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# IB Chemistry Order Reaction, Rate Law and Half life

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IB Chemistry Order Reaction, Rate Law and Half life

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### IB Chemistry Order Reaction, Rate Law and Half life

1. 1. http://lawrencekok.blogspot.com Prepared by Lawrence Kok Tutorial on Rate Law, Rate Expression, Order of Reaction, Initial Rate and Half Life .
2. 2. Reaction Rates / Kinetics • Chemical rxn, reactant consumed, product formed • Amt reactant decrease ↓ , Amt product increase ↑ • Rate follow stoichiometric principles A → B • For every ONE A breakdown = ONE B will form • Rate decomposition A = Rate formation of B dt Bd dt Ad ][][    2NO2 → N2O4 • Two mole NO2 decompose = One mole N2O4 form • NO2 used up is twice as fast as N2O4 produced dt ONd dt NOd ][1 2 ][1 422    dt HId dt Id dt Hd 2 ][][1][1 22      H2 + I2 → 2HI • One mole H2 decompose = TWO mole HI form • Rate H2 and I2 decomposition the same but only half the rate HI formation Amt/Conc Amt/Conc Time Time Reactants Product (Reactants) • X decrease/consume ↓ over time (Products) • Y increase/form ↑over time Rate of Decrease of X • Decrease ↓ Conc X /time • Decrease ↓ Vol X /Time • Decrease ↓ Abs X /Time Rate of Increase of Y • Increase ↑ Conc Y /time • Increase ↑ Vol Y /Time • Increase ↑ Abs Y /Time Amt/Conc/Vol/Abs X Time Amt/Conc/Vol/Abs Y X Y Time Gradient= rate change at time,t Gradient= rate change at time,t Instantaneous rate time, t1 Initial rate, t = 0 Initial rate, t = 0 Instantaneous rate time, t1 X → Y Click here notes
3. 3. Graphical Representation of Order :ZERO, FIRST and SECOND order ZERO ORDER FIRST ORDER SECOND ORDER Rate – 2nd order respect to [A] Conc x2 – Rate x 4 Unit for k Rate = k[A]2 Rate = kA2 k = M-1s-1 Rate Conc reactant Rate Conc reactant Conc reactant Conc Conc Conc Time Time Time Time Conc reactant Rate Time ln At Time 1/At ktAA ot  ][][ Rate = k[A]0 Rate independent of [A] Unit for k Rate = k[A]0 Rate = k k = Ms-1 Rate vs Conc – Constant Conc vs Time – Linear Rate = k[A]1 Rate - 1st order respect to [A] Unit for k Rate = k[A]1 Rate = kA k = s-1 Rate vs Conc - proportional Conc vs Time ktAA eAA ot kt ot    ]ln[]ln[ ][][ [A]t [A]o kt AA ot  ][ 1 ][ 1 ln Ao 1/Ao Conc at time t Conc at time t
4. 4. Order of rxn found using THREE mtds Initial Rate mtd (Multiple Single Runs) Conc Vs Time Mtd (Half Life) Conc Vs Time Mtd (Whole Curve/Tangent) Multiple Single Runs Vary/Keep certain conc fixed Wasteful as multiple runs needed Monitor decrease in conc reactant Using Half Life to determine order Monitor decrease conc of single reactant Using gradient/ tangent at diff conc Conc x2 – rate x2 - 1st order Conc x2 – rate x4 – 2nd order Conc x2 – rate 0 – zero order Convert Conc Vs Time to Rate vs Conc Rate Vs Conc – Linear – 1st Order Initial Rate taken, time 0 Draw tangent at time 0 Half Life directly prop to Conc Half Life inversely prop to Conc Expt Conc A Conc B Initial rate 1 0.01 0.02 2 2 0.01 0.04 4 3 0.02 0.02 4 Conc Time Expt 2 Expt 1 Conc reactant Time Zero order Conc reactant Time Half Life constant 1st order 2nd order Conc reactant Time Gradient at diff conc Conc Rate
