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Unformatted text preview: ChE 101 2012 Problem Set 2 Solutions 1. Consider the following reversible elementary reaction: A 2B The rate coefficient of the forward reaction is k f = 25 hr 1 , and the rate coefficient of the reverse reaction is k b = 4 M 1 hr 1 . The aqueous phase reaction is carried out in a constant volume batch reactor. The reactor is initially charged with 1 mole of pure A and 1 mole of pure B at 1 atm. The reactor volume is 1 L. (a) Write the rates of reaction for A and B. r A = k b c 2 B k f c A = 4 c 2 B 25 c A r B = 2 k f c A 2 k b c 2 B = 50 c A 8 c 2 B (b) Write r A as a function of the conversion of A. From part (a), we know that r A = 4 c 2 B 25 c A (1) Let X be the fractional conversion of A. Then we can make the following table: Species Initial (mol) Change (mol) Final (mol) A 1X 1X B 1 +2X 1+2X Total 2 X 2+X We are given that the initial reactor volume is V = 1 L, so we can write c A = N A V = 1 X c B = N B V = 1 + 2 X By substitution then, we obtain r A = 4(1 + 2 X ) 2 25(1 X ) 1 (c) What is the equilibrium conversion of A? Recall that K eq = Q S j =1 a j j = exp( G R RT ), where a j is the activity of species j and j is the stoichiometric coefficient of species j . Furthermore, K eq = k f k b . For ideal liquid solutions, we write a j = c j , where c j is the concentration of species j in the reactor. Thus, we can write K eq = k f k b = 25 4 K eq = S Y j =1 c j j = c 2 B c A = (1 + 2 X eq ) 2 1 X eq where X eq is the equilibrium conversion of A. Equating expressions, we find that (1 + 2 X eq ) 2 1 X eq = 25 4 X eq = 0 . 4375 (d) Find the conversion of A as a function of time. Recall from part (b) that r A = 4(1 + 2 X ) 2 25(1 X ) Since r A = dc A dt = d (1 X ) dt = dX dt . we can rewrite this expression as dX dt = 4(1 + 2 X ) 2 25(1 X ) dX dt = 4(1 + 2 X ) 2 + 25(1 X ) 2 Separating variables and integrating from X = 0 and t = 0, we obtain dX dt = 4(1 + 2 X ) 2 + 25(1 X ) dX dt = 16 x 2 41 x + 21 dX dt = ( x + 3)(16 x 7) dX ( x + 3)(16 x + 7) = dt Z X X =0 dX ( x + 3)(16 x + 7) = Z t t =0 dt 1 55 Z X X =0 dX x + 3 + 16 55 Z X X =0 dX 16 x + 7 = t 1 55 (ln  16 X 7   ln  X + 3   ln( 7 3 )) = t 1 55 (ln(7 16 X ) ln( X + 3) ln( 7 3 )) = t ln( 7 16 X X + 3 ) = 55 t + ln( 7 3 ) 7 16 X X + 3 = 7 3 exp( 55 t ) 55 X + 3 16 = 7 3 exp( 55 t ) X + 3 = 55 16 + 7 3 exp( 55 t ) X = 165 48 + 7exp( 55 t ) 3 . Note that 16 X 7 = 0 for X = 0 . 4375, the equilibrium conversion. Since the conversion will never exceed the equilibrium conversion over the range of interest,  16 X 7  = 7 16 X for all possible X in this reaction. Also note that  X + 3  = X + 3 for all X of interest. (e) How long does it take for the system to reach 90% of the equilibrium conversion?...
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This note was uploaded on 03/21/2012 for the course CHE 101 taught by Professor Arnold during the Winter '11 term at Caltech.
 Winter '11
 ARNOLD
 pH, Reaction

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