Chem 104 1st Edition Lecture 25 Outline of Last Lecture I. Rate lawa. Graphical methodsb. Integrated rate lawi. Zero order1. Graphical representationii. Half lifeOutline of Current LectureI. Review of important equationsCurrent 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: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=−4.04 5ln ¿¿e−4.04=.0175
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