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[answered] Biochemistry 5721 (AU 2016) Extra credit problem [up to 50


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Biochemistry 5721 (AU 2016) Extra credit problem [up to 50 points will be added to your score for PS 7] Due: Friday, November 18, 2016 1. In this problem you will hopefully derive the rate equations for a 2nd order reaction:

 

A + B ? products described by the rate law: = ? A

 

= A B a) (5 points) Use the following expressions for [A] and [B]:

 

A = A ! ? B = B ! ? where [A] and [B] are the concentrations of A and B at time t, [A]0 and [B]0 are the initial concentrations of A and B, and x is the extent of the reaction, to show that the above expression for the reaction rate can be rewritten in terms of [A]0, [B]0 and x as follows: A !? B ! = ? b) (10 points) Now, show that:

 

1

 

A ! ? B ! ? 1

 

1

 

?

 

?

 

A !?

 

B !?

 

B = 1

 

!? A ! c) (15 points) Integrate the expression from part (a) between the appropriate limits for x and t

 

? i.e.,:

 

! ! A !? B !

 

! = ? ! Use the result in part (b) to simplify the integration of the left hand side, and derive the final result: 1

 

B !? A ! ln 1 B A

 

A B !

 

! = d) (20 points) The derivation of the integrated rate equation in part (c) implicitly assumed that [B]0 >[A]0. However, this expression is not applicable in the case when [B]0 = [A]0 , because the denominator goes to zero. Show that when [A]0 = [B]0 the rate equation obtained in part (c) reduces to: 1

 

1

 

=

 

+ [A] [A]! Hint: To derive this result, let [B] = [A] + x and [B]0 = [A]0 + x, and apply L?Hopital?s rule (check your Calculus textbook and/or see below) in the limit of x ? 0. ()

 

?() = lim

 

; ! =

 

, ! = !?! ()

 

!?! ?() lim 2

 


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