Chem 104 1st Edition Lecture 22 Outline of Last Lecture I. Review of practice examOutline of Current LectureI. Kineticsa. Rate lawi. Remartson rate lawii. Reaction orderiii. Finding the rate law1. Zero orderiv. Proving through algebraCurrent LectureI. Rate lawa. 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 ¿b. 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 chapterc. 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]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 0−7d. 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:e. 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.f. How to prove mathematicallyi. (This is copied directly from his slides)Experiment 2Experiment 1when 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=
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