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Lecture 14 Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place. Lecture 14 – Thursday 2/28/2013 Pseudo Steady State Hypothesis (PSSH) Net rate of reaction of active intermediates is zero Hall of Fame Reaction: 2NO +O2 2NO2 Introduction to Enzyme Kinetics Begin non-Isothermal reactor design 2 Active Intermediates and PSSH An active intermediate is a molecule that is in a highly energetic and reactive state It is short lived as it disappears virtually as fast as it is formed. That is, the net rate of reaction of an active intermediate, A*, is zero. The assumption that the net rate of reaction is zero is called the Pseudo Steady State Hypothesis (PSSH) 3 Active Intermediates and PSSH 4 Example The rate law for the reaction A BC is found from experiment to be 2 A kC rA 1 k C A How did this rate law come about? Suggest a mechanism consistent with the rate law. 5 Example For reactions with active intermediates, the reaction coordinated now has trough in it and the active intermediate, A*, sits in this trough 6 Example - Solution 7 1 A A A * A 2 A * A A A r2 A* k2C A*C A 3 A * B C r3 A* k3C A* k1 k2 k3 r1 A* k1C 2 A Rate Laws: k3 is defined w.r.t. A* Reaction (1) r1A* k1C Reaction (2) r2 A* k2C AC A* (2) Reaction (3) r3 A* k3CA* (3) 2 A (1) But C*A cannot be measured since it is so small Relative Rates: 8 r1A r1A* , r3B r3 A* Net Rates: Rate of Formation of Product rB r3B r3 A* k3CA* Pseudo Steady State Hypothesis r*A = 0 rA* rA* r1A* r2 A* r3 A* (5) k1CA2 k2CACA* k3CA* 0 (6) Solving for C A* 9 (4) 2 A k1C C A* k3 k 2C A Substituting for C A* in Equation (4) the rate of formation of B is k1k3C A2 (8) rB k3 k 2 C A Relative rates overall A BC rA rB 1 1 k1k3C A2 rA rB k3 k 2 C A (9) 10 For high concentrations of A, we can neglect k 3 in the denominator with respect to k2C A k 2C A k3 and the rate law becomes k1k3 rA C A kCA k2 (10) (apparent first order) 11 For low concentrations of A, we can neglect k 2C A in the denominator with respect to k3. k3 k 2C A and the rate law becomes k3k1 2 rA C A k1C A2 k3 (apparent second order) (11) Dividing by k3 and letting k’=k2/k3 and k=k1 we have the rate law we were asked to derive kCA2 rA 1 k C A (12) Active Intermediates Why do so many reactions follow elementary rate laws? What about rA kCA kC A rA 1 k ' CI (1) ( I Inert ) k1 A I A* I (2) A * I A I k2 13 k3 (3) A BC k1k3C A - rA k 2 k 2C I Active Intermediates/Free Radicals and PSSH Hall of Fame Reaction The reaction: 2NO +O2 2NO2 has an elementary rate law 2 rNO2 kCNO CO2 However… Look what happens to the rate as the temperature increases: -rNO2 14 T Why does the rate law decrease with increasing temperature? Mechanism: k1 NO O2 NO3* (1) k2 NO3* NO O2 (2) k3 NO3* NO 2 NO2 (3) Write rate of formation of product rNO2 Note: k3 is defined w.r.t. NO3* 15 Define k with respect to NO3* Assume that all reactions are elementary reactions, such that: r1NO3 = k1CNOCO2 = k1 [ NO][O2 ] r2 NO* = -k2CNO* = k2 éë NO ùû 3 3 r3NO* = -k3CNO* CNO = k3 éë NO3* ùû [ NO] * 3 3 3 rNO = 2 é -r3NO* ù ë 2 3 û 16 The net reaction rate for NO3* is the sum of the individual reaction rates for NO3*: r1NO* = r1 Þ r1NO* = k1[ NO] [O2 ] 3 3 -r2NO* = r2 Þ r2NO * = -k2 [ NO 3 * 3 3 ] r3 NO* r3 r3 NO* k3 NO3* NO 3 3 rNO* = r1NO* + r2NO* + r3NO* 3 3 3 3 rNO* k1 NO O2 k2 NO3* k3 NO3* NO 3 17 Pseudo Steady State Hypothesis (PSSH) The PSSH assumes that the net rate of species A* (in this case NO3*) is zero. rNO* 0 3 0 k1 NO O2 k2 NO3* k3 NO3* NO 0 = k1 [ NO] [O2 ] - éë NO3* ùû ( k2 + [ NO]) k1 [ NO ][O2 ] éë NO ùû = k2 + k3 [ NO ] * 3 18 Pseudo Steady State Hypothesis (PSSH) [3∗ ] 1 2 = 2 + 3 rNO2 2r3 NO* 2 NO3* NO 3 2 19 k1 k 3 NO 2 O2 =2 k 2 + k 3 NO Pseudo Steady State Hypothesis (PSSH) k2 >> k3 [ NO] rNO2 k1k3 A1 A3 2 NO O2 2 2 e k2 A2 E2 E1 E3 RT NO O2 2 E2 E1 E3 This result shows why the rate decreases as temperature increases. 20 -rNO2 T End Lecture 14 21