5. 5. Rxn : A + B → AB Find order A (fix conc B ) Let Rate = k[A]x[B] y Rate = k[A]2 [B]1 2nd order respect to A 1st order respect to B Using Initial rate for order of rxn Find order B (fix conc A) Let Rate = k[A]x[B] y 2 806.0log652.0lg 806.0652.0 0713.0 0575.0 1026.1 1021.8 2. 1. 2. 1. 2 3                      x x Conc Conc Rate Rate x x x 1 649.0log652.0lg 649.0652.0 0333.0 0216.0 1026.1 1021.8 3. 1. 3. 1. 2 3                      y y Conc Conc Rate Rate y y y Expt Conc A Conc B Initial rate 1 0.0575 0.0216 8.21 x 10-3 2 0.0713 0.0216 1.26 x 10-2 3 0.0575 0.0333 1.26 x 10-2 Expt Conc F2 Conc CIO2 Initial rate 1 0.10 0.01 1.2 x 10-3 2 0.10 0.04 4.8 x 10-3 3 0.20 0.01 2.4 x 10-3 Rxn : F2 + 2CIO2 → 2FCIO2 Find order CIO2 (fix conc F2 ) Let Rate = k[F2]x [CIO2] y Find order F2 (fix conc CIO2) Let Rate = k[F2]x [CIO2] y 1st order respect to CIO2 1st order respect to F2 Rate = k [CIO2]1 [F2]1 1 44 01.0 04.0 102.1 108.4 1. 2. 1. 2. 3 3                     y Conc Conc Rate Rate y y y 1 22 10.0 20.0 102.1 104.2 1. 3. 1. 3. 3 3                     x Conc Conc Rate Rate x x x To calculate k Expt 1 : Ini rate = 1.2 x 10-3, [F2] = 0.10M, [CIO2] = 0.01M Rate = k[F2]1[CIO2]1 1.2 x 10-3 = k[0.10]1[0.01]1, k = 1.2 M-1s-1 To calculate k Expt 1 : Ini rate = 8.21 x 10-3, [A] = 0.0575, [B] = 0.0216 Rate = k[A]2[B]1 8.21 x 10-3 = k[0.0575]2[0.0216]1, k = 115
6. 6. Rxn : 2CIO2 + 2OH- → CIO3 - + CIO2 - + H2O Find order CIO2 (fix conc OH- ) Let Rate = k[CIO2]x[OH- ]y Expt 1 : Ini rate = 8 x 10-3 , [CIO2] = 0.025M, [OH- ] = 0.046M Rate = k[CIO2]2[OH- ] 1 8 x 10-3 = k[0.025]1[0.046]1, k = 278.3M-1s-1 Find order OH- (fix conc CIO2 ) Let Rate = k[CIO2]x[OH- ]y 2nd order respect to CIO2 1st order respect to OH- Rate = k[CIO2]2[OH- ]1 Using Initial rate for order of rxn To calculate k 2 4.1log96.1lg 4.196.1 025.0 035.0 1000.8 1057.1 1. 2. 1. 2. 3 2                      x x Conc Conc Rate Rate x x x 1 22 046.0 092.0 1057.1 1014.3 2. 3. 2. 3. 2 2                     y Conc Conc Rate Rate y y y Expt Conc OH Conc CIO2 Initial rate 1 0.046 0.025 8 x 10-3 2 0.046 0.035 1.57 x 10-3 3 0.096 0.035 3.14 x 10-3 Rxn : Br2 + 2NO → 2NOBr Find order Br2 (fix conc NO ) Let Rate = k[Br2]x[NO]y Find order NO (fix conc Br2 ) Let Rate = k[Br2]x[NO]y Expt Conc Br2 Conc NO Initial rate 1 0.1 0.1 12 2 0.2 0.1 24 3 0.1 0.2 48 1 2 1 2 1 2.0 1.0 24 12 2. 1. 2. 1.                       x Conc Conc Rate Rate x x x 2 2 1 4 1 2.0 1.0 48 12 3. 1. 3. 1.                       y Conc Conc Rate Rate y y y 2nd order respect to NO1st order respect to Br2 Rate = k[Br2]1[NO]2 Expt 1 : Ini rate = 12Ms-1 , [Br2] = 0.1M, [NO] = 0.1M Rate = k[Br2]1[NO]2 12 = k[0.1]1[0.1]2 , k = 12,000 M-1min-1 To calculate k
