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Unformatted text preview: ChE 101 2012 Set 7 Solutions 1. The steadystate approximation is not valid until the reactive intermediates approach their steadystate concentration. The time required for this to occur is called the relaxation time t r . Past the relaxation time, the steadystate 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 steadystate 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 pseudosteadystate 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 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 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 (c) Plot the time dependence of for k 1 k 2 = 0 . 5 , . 1 , . 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 = linspace (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 ) ) ; figure plot ( t , e1 , t , e2 , t , e3 ) leg = legend ( ' \ epsilon = 0.5 ' , ' \ epsilon = 0.1 ' , ' \ epsilon = 0.05 ' ) ; set ( leg , ' FontSize ' , 14) set ( leg , ' Box ' , ' on ' ) set ( leg , ' location ' , ' EastOutside ' ) ; xlabel ( ' t ( sec ) ' , ' FontSize ' , 16) ylabel ( ' \ epsilon ' , ' 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 ?...
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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
 Mass Balance

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