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Horvath Practice Problems 21

# Horvath Practice Problems 21 - − ν i k 1 I |{z y = −...

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With a computer and a program like Microsoft Excel it is easy to work with the data as is. Sometimes, however, it is convenient to cast the data into linear form. For a fi rst order reaction this is done by taking the natural log of the integrated rate law ln[ I ] = ln ¡ [ I 0 ] e ν i kt ¢ (4.2) ln[ I ] = ln[ I 0 ] + ln ¡ e ν i kt ¢ ln[ I ] | {z } y = ν i kt |{z} mx + ln[ I 0 ] | {z } b . So, by plotting the natural log of experimental data versus time, fi rst order reactions will appear as straight lines with a slope equal to ν i k. The second and higher order integrated rate laws given in the table about are already in linearized form. If a reaction is described by a second order rate law then when one over the data is plotted one gets a straight line with slope
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Unformatted text preview: − ν i k : 1 [ I ] |{z} y = − ν i kt | {z } mx + 1 [ I ] |{z} b . (4.3) The table lists the integrated rate laws for elementary reactions. Over all reactions also follow these rate laws but now one can not express the slopes of the linearized equations as a product of the rate constant and the stoichiometric factor. Instead one uses an observed rate constant, k obs . Thus overall reactions that appear to be f rst order follow the linearized integrated rate law, ln[ I ] = − k obs t + ln[ I ] (4.4) and second order reactions go as 1 [ I ] = k obs t + 1 [ I ] (4.5) 20 INTEGRATED RATE LAWS...
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