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Set2Solutions - ChE 101 2012 Problem Set 2 Solutions 1...

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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 1 -X 1-X 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
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(c) What is the equilibrium conversion of A? Recall that K eq = Q S j =1 a ν j j = exp( - Δ G R RT ), where a j is the activity of species j and ν 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 ν 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
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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 + 7 exp( - 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? We want to find the amount of time it takes to reach X = 0 . 9 × 0 . 4375 = 0 . 39375. We can find this now using an equation we derived in part (d). t 90% = - 1 55 (ln(7 - 16 × 0 . 39375) - ln(0 . 39375 + 3) - ln( 7 3 )) = 0 . 0441074 hr = 2 . 65 min = 159 sec 2. Consider the gas phase following reaction, which is performed in a constant volume batch reactor at 500 C at 30-60 MPa: A + 3 B 2 C 3
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The standard state thermodynamic data on this reaction are Δ H R, 298 = - 92 . 4 kJ/mol and Δ S R, 298 = - 199 J/(mol · K).
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