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Second and Zero rate orders. Chapter 14 part IV. Second Order Rate Laws. Butadiene forms its dimer 2C 4 H 6 (g) - > C 8 H 12 (g). 2nd Order Data. [C 4 H 6 ] vs time. The general equation is : aA --> products The second order rate law is : Rate = -∆[A]/∆t = k[A] 2
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Second and Zero rate orders Chapter 14 part IV
Second Order Rate Laws • Butadiene forms its dimer • 2C4H6(g) - > C8H12(g)
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. Second order Rate laws
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
It is important that one recognizes the difference in the half life between 1st & 2nd order reactions. For the first order reaction t1/2 is dependent only on k. For the 2nd order reaction, t1/2 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 t1/2 is double the preceding one. Second order half life
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.
Zero order • The half life of a zero order reaction is expressed: • [A] = [A]0/2 at t=t1/2 • If [A] = -kt + [A]t=0 • Then [A]0/2 = -kt1/2 + [A]t=0 • t1/2 = [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.
Rate laws with more than one reactant • BrO3-(aq)+ 5Br-(aq) +6H+(aq) --> 3Br2(l) + 3H2O(l) • From the experimental evidence we know the rate law to be: • Rate = -∆[BrO3-]/∆t = k[BrO3-][Br-][H+]2 • All reactants are expressed in the rate law. • The overall rate order here is 4