7. 7. Conc Vs Time / Conc Vs Rate for Order Rxn: 2A → B + C Plot Conc A vs Time for order, initial rate and rate constant, k Rxn: 2N205 → 4N02 + 02 Plot Rate vs Conc for order and rate constant, k Conc vs Time Mtd • Half Life A -constant = 80s • 1st order respect to [A] • Formula for 1st order half life Conc vs Rate Mtd • Straight Line – 1st order respect to [N205] • Rate = k[N205 ], k = gradient = 7.86 x 10-6 s-1 Time 0 40 80 120 160 200 240 Conc 0.8 0.58 0.40 0.28 0.20 0.14 0.10 Conc Time 80s 80s 80s 13 2/1 1066.8 80 693.0 693.0    sk k t Conc Rate /10-5 0.94 1.26 1.40 1.52 1.79 1.93 2.00 2.10 2.21 2.26 Conc Rate rate constant Rate Law / Rate Expression Rxn: aA + bB → cC + dD • Stoichiometry eqn : Show mole ratio of reactant/product • Rate eqn : Eqn relate rate with conc of reactant : How conc reactant affect rate Rxn eqn = k[A]x [B]y x = order respect to [A] y = order respect to [B] (x +y) = overall order k = rate constant Order must be determined experimentally , NOT derived from stoichiometry coefficients Gradient = k
8. 8. Using Initial rate and Half Life for order Hydrolysis of ester by OH- : Ester + OH- → X + Y Rxn done using two diff OH- conc. Run 1 – [OH- ] – 0.20M Run 2 – [OH-] – 0.40M Plot Conc ester vs Time. Find order and initial rate Find order for OH- (fix conc ester) Let Rate = k[OH-]x [ester] y Find order for ester (Using Half Life ) Using expt 2 : Conc ester vs time Half Life Ester t1/2 = 12 m(constant) 1st order respect to ester Rate = k[OH-]1 [ester]1 For EXPT 2 : • Ini rate = 8.00, [OH-]= 0.4M, [ester] = 100M • Rate = k[OH-]1 [ester]1 • 8.00 = k[0.4]1[100]1 • k = 0.2M-1min-1 Half life : 100 → 50→ 25 (12 min) • Ini Rate expt 1 – Gradient time 0 = 4.00 • Ini Rate expt 2 – Gradient time 0 = 8.00 1st order respect to OH - Conc ester Time Expt 2- [OH] = 0.40M Expt 1 - [OH] = 0.20M Compare Expt 1 and 2 1 2 1 2 1 40.0 20.0 00.8 00.4 2. 1. 2. 1.                       x Conc Conc Rate Rate x x x Conc ester Time Expt 1 - [OH] = 0.20M Expt 2- [OH] = 0.40M Gradient, rate = 4.00 Gradient, rate = 8.00 12 m 12 m
9. 9. RBr + OH- → ROH + Br- Rxn done using TWO diff conc OH- Expt 1 – [OH- ] – 0.10M Run 2 – [OH- ] – 0.15M Plot Conc RBr vs time. Find order and initial rate Determine order for OH- (fix conc RBr) Let Rate = k[OH-]x [RBr] y Find order RBr (using half life) Using expt 2 : Conc vs time Half Life RBr t1/2 = 78 m Rate = k[OH-]1 [RBr]1 • For expt 1 Initial rate = 5.25, [OH-] = 0.10M, [RBr] = 0.01M • Rate = k[OH-]1 [RBr]1 • 5.25 = k[0.10]1[0.01]1 • k = 5250 M-1min-1 Half life : 0.01 → 0.005 → 0.0025 = 78 m Ini Rate expt 1 – Gradient time 0 = 5.25 Ini Rate expt 2 – Gradient time 0 = 8.00 1st order with respect to OH - Rate = k[OH-]1 [RBr]1 Using Initial rate and Half Life for order 1 65.065.0 15.0 10.0 00.8 25.5 2. 1. 2. 1.                 x Conc Conc Rate Rate x x x Expt 1 Expt 2 Time/m [RBr]/M in [OH] = 0.10 [RBr]/M in [OH] = 0.15 0 0.0100 0.0100 40 0.0079 0.0070 80 0.0062 0.0049 120 0.0049 0.0034 160 0.0038 0.0024 200 0.0030 0.0017 240 0.0024 0.0012 Expt 1 - [OH] = 0.20M Expt 2- [OH] = 0.15M Time Conc RBr 78s 78s Gradient, rate = 8.00 Gradient, rate = 5.25 1st order with respect to RBr
