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1 compare rates of product formation with and without

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Unformatted text preview: l state at t = 0 ([S]=[So] and [P]=0) to the final state.  ­([S] ­[So])=[P] Rearranging in terms of [S]= [So]  ­[P]. [P]= [So]  ­[S] is also correct. c. No Inhibition d [ P ] k2 [ E ][ S ] k2 [ E ]([ So ] ! [ P ]) = = dt K m + [ S ] K m + ([ So ] ! [ P ]) Product Inhibition k2 [ E ]([ So ] ! [ P ]) k2 [ E ]([ So ] ! [ P ]) d [P] = = dt K m (1 + + [ P ] / K I ) + ([ So ] ! [ P ]) K m + [ So ] + [ P ]( K m / K I ! 1) Compare rates of product formation with and without production inhibition [So]=Km and KI=0.1Km For [P]= 0.1Km No inhibition d [ P ] k2 [ E ][ S ] k2 [ E ]([ So ] ! [ P ]) k2 [ E ]( K m ! 0.1K m ) 0.9 = = = = k2 [ E ] = 0.4737k2 [ E ] dt K m + [ S ] K m + ([ So ] ! [ P ]) K m + ( K m ! 0.1K m ) 1.9 Inhibition k2 [ E ]([ So ] ! [ P ]) k2 [ E...
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This document was uploaded on 01/31/2014.

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