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Chem 300 - Ch 28/#2 Today’s To Do List

Chem 300 - Ch 28/#2 Today’s To Do List. 1 st -Order Reaction Kinetics ½-life & Reaction Order 2 nd -Order Reactions Reversible Reactions. Analysis of the Rate Law. In general: For a reaction: A + B  products Rate law: v(t) = -d[A]/dt = k [A] m(A) [B] m(B)

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Chem 300 - Ch 28/#2 Today’s To Do List

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  1. Chem 300 - Ch 28/#2 Today’s To Do List • 1st-Order Reaction Kinetics • ½-life & Reaction Order • 2nd-Order Reactions • Reversible Reactions

  2. Analysis of the Rate Law • In general: • For a reaction: A + B  products • Rate law: v(t) = -d[A]/dt = k [A]m(A) [B]m(B) • Find time-dependence of [A] . • Integrate the rate-law. • Consider 1st & 2nd –order rate-laws.

  3. 1st-Order Reactions • v(t) = - d[A]/dt = k [A] • (d [A]) / [A] = - k dt • d [A] /[A] = - k  dt • From [A]0 [A] & t = 0  t = t • ln [A] – ln [A]0 = -kt • [A] = [A]0 e-kt

  4. 1st Order Reactionk = (dot)0.0125 s-1, 0.025, 0.050, 0.10

  5. N2O5 2 NO2 + ½ O2 @ 318Kslope = -k = 3.04 x 10-2 min-1

  6. Reaction ½-Life (t1/2) • t1/2 = Time required for ½ of reactant to disappear (when [A] = ½ [A]0) • For 1st-order reaction: • ln [A] – ln [A]0 = -kt • ln (½ [A]0) – ln [A]0 = -k t1/2 • ln (½) + ln [A]0) – ln [A]0 = -k t1/2 • ln ½ = - ln2 = -k t1/2 • t1/2 = 0.693/ k • Independent of [A]0

  7. N2O5 2 NO2 + ½ O2 @ 318K

  8. Wide range of Reaction Speeds (k-values) • cyclopropene  propyne • k = 5.7 x 10-4 /s at 500 K • cyclobutane  2-ethene • k = 1.8 x 10-12 /s at 500 K

  9. 2nd-Order Reactions • - d[A]/dt = k [A]2 • Separate variables: • (d [A]) / [A]2 = - k dt • Integrate: • 1/[A] = 1/[A]0 + kt • Plot 1/[A]vs t gives a straight line. • Slope = - k intercept = 1/[A]0

  10. NOBr  NO + ½ Br22nd-Order Reaction

  11. 2nd-Order ½-Life • 1/[A] = 1/[A]0 + kt • Substitute [A] = ½ [A]0 at t = t1/2 • 2/[A]0 = 1/[A]0 + kt1/2 • Rearrange: • t1/2 = 1/k[A]0 • Depends upon [A]0 (initial conc.)

  12. 2nd-Order Rate Law • A + B  products • -d[A]/dt = -d[B]/dt = k[A][B] • The integrated form: • kt = {1/([A]0 – [B]0)} ln[A][B]0/[B][A]0

  13. Reversible Reactions • A = B • k1 = forward reaction • k-1 = reverse reaction • At equilibrium: -d[A]/dt = d[B]/dt = 0 • Rate forward = k1[A] • Rate reverse = k-1[B] • k1[A]eq = k-1[B]eq • k1/k-1 = [B]eq/[A]eq = Keq

  14. Next Time • Relaxation Methods & Fast Reactions • Temperature Dependence • Transition-State Theory

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