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Lecture 27

# Chem 402 Lecture 27: L27 3:31:17 Premium

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Department
University College - Chemistry
Course
University College - Chemistry Chem 402
Professor
Barnes Alexander
Semester
Spring

Description
31 March 2017 L27: 2 Order Rate Laws and Kinetic Parameters I. 2 Order Rate Laws A. From Chem 112 1. 2A  Products 1 𝑑 𝐴 a. Differential Rate Law: 𝑟𝑎𝑡𝑒 = − = 𝑘 𝐴 ]2 2 𝑑𝑡 • Integrate:∫ [𝐴]𝑑 𝐴]= 2𝑘 ∫𝑡 𝑑𝑡 [𝐴]0[𝐴] 0 1 1 b. Integrated Rate Law: [𝐴]= 𝐴 0 + 2𝑘𝑡 1 c. Half life: 𝑡 = [ ] 2 2 𝐴 0 2. A + B  Products 𝑑 𝐴] 2 a. Differential Rate Law: − 𝑑𝑡 = 𝑘 𝐴 𝐵 = 𝑘 𝐴 [ ] (if [A] = [B]) [𝐴]𝑑 𝐴] 𝑡 • Integrate:∫[ ] 2 = −𝑘 ∫ 𝑑𝑡 𝐴 0[𝐴] 0 b. Integrated Rate Law: 1 = 1 + 𝑘𝑡 [𝐴] 𝐴 0 1 c. Half life: 𝑡2= [𝐴 0 d. Only works if [A] = [B] during reaction 3. Plot of 1/[A] vs. t for 2 Order reactions 1/[A] 1/[A]0 t a. Linear slope of k or 2k b. Y-intercept of 1/[A] 0 nd B. Deriving the 2 Order Rate Law 1. If [A] ≠ [B]: 𝑑 𝐴] a. – = 𝑘 𝐴 [𝐵] [𝐴]𝑑 𝐴 ] 𝑡 b. ∫ = −𝑘 ∫ 𝑑𝑡 = −𝑘𝑡 [𝐴0 𝐴 [𝐵] 0 𝑑 𝐴] • Problem: 𝐴 [𝐵] has 2 variables that change, but only one variable of integration • Must change variable needed 2. Look at the stoichiometry a. [A] =[A] –0x • Actual = initial – how much has gone (the progress of reaction) 𝑑 𝐴] • = −1 or 𝑑 𝐴 = −𝑑𝑥 𝑑𝑥 b. [B] = [B] –0x 3. Re-write 𝑥 𝑑𝑥 a. ∫ = 𝑘𝑡 0 ( 𝐴0−𝑥)( 𝐵 0𝑥) • Lose the negative because subbing in d[A] = -dx 4. Must split the integrand into partial fractions  method of partial fractions 1 𝐿 𝑀 a. ( 𝐴0−𝑥)( 𝐵 0𝑥) = ( 𝐴0−𝑥) + ( 𝐵 0𝑥) • Must find L and M to make the RHS = LHS, x by denominator b. 1 = 𝐿 𝐵([ ] − 𝑥 + 𝑀( 𝐴 [ ] − 𝑥) 0 0 c. Gather and equate like powers of x: (set up equations LHS = RHS) • X  1= 𝐵 𝐿 + 𝐴 𝑀 [ ] 1 0 0 • X  0 = −𝐿 − 𝑀 d. Solve for L and M 1 1 • 𝐿 = , 𝑀 = −𝐿 = 𝐵 0 𝐴 ]0 [𝐴 0 𝐵 0 5. Substitute for integrand 𝑥 𝑑𝑥 1 𝑥 𝑑𝑥 1 𝑥 𝑑𝑥 a. ∫ = ( ∗ ∫ ) + ( ∗ ∫ ) 0 ( 𝐴0−𝑥)( 𝐵 0𝑥) [𝐵 0 𝐴 ]0 0 [𝐴0−𝑥 [𝐴 0 𝐵 0 0 𝐵 0𝑥 𝑑𝑥 1 • Using the standard i
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