lecture_13_2009

lecture_13_2009 - 68 Integrated Rate Laws We observed the...

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68 Integrated Rate Laws We observed the time dependences of zero, first, and second-order kinetics and saw the half-life relations that occur in such systems. Now, we would like to develop integrated (rather than differential rate law expressions. Here’s what we would like to do: Differential rate law -d[reactant]/dt = k[reactant] n Integrated rate laws give us [reactant] vs. time. We have to do a little calculus to get these expressions: zero-order kinetics gives us that info directly. Let’s look at first-order kinetics: -d[A]/dt = k[A]. Note that the slope is proportional to the concentration . This is characteristic of an exponential decay We can write [A(t)] = [A(0)]e -kt . You can show by direct substitution that this solution is correct. Now ln[A(t)] = ln[A(0)] – kt. So, a plot of ln of measured concentrations vs. time is linear, with slope –k. Now, let’s define the time required for the initial [A(0)] concentration to fall to ½ [A(0)] as t 1/2 By direct substitution, ln { ½ [A(0)]} – ln [A(0)]} = -kt 1/2 ln [A(t)] t Slope = -k

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69 By the properties of logarithms (exponents), ln { ½ [A(0)]} = ln (½) + ln [A(0)] = -ln2 + ln [A(0)] Sometimes it is easy to determine the half-life from the data, so the calculation of k follows. At other times, a plot of ln [A] gives k, and then t 1/2 follows. It is also easy to show that every half-life is equal. This is an especially nice characteristic of first-order kinetics. Now, second-order kinetics:
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This note was uploaded on 02/13/2010 for the course CHM 29275 taught by Professor Farrar during the Spring '10 term at Rochester.

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lecture_13_2009 - 68 Integrated Rate Laws We observed the...

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