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**CHEM1612 - Pharmacy Week 11: Kinetics - Half Life**

Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone:

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**Unless otherwise stated, all images in this file have been reproduced from:**

Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, John Wiley & Sons Australia, Ltd ISBN:

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**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.

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**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 m = 0, reaction is zero order in A. Rate = k [A]0 2 4 1

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**Using Data to Determine Order**

Consider the reaction: ClO3-(aq) I-(aq) H+(aq) Cl-(aq) I3-(aq) H2O(l) Expt. [ClO3-] / M [I-] / M [H+] / M Initial Rate / M s-1 Determine the rate law for the above reaction. General expression: Rate = k [ClO3-]x[I-]y[H+]z; must find x, y, and z

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**Using Data to Determine Order in ClO3-**

Expt. [ClO3-] / M [I-] / M [H+] / M Initial Rate / M s-1 General expression: Rate = k [ClO3-]x [I-]y [H+]z Determine x: Compare experiment 1 and 2

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**Using Data to Determine Order in I-**

Expt. [ClO3-] / M [I-] / M [H+] / M Initial Rate / M s-1 General expression: Rate = k [ClO3-]x [I-]y [H+]z Determine y: Compare experiment 2 and 3

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**Using Data to Determine Order in H+**

Expt. [ClO3-] / M [I-] / M [H+] / M Initial Rate / M s-1 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

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**Determine the Rate Constant**

ClO3-(aq) I-(aq) H+(aq) Cl-(aq) I3-(aq) 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 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)

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**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) Expt. [CH3CHO] Rate of loss (mol/L) (mo/L.sec) 1 0.10 0.020 2 0.20 0.081 3 0.30 0.182 4 0.40 0.318

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**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 = / (0.3)2 = 2.0 (L / mol·s) Using k you can calculate the rate at other values of [CH3CHO] at same T.

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Homework Determine the rate equation 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 Homework Rate = k[NO2]2 k = 0.5 s-1 M-1

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**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 elemental iodine 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

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**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:

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**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.

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**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

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**Zero Order Reactions [A]t = [A]0 – k t Integrating gives**

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

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**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.

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**What happened to Giant Kangaroos?**

Radiocarbon Dating What happened to Giant Kangaroos? 14C C occurs at a known rate. How old is bone if 14C conc. only 9% of natural conc. and k = 1.21 x 10–4 yr –1 ? Consider the example of the bones unearthed in Australia. If it were found that the concentration of 14C were only 9% of that found in living tissue, then would only need rate constant for decay to 12C to calculate the time taken The rate constant for such decay has been measured accurately using samples with known age and is 1.2 x 10-4 yr-1 Note 1: 9% of original concentration: this is the same as 9 parts in 100 (or 0.09) Note 2: equation requires [C]o / [C] cf. [C]/[C]o = Therefore, need to invert both sides of equation to subsitute it into equation (write it on the board) Exponential decay Need to draw a graph on the board which shows the rate of decay of 14C with time. Just shown [14C] vs. t as an exponential decay (see fig on p 526 Chang) Note t = k-1ln([Ao]/[A]), when [A] = 0.5[Ao], then it is easy to show that t = 0.693/k 1/2 [A] 1/2 t1/2 t1/2 t

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Half-life Half-life: time required for concentration of reactant to decrease to half of its original concentration. Expression for half-life: t1/2 = / k Half lives apply to first order reactions and are a useful measure to gauge the speed of a reaction It takes 5730 years for the concentration of 14C in the bones to decrease to half of the original concentration Plutonium-239 is used as a nuclear fuel and is produced in breeder reactors (reactors that produce nuclear fuel). Plutonium-239 is one of the most toxic substances known emitting alpha particles (highly energetic and damage DNA on collision). The long half life is precisely why it cannot be allowed to permeate through the environment and why radioactive waste sites need to be underground in extremely stable regions (geologically) Derivation for half-life: simply take t = (1/k)ln{[A]o / [A]} and insert [A] = 0.5[A]o Half life: takes the same amount of time for a 2.0M solution to fall to 1.0M as it does for a 0.004M to fall to 0.002M If we know the half life, then can calculate the rate constant for a first order reaction Consider how long required for Sr to decay to 10% of original? t = ln(10)/ = 93 years Plutonium: t1/2 = 24,400 yrs = 0.693/k, k = 2.84 x 10-5 yr-1 Say, 10% left, require 81,000 years (1000 generations) Half-life for a first-order reaction is independent of initial concentration.

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Half-life Example 90Sr was released during the Chernobyl explosion in 1986. Given t1/2 (90Sr) = 28.1 y, what percentage remains today? The example chosen is one which is particularly topical in the northern hemisphere The reactions encountered in the laboratory usually have rate constants of the order of seconds. However, for nuclear decay the rate constants are the order of years. This is not an SI unit, however as long as one is consistent, the times calculated are in years (ie., it is OK) Mathematical point: if A = ln(B), then B = exp(A) - simply use calculator

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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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Chemical Kinetics Chapter 13

Chemical Kinetics Chapter 13

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