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CHEM1612 - Pharmacy Week 11: Kinetics - Half Life

CHEM1612 - Pharmacy Week 11: Kinetics - Half Life. Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone: 9351 4196 E-mail: siegbert.schmid@sydney.edu.au. Unless otherwise stated, all images in this file have been reproduced from:

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CHEM1612 - Pharmacy Week 11: Kinetics - Half Life

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  1. CHEM1612 - PharmacyWeek 11: Kinetics - Half Life Dr. SiegbertSchmid School of Chemistry, Rm 223 Phone: 9351 4196 E-mail: siegbert.schmid@sydney.edu.au

  2. Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille,Chemistry, John Wiley & Sons Australia, Ltd. 2008      ISBN: 9 78047081 0866

  3. Rate Law and Reaction Order For the general reaction: a A + b B + c C … d D + e E …. rate = k [A]m[C]n • k rate constant (depends only on temperature) • m is the order of the reaction with respect to A (or “in” A), • n is the order of the reaction with respect to C • Overall order of the reaction is = m + n • Reaction orders cannot be deduced from the balanced reaction, but only by experiment.

  4. Interpreting Rate Laws Rate = k [A]m[B]n[C]p • If m = 1, reaction is 1st order in A Rate = k [A]1 If [A] doubles, then rate goes up by a factor of ? • If m = 2, reaction is 2nd order in A. Rate = k [A]2 If [A] doubles, then rate goes up by a factor of ? • If m = 0, reaction is zero order in A. Rate = k [A]0 If [A] doubles, then rate goes up by a factor of ? 2 4 1

  5. Using Data to Determine Order Consider the reaction: ClO3-(aq) + 9 I-(aq) + 6 H+(aq) Cl-(aq) + 3 I3-(aq) + 3 H2O(l) Expt.[ClO3-] / M [I-] / M [H+] / M Initial Rate / M s-1 1 0.10 0.10 0.10 0.05 2 0.20 0.10 0.10 0.10 3 0.20 0.20 0.10 0.20 4 0.20 0.20 0.20 0.80 Determine the rate law for the above reaction. General expression: Rate = k [ClO3-]x[I-]y[H+]z; must find x, y, and z

  6. Using Data to Determine Order in ClO3- Pick expts where only one conc. changes Expt. [ClO3-] / M [I-] / M [H+] / M Initial Rate / M s-1 1 0.10 0.10 0.10 0.05 2 0.20 0.10 0.10 0.10 3 0.20 0.20 0.10 0.20 4 0.20 0.20 0.20 0.80 General expression: Rate = k [ClO3-]x [I-]y [H+]z Determine x: • Compare experiment 1 and 2

  7. Using Data to Determine Order in I- Expt. [ClO3-] / M [I-] / M [H+] / M Initial Rate / M s-1 1 0.10 0.10 0.10 0.05 2 0.20 0.10 0.10 0.10 3 0.20 0.20 0.10 0.20 4 0.20 0.20 0.20 0.80 General expression: Rate = k [ClO3-]x [I-]y [H+]z Determine y: • Compare experiment 2 and 3 Pick expts where only one conc. changes

  8. Using Data to Determine Order in H+ Expt. [ClO3-] / M [I-] / M [H+] / M Initial Rate / M s-1 1 0.10 0.10 0.10 0.05 2 0.20 0.10 0.10 0.10 3 0.20 0.20 0.10 0.20 4 0.20 0.20 0.20 0.80 General expression: Rate = k [ClO3-]x [I-]y [H+]z Determine z: • Compare experiment 3 and 4 So overall the rate law is: Rate = k [ClO3-] [I-] [H+]2 Pick expts where only one conc. changes

  9. Determine the Rate Constant ClO3-(aq) + 9 I-(aq) + 6 H+(aq) Cl-(aq) + 3 I3-(aq) + 3 H2O(l) rate= k [ClO3-][I-][H+]2 The order is 1 with respect to [ClO3-] The order is 1 with respect to [I-] The order is 2 with respect to [H+] The overall order of the reaction is (1+1+2)= 4. Calculate the rate constant, k e.g. Experiment1 0.05 M s-1 = k (0.10 M)(0.10 M)(0.10 M)2 k = 5.0 x 102 M-3 s-1 (note: units depend on rate law) [ClO3-] / M [I-] / M [H+] / M Initial Rate / M s-1 0.10 0.10 0.10 0.05

