Presentation on theme: "CHEM1612 - Pharmacy Week 11: Kinetics - Half Life Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone: 9351 4196"— Presentation transcript:
CHEM Pharmacy Week 11: Kinetics - Half Life Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone:
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Lecture 31-3 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. Rate Law and Reaction Order
Lecture 31-4 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 ? Interpreting Rate Laws
Lecture 31-5 Consider the reaction: ClO 3 - (aq) + 9 I - (aq) + 6 H + (aq) Cl - (aq) + 3 I 3 - (aq) + 3 H 2 O(l) Expt.[ClO 3 - ] / M[I - ] / M[H + ] / M Initial Rate / M s Determine the rate law for the above reaction. General expression: Rate = k [ClO 3 - ] x [I - ] y [H + ] z ; must find x, y, and z Using Data to Determine Order
Lecture 31-6 Using Data to Determine Order in ClO 3 - Expt. [ClO 3 - ] / M[I - ] / M[H + ] / M Initial Rate / M s General expression: Rate = k [ClO 3 - ] x [I - ] y [H + ] z Determine x: Compare experiment 1 and 2
Lecture 31-7 Using Data to Determine Order in I - Expt. [ClO 3 - ] / M[I - ] / M[H + ] / M Initial Rate / M s General expression: Rate = k [ClO 3 - ] x [I - ] y [H + ] z Determine y: Compare experiment 2 and 3
Lecture 31-8 Using Data to Determine Order in H + Expt. [ClO 3 - ] / M[I - ] / M[H + ] / M Initial Rate / M s General expression: Rate = k [ClO 3 - ] x [I - ] y [H + ] z Determine z: Compare experiment 3 and 4 So overall the rate law is: Rate = k [ClO 3 - ] [I - ] [H + ] 2
Lecture 31-9 ClO 3 - (aq) + 9 I - (aq) + 6 H + (aq) Cl - (aq) + 3 I 3 - (aq) + 3 H 2 O(l) rate= k [ClO 3 - ][I - ][H + ] 2 The order is 1 with respect to [ClO 3 - ] 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 10 2 M -3 s -1 (note: units depend on rate law) Determine the Rate Constant
Lecture Deriving Rate Law and Rate Constant From the following experimental data derive rate law and k for the reaction: CH 3 CHO(g) CH 4 (g) + CO(g) Expt.[CH 3 CHO]Rate of loss (mol/L)(mo/L.sec)
Lecture Rate law is: rate = k [CH 3 CHO] 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: 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 [CH 3 CHO] at same T. Deriving Rate Law and Rate Constant
Lecture Determine the rate equation and value of the rate constant for this reaction NO 2 (g) + CO (g) NO (g) + CO 2 (g) [NO 2 ] / M[CO] / M Initial Rate / M s Homework Rate = k[NO 2 ] 2 k = 0.5 s -1 M -1 Homework
Lecture The Iodine Clock Mix different amounts of HIO 3 + NaHSO 3 + 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
Lecture 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:
Lecture Integrating gives [A] t / [A] 0 = fraction remaining after time t has elapsed. Called the integrated first-order rate law. First Order Reactions [A] t =conc. of A at time = t during exp [A] 0 =conc. of A at time t = 0 k = rate constant t = time
Lecture 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
Lecture 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
Lecture Identifying reaction order FIRST ORDER SECOND ORDERZERO ORDER [A] t = [A] 0 – k t ln[A] t = -k t + ln[A] 0 Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.
Lecture C 12 C occurs at a known rate. How old is bone if 14 C conc. only 9% of natural conc. and k = 1.21 x 10 –4 yr –1 ? Radiocarbon Dating What happened to Giant Kangaroos?
Lecture Half-life for a first-order reaction is independent of initial concentration. Expression for half-life: t 1/2 = / k Half-life Half-life: time required for concentration of reactant to decrease to half of its original concentration.
Lecture Sr was released during the Chernobyl explosion in Given t 1/2 ( 90 Sr) = 28.1 y, what percentage remains today? Half-life Example
Lecture 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