Set7Solutions - ChE 101 2012 Set 7 Solutions 1 The...

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ChE 101 2012 Set 7 Solutions 1. The steady-state approximation is not valid until the reactive intermediates approach their steady-state concentration. The time required for this to occur is called the relaxation time t r . Past the relaxation time, the steady-state approximation is normally satisfactory. Consider the reaction A k 1 -→ B k 2 -→ C Here B is the reactive intermediate. Let’s assume c B differs from its steady-state approxima- tion concentration c * B by an amount : c B = c * B (1 + ) (a) Assuming the feed is pure A , derive an expression for as a function of time. Assuming power law kinetics, we can write dc B dt = k 1 c A - k 2 c B Moreover, dc B dt = d dt ( c * B (1 + )) = c * B d dt + (1 + ) dc * B dt From the pseudo-steady-state approximation, we know that dc B dt = 0 = k 1 c A - k 2 c * B c * B = k 1 k 2 c A 1
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Note that dc A dt = - k 1 c A , so we can write dc * B dt = - k 2 1 k 2 c A dc B dt = c * B d dt + (1 + ) dc * B dt = k 1 k 2 c A d dt - k 2 1 k 2 (1 + ) c A k 1 c A - k 2 c B = k 1 k 2 c A d dt - k 2 1 k 2 (1 + ) c A k 1 c A - k 2 c * B (1 + ) = k 1 k 2 c A d dt - k 2 1 k 2 (1 + ) c A [ k 1 c A - k 2 k 1 k 2 c A (1 + ) = k 1 k 2 c A d dt - k 2 1 k 2 (1 + ) c A ] × 1 k 1 c A [ - = 1 k 2 d dt - k 1 k 2 (1 + )] × k 2 - k 2 = d dt - k 1 - k 1 0 = d dt + ( k 2 - k 1 ) - k 1 Thus we have the initial value problem d dt + ( k 2 - k 1 ) - k 1 = 0 ( t = 0) = - 1 since we know c B ( t = 0) = c * B (1 + ) = 0 Separating and integrating yields d dt = k 1 + ( k 1 - k 2 ) d k 1 + ( k 1 - k 2 ) = dt Z - 1 d k 1 + ( k 1 - k 2 ) = Z t 0 dt 1 k 1 - k 2 ln( k 1 + ( k 1 - k 2 ) ) | - 1 = t ln( k 1 + ( k 1 - k 2 ) ) - ln( k 2 ) = ( k 1 - k 2 ) t k 1 + ( k 1 - k 2 ) ) = k 2 exp[( k 1 - k 2 ) t ] = - k 1 k 1 - k 2 + k 2 k 1 - k 2 exp[( k 1 - k 2 ) t ] = - 1 ¯ K - 1 [ ¯ K - exp[( ¯ K - 1) k 2 t ]] where ¯ K = k 1 k 2 . (b) What is lim t + ? Since B is a reactive intermediate, ¯ K = k 1 k 2 < 1. Thus lim t + = lim t + ( - 1 ¯ K - 1 [ ¯ K - exp[( ¯ K - 1) k 2 t ]]) = ¯ K 2
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(c) Plot the time dependence of for k 1 k 2 = 0 . 5 , 0 . 1 , 0 . 05 and k 2 = 0 . 1s - 1 . The plot was generated in MATLAB using the following script: K = [ 0 . 5 ; 0 . 1 ; 0 . 0 5 ] ; k2 = 0 . 1 ; t = l i n s p a c e (0 ,100 ,10000) ; e1 = - 1/( K (1) - 1) * ( K (1) - exp ( ( K (1) - 1) * k2 * t ) ) ; e2 = - 1/( K (2) - 1) * ( K (2) - exp ( ( K (2) - 1) * k2 * t ) ) ; e3 = - 1/( K (3) - 1) * ( K (3) - exp ( ( K (3) - 1) * k2 * t ) ) ; f i g u r e plot ( t , e1 , t , e2 , t , e3 ) leg = legend ( ' \ e p s i l o n = 0.5 ' , ' \ e p s i l o n = 0.1 ' , ' \ e p s i l o n = 0.05 ' ) ; s e t ( leg , ' FontSize ' , 14) s e t ( leg , ' Box ' , ' on ' ) s e t ( leg , ' l o c a t i o n ' , ' EastOutside ' ) ; x l a b e l ( ' t ( sec ) ' , ' FontSize ' , 16) y l a b e l ( ' \ e p s i l o n ' , ' FontSize ' , 16) (d) The relaxation time is the time required for a quantity to decay to a fraction 1 e of its original value. If k 1 k 2 << 1, what is t r ? When k 1 k 2 = ¯ K << 1, = - 1 ¯ K - 1 [ ¯ K - exp[( ¯ K - 1) k 2 t ]] exp( - k 2 t ) Thus, the relaxation time when ¯ K << 1 is t r = 1 k 2 2. The pyrolysis of acetaldehyde is believed to take place according to the following sequence CH 3 CHO k 1 -→ CH 3 · + CHO · CH 3 · + CH 3 CHO k 2 -→ CH 3 · + CO + CH 4 CHO · + CH 3 CHO k 3 -→ CH 3 · +2 CO + H 2 2 CH 3 · k 4 -→ C 2 H 6 (a) Identify the initiation, propagation, and termination steps. (Initiation) CH 3 CHO k 1 -→ CH 3 · + CHO · (Propagation) CH 3 · + CH 3 CHO k 2 -→ CH 3 · + CO + CH 4 (Propagation) CHO · + CH 3 CHO k 3 -→ CH 3 · +2 CO + H 2 (Termination) 2 CH 3 · k 4 -→ C 2 H 6 (b) Derive the rate expression for the rate of disappearance of acetaldehyde, - r Ac .
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