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Solve the rate equation for the following reaction: ch3coch3 + Br2 + H+---------

ID: 75531 • Letter: S

Question

Solve the rate equation for the following reaction: ch3coch3 + Br2 + H+------------> Ch3COCH2Br +Hbr Ch3COCH2Br          Br2                   H+             rate .3                           .050                   .050               5.7x 10^-5 .3                           .010                   .050               5.7x 10^-5 .3                           .050                   .010               1.2x 10^-5 .4                           .050                   .050               7.6x 10^-5 I was going to use the 3rd (1.2 x 10^-5) as the defaultRATE... so that everthing would have been a rate (1, 2, or 4) all overrate 3.... please help! Solve the rate equation for the following reaction: ch3coch3 + Br2 + H+------------> Ch3COCH2Br +Hbr Ch3COCH2Br          Br2                   H+             rate .3                           .050                   .050               5.7x 10^-5 .3                           .010                   .050               5.7x 10^-5 .3                           .050                   .010               1.2x 10^-5 .4                           .050                   .050               7.6x 10^-5 I was going to use the 3rd (1.2 x 10^-5) as the defaultRATE... so that everthing would have been a rate (1, 2, or 4) all overrate 3.... please help!

Explanation / Answer

Compare second and first experiment. You see that changing Br2 doesnothing to rate. Rate is 0th order with Br2 Compare third and first experiment. Dividing conc of H+ by fivedivides rate by five. Rate is 1st order with H+ Compare fourth and first. Increasing conc of Ch3COCH2Br by a factorof 4/3 increases rate by 4/3 (7.6/5.7 ~ 4/3). rate is first orderwith CH3COCH2Br rate = k*[CH3COCH2Br][H+] Sub in concentrations and rate if you want to find k: k = rate/{[CH3COCH2Br][H+]} = 7.6e-5/(0.4*0.05) = 0.0038L/mol/s rate = 0.0038*[CH3COCH2Br][H+]

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