Chem 104 1st Edition Lecture 26 Outline of Last Lecture I. Review of important equationsOutline of Current LectureI. Review of important equationsII. Effect of temperature on ratea. Suvante ArrheniusIII. Activation energy and activated complexCurrent LectureI. 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: 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 sII. 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 change of reaction. So temperature affects kinetics.a. Svante Arrhenius equation models temperatures effect on the rate:k =A (−EaeRT) Or k =A∗e−EaRTEa= activation energyb. When you have a molecule: A+B→ProductMolecules have to go through some difficulties, to get to the result. This is similarto a freshman in college. At the start you enter your dorm room and classes and then in 4 years you get a degree, or a “product”. The process in between are the difficulties. Sometimes you get to a certain point and drop out. With molecules, this is when there isn’t enough energy to complete a reaction and make product. The energy needed to continue is the activation energy.So….lnk=(−EaR)(1T)+lnAT = temperature in kelvinsR = gas constant (8.314 J/mol)This is just like the equation for heat of vaporization:−∆ HvapR=ln (P2P1)1T2−1T1So set it up as such:−EaR=ln k2−ln k11T2−1T1When you reach the top of activation energy this is the activated complex:Example:Determine the activation energy and frequency factor for O3→O2+O given:K2=4.83*104K1=3.37*103T2=700T1=600−Ea8.314=ln (4.83∗1 043.37∗1 03)1700−1600−Ea=−575393.012=5.57∗1
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