applying calculus to show the relationship between the concentration of A and the time of the reaction. This represented as:A ¿n−∆[A]∆ t=k ¿a. The types of rate laws vary depending on what you need to solve in what order. For the zero order, n=0. As though solving for the zero order, you can alter the integrated rate law to:A ¿0A ¿t=− kt +¿¿This is basically y=mx+b. So at any given time you can figure out the cause of the reactants since the constant won’t change. You can calculate the initial concentration like you would a y-intercept. This is done by looking at a graph. K would also represent the slope.b. Half lifei. Half-life is measuring the time for the initial concentration to be half. This can be represented by the equation where the unit of k is M/s:A ¿0¿¿t1 /2=¿Lecture 25I. Below are equations and examples that are important to know and memorize for ourupcoming test:rate(by definition)=−∆[A]∆ tA ¿nrate(by rate law)=k ¿The integrated rate law:A ¿n−∆ [ A ]∆ t=k ¿Case #1:N= 0 (zero order)1.A ¿0A ¿t=− kt +¿¿2. Slope is decreasing3. Half life:A ¿0¿¿t1 /2=¿The unit of k = M/s (molar per second)Case #2:N=1 (first order)1.− ∆ [ A ]∆ t=k [ A ]The P solution:A ¿0A ¿t=− kt +ln ⁡¿ln ⁡¿2. Slope is decreasing3. Half life:t1 /2=.693kCase #3:N=21. The rate = k[A]2A ¿t¿A ¿0¿¿1¿2. Slope is increasing3. Half life:A ¿0k ¿t1 /2=1¿Example:For the reaction SO2→SO2+Cl2 is first order with a rate constant of 2.90*10-4 s-1 at a given set of conditions. Find the [SO2Cl2] at 865s when [SO2Cl2]initial= 0.0225 MA ¿t=− kt +ln ⁡[ A]ln ⁡¿S O2C l2¿865 s=k(865 s)+ln ⁡[0.0225]ln ⁡¿S O2C l2¿865 s=(2.90∗1 0−4)(865)+ln ⁡[0.0225]ln ⁡¿S O2C l2¿865 s=− 4.045ln ⁡¿¿e−4.04=.0175 MLecture 264. Half-life:A ¿0k ¿t1 /2=1¿Example: Molecule 1 dissociates at 625 k with first order rate constant of 0.271s-1. What is the half life?t1 /2=.369k=.369.271=2.55 sExample: Reaction Q→2R is second order in Q. If initial [Q] =0.010 M and k=1.8 M-1s-1. Find length of time for [Q] =1/2[Q] in it.So essentially find the half-life:Q ¿0k ¿t1 /2=1¿t1 /2=1(1.8)(0.010)=55.555=56 sIV. Effect of temperature on rate:Let’s say a closed container contains molecules. So they have to collide, basically that’s a reaction. But what if I add heat? The molecules move more rapidly, faster, and cause a higher