  10. Deriving Rate Law and Rate Constant From the following experimental data derive rate law and k for the reaction: CH3CHO(g)  CH4(g) + CO(g)

  11. Deriving Rate Law and Rate Constant Rate law is: rate = k [CH3CHO]2 Here the rate goes up by a factor of four when initial concentration doubles. Therefore, this reaction is second order. Now determine the value of k. From experiment #3 data: 0.182 mol/L·s = k (0.30 mol/L)2 k = 0.182 / (0.3)2 = 2.0 (L / mol·s) Using k you can calculate the rate at other values of [CH3CHO] at same T.

  12. Homework Determine the rate law and value of the rate constant for this reaction: NO2 (g) + CO (g)  NO (g) + CO2 (g) [NO2] / M [CO] / M Initial Rate / M s-1 1 0.10 0.10 0.0051 2 0.40 0.10 0.080 3 0.10 0.20 0.0052 Homework Rate = k[NO2]2 k = 0.5 s-1 M-1

  13. The Iodine Clock • Mix different amounts of HIO3 + NaHSO3 + starch. • Concentration of reactants is: [beaker I] > [beaker II] >[beaker III]. • The following reactions take place consecutively in each beaker: • Starch forms a blackish blue complex with iodine. As the final reaction is the fastest, the colour of the complex only becomes apparent once the sulphite is fully consumed. • The reaction is slowest in the solution with the lowest concentration, as the reaction time is dependent on the concentration. + starch Blue-Black complex

  14. Concentration - Time Relationships How long will it take for x moles of A to be consumed? What is the concentration of reactant as function of time? For FIRST ORDER REACTIONS the rate law is For infinitesimal differences, Δ /Δt becomes a differential, so Integrating both of the expression , gives:

  15. First Order Reactions Integrating gives [A]t =conc. of A at time = t during exp [A]0 =conc. of A at time t = 0 k = rate constant t = time Called the integrated first-order rate law. [A]t / [A]0 = fraction remaining after time t has elapsed.

  16. Second Order Reactions • For a second-order reaction, with one reactant A only: Integrating both sides gives: [A]t =conc. of A at time = t during exp [A]0 =conc. of A at time t= 0 k = rate constant t= time

  17. Zero Order Reactions • For a zero order reaction: Integrating gives [A]t = [A]0 – k t [A]t =conc. of A at time = t during exp [A]0 =conc. of A at time t= 0 k = rate constant t= time

  18. Identifying reaction order FIRST ORDER SECOND ORDER ZERO ORDER [A]t = [A]0 – k t ln[A]t = -k t + ln[A]0 Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.

  19. What happened to Giant Kangaroos? Radiocarbon Dating • 14C 12C occurs at a known rate. • How old is a bone if the 14C concentration is only 9% of the natural concentration and k = 1.21 x 10–4y –1 ? 9% concentration: [14C] / [14C]o = 0.09 Now: ln[14C] / [14C]o= -kt Therefore: ln(0.09) = -(1.21 x 10–4) x t Solving for t: -2.407 / -1.21 x 10–4 = t, or t = 20,000 y

  20. Half-life • Half-life: time required for concentration of reactant to decrease to half of its original concentration. • Half-life for a first-order reaction is independent of initial concentration. Expression for half-life: t1/2 = 0.693 / k

  21. Half-life Example • 90Srwas released during theChernobyl explosion in1986. Givent1/2 (90Sr) = 28.1 y, what percentage remains today? • Reaction is of type: A B (nuclear decay follows 1storder reaction) t1/2 = 0.693 / k Thus, k = 0.693 / 28.1 = 0.02466 y–1 Now, ln([A] / [A]o) = -kt Therefore, [A] / [A]o= exp(-0.02466 x 25) [A] / [A]o= 0.54, thus 54% remain today 2011 - 1986 years

  22. Summary CONCEPTS • Reaction rate as change in concentration of reactants and products • Rate law • Reaction order • Integrated rate laws for first, second, and zero order reactions • Half-life relations CALCULATIONS • Calculate average and instantaneous rates from data (with units) • Derive rate law and rate constant from experimental data (with units) • Use integrated rate laws to calculate [A] at given times • Calculation using the half-life relations

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