10. 10. Ester + H2O → CH3CO2H + C2H5OH Rxn done using TWO diff HCI conc Expt 1 : [HCI] – 0.10M Expt 2 :[HCI] – 0.20M Plot Conc Estervs time. Find order and rate of rxn Find order HCI (fix conc Ester) Rate = k[HCI]1[Ester]1 1st order respect to HCI Using Initial rate and Half Life for order Expt 1 Expt 2 Time/m [Ester]/M in [HCI] = 0.1 [Ester]/M in [HCI] = 0.2 0 0.200 0.200 25 0.152 0.115 50 0.115 0.067 75 0.088 0.038 100 0.067 0.022 120 0.051 0.013 Time Conc Ester Gradient, rate = 1.9 Conc Ester Time Gradient, rate = 3.8 Find order Ester (use half life) Half life Ester -> 0.200 → 0.100 → 0.050 = 31 m 31 m 31 m 1st order respect to Ester 1 5.05.0 2.0 1.0 8.3 9.1 2. 1. 2. 1.                 x Conc Conc Rate Rate x x x Ini rate Expt 1 – Gradient time 0 = 1.90 Ini rate Expt 2 – Gradient time 0 = 3.80 Expt 1 Expt 2 Half life is 31 min (constant) Ini rate Expt 1 – Gradient time 0 = 1.90 Ini rate Expt 2 – Gradient time 0 = 3.80
11. 11. C3H8 + 5O2 → 3CO2 + 4H2O2H2 + O2 → 2H2O Rate O2 decrease ↓ is 0.23Ms-1 , what is rate of H2O formation/increases ↑ Rate C3H8 decrease ↓ is 0.30Ms-1 , what is the rate of 02 decrease ↓ Rxn Rates / Kinetics 122 22 2 46.0)23.0(2 ][ 2 ][ 2 ][1 1 ][1 2 ][ 1 ][1 2 ][1 22     Ms dt Od dt OHd dt OHd dt Od dt OHd dt Od dt Hd 1832 832 283 5.1)30.0(5 ][ 5 ][ 1 ][1 5 ][1 3 ][1 5 ][1 1 ][1 2     Ms dt HCd dt Od dt HCd dt Od dt COd dt Od dt HCd Benzenediazonium chloride, unstable, decomposes to produce N2 gas shown below C6H5N2 +CI- + H2O → C6H5OH + N2 + HCI Vol of N2 was collected over time Vol of gas produced N2 in time t is proportional to amt C6H5N2 +CI- used up V∞ α [C6H5N2 +CI- ] at start (V∞ - Vt ) α [C6H5N2 +CI- ] remaining at time t Plot of (V∞ - Vt ) vs time = Plot of conc vs time Time/t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ∞ Vt = Vol N2 0 14 28 41 54 65 76 87 96 104 112 120 127 133 139 219 (V∞ – Vt)/ cm3 219 205 191 178 165 154 143 132 123 115 107 99 92 86 139 0 Find rate at diff conc Time Plot of (V∞ - Vt ) vs time = Conc vs time(V∞ - Vt ) Time Conc V∞ - Vt Rate/ Slope 0 219 16.5 4 165 12.1 7 132 10.0 14 80 6.22 21 47 3.84 (V∞ - Vt ) Rate Plot Rate vs Concslope = rate
12. 12. Acknowledgements Thanks to source of pictures and video used in this presentation Thanks to Creative Commons for excellent contribution on licenses http://creativecommons.org/licenses/ Prepared by Lawrence Kok Check out more video tutorials from my site and hope you enjoy this tutorial http://lawrencekok.blogspot.com