1. Second and Zero rate orders Chapter 14 part IV

2. Second Order Rate Laws  Butadiene forms its dimer  2C 4 H 6 (g) - > C 8 H 12 (g)

3. 2nd Order Data [C 4 H 6 ] (mol/L) Time (sec) 0.010000 0.006251000 0.004761800 0.003702800 0.003133600 0.002704400 0.002415200 0.002086200

4. [C 4 H 6 ] vs time

5. Second order Rate laws  The general equation is :  aA --> products  The second order rate law is :  Rate = -∆[A]/∆t = k[A] 2  The integrated rate law is :  1/[A] = kt + 1/[A] t=0  The plot of 1/[A] versus time is a straight line with slope = k  Using this equation, one can calculate [A] at any time provided k and [A] at t=0 are known.

6. Second order half life  When the half life of the second order has elapsed, [A] = [A] 0 /2.  So: 1/([A] 0 /2) = kt 1/2 + 1/[A] t=0  And 2/[A] 0 - 1/[A] 0 = kt 1/2  1/[A] 0 =kt 1/2  The expression for second order half life is:  t 1/2 = 1/k[A] 0

7. Second order half life  It is important that one recognizes the difference in the half life between 1st & 2nd order reactions.  For the first order reaction t 1/2 is dependent only on k.  For the 2nd order reaction, t 1/2 is dependent on both k and [A] t=0.  Note that each successive half life in a 2nd order reaction is longer than the first.  In fact each successive 2nd order t 1/2 is double the preceding one.

8. Determining Rate order Time (sec) [C 4 H 6 ] (mol/L) ln[C 4 H 6 ] 1/[C 4 H 6 ] 00.01000-4.60517100.0 10000.00625-5.07517160.0 18000.00476-5.34751210.1 28000.00370-5.59942270.3 36000.00313-5.76672319.5 44000.00270-5.9145370.4 52000.00241-6.02813414.9 62000.00208-6.17539480.8

9. 2 nd order reaction using integrated law for first order

10. 2 nd order reaction and 2nd order integrated rate law

12. Zero Order Rate laws  Most reaction are 1st or second order.  However zero order has a rate law of:  Rate = k[A] 0 = k(1) = k  For zero order the rate is constant and does not depend on the [reactant].  Zero order is concentration independent.  The integrated rate law is:  [A] = -kt + [A] t=0  The plot of [A] versus time is a straight line, with a slope of -k.

13. Zero order  The half life of a zero order reaction is expressed:  [A] = [A] 0 /2 at t=t 1/2  If [A] = -kt + [A] t=0  Then [A] 0 /2 = -kt 1/2 + [A] t=0  t 1/2 = [A] t=0 /2k  Zero order reactions are most likely to occur in the presence of a metal surface or enzyme, some type of catalysis.

14. Rate laws with more than one reactant  BrO 3 - (aq) + 5Br - (aq) +6H + (aq) --> 3Br 2 (l) + 3H 2 O (l)  From the experimental evidence we know the rate law to be:  Rate = -∆[BrO 3 - ]/∆t = k[BrO 3 - ][Br - ][H + ] 2  All reactants are expressed in the rate law.  The overall rate order here is 4

15. Rate Laws: Summary Zero order 1st order 2nd order Rate law Rate=kRate=k[A] Rate=k[A] 2 Integrated rate law [A]=-kt+[A] 0 ln[A]=-kt+ln[A] 0 1/[A]=kt+1[A] 0 Straight line plot [A] vs t ln[A] vs t 1/[A] vs t straight line slope Slope = -k Slope = k Half life t 1/2 =[A] 0 /2k t 1/2 =0.693/k t 1/2 =1/k[A] 0

16. Identify the rate order!