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MIT 10 37 - Study Guide

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10.37 Exam 2 25 April, 2007 100 points Problem 1: 35 points A protein and ligand bind reversibly with Kd = 10 nM . The association rate constant k = 2x104M−1s-1. The two species are mixed at an initial protein concentration of 3 nM on and an initial ligand concentration of 0.2 nM. a) At equilibrium, what fraction of the ligand will be complexed with protein? (15 points) b) At what time will the fraction of ligand in complex reach 95% of the equilibrium value? (20 points) Justify any assumptions you make to simplify equations. P + L ←→ C koff [P]eq [L]eqKd = = kon [C]eq Using a batch reactor mole balance and looking at the reaction stoichiometry, it is easy to see that given the initial conditions any unit of complex formed takes away a unit of protein and ligand: [P] + [C] = [P]0 [L] + [C] = [L]0 Using these in the equilibrium equation we can get a quadratic equation in [C]eq. ([P]0 − [C]eq )([L]0 − [C]eq )Kd = [C]eq 0 = [C]eq 2 −([L]0 + [P]0 + Kd )[C]eq + [P]0[L]0 = 0 [C] =([L]0 + [P]0 + Kd )± ([L]0 + [P]0 + Kd )2 − 4[P]0[L]0 = 0eq 2 Reject the positive root, it is too large (larger than the initial amount of ligand and protein). [C]eq = 0.0456nM [C]eq = 0.228 = 23% [L]0 At this point, it is interesting to look at different approximations to the expression. Good Approximation: [P]0 >> [C] This leads to ([P]0 )([L]0 − [C]eq )Kd ≈ [C]eq Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].[C] = [L]0[P]0 0.462nMeq Kd +[P]0 [C]eq = 0.231 = 23% [L]0 Bad Approximation: [L]0 >> [C]([P]0 −[C]eq )([L]0 )Kd ≈ [C]eq [C]eq = [L]0[P]0 = 0.0588 Kd + [L]0 [C]eq = 0.294 = 29% [L]0 The error in [C]eq of the bad approximation is about 30% of the true answer, whereas the good approximation is only off by about 1%. By noticing that the “good” approximation is a good approximation, the dynamic equation becomes easier to solve. (As an aside, an even better approximation would be just to neglect the second order term that is O([C]eq2).) Start with the full dynamic equation: d[C] = kon [L][P] − koff [C] = kon ([L]0 −[C])( [P]0 −[C])− koff [C]dt Make an appropriate approximation: ([P]0 −[C])≈ [P]0 d[C] ≈ kon [L]0[P]0 − kon [P]0[C] − koff [C]dt Rearrange and solve using the integrating factor: d[C] +[C](kon [P]0 + koff )= kon [L]0[P]0dt d ([C]exp[(kon [P]0 + koff )t])= kon [L]0[P]0 exp[(kon [P]0 + koff )t]dt on 0 0 on 0 off([C ] exp [(kon [P]0 + koff ) t])= k [L] [P] exp ((k [P] + k )t )+ I.C.(kon [P]0 + koff ) [C] =([[ PL ]]00[ + PK ]0 d )+ I.C.exp[−(kon [P]0 + koff )t] Using the initial condition, [C](t = 0) = 0, we find the integration constant to be: [L]0[P]0I.C. = − ([P]0 + Kd ) Hence, Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].[C](t) = [L]0[P]0 {1− exp[−(kon [P]0 + koff )t]}([P]0 + Kd ) The equilibrium value is clearly the value when t gets large. [C](t) = [C]eq {1− exp(− (kon [P]0 + koff ) t)} In order to find the point at 95% of the equilibrium value, rearrange and solve for the time when [C]/[C]eq=0.95: [C](t*) = 0.95 ={1− exp[−(kon [P]0 + koff )t *]}[C]eq 0.05 = exp[−(kon [P]0 + koff )t *] − ln(0.05) − ln(0.05) 3.00 t* = = = =11500s −5 −1 −1(kon [P]0 + koff )(kon [P]0 + konKd ) 2x10 nM s [3nM +10nM ] t* = 11500s ≈ 3.2h Problem 2: 30 points A surface-catalyzed reaction follows Rideal-Eley kinetics as follows: kA→A + S← AS k− A k1AS + A  → A2 + S Where A and A2 are in the gas phase, S is a reactive site on the surface, and AS is a molecule of A adsorbed to a reactive site. Assuming that: • adsorption of A is at rapid equilibrium • reaction of AS with A is rate-limiting • desorption of A2 is very rapid Derive the steady-state rate law for production of A2 as a function of the concentration of A and the total initial reactive site density So . Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].r= k C C − CAS  where K = kA Ad A A v A  KA  k− A rs = k1CASCA Adsorption is at rapid equilibrium, so rAd ≈ 0kA CACv = CAS ⇒ CAS = K ACACvKA Overall site balance in terms of So: SSo = Cv + CAS = Cv + K ACACv ⇒ Cv = o 1+ K ACA Given that the surface reaction is the rate limiting step, and the stoichiometric coefficient is +1 for A2, the rate of production of A2 is: k K S C2 rA '2 = rs = k1CASCA = k1K ACvCA 2 = 1 A o A 1 + KACA Problem 3: 35 points It is desired to make a product X-Y via this reaction: X-OH + Y-H → X-Y + H2O An equimolar feed of liquid X-OH and Y-H at 25oC are fed to a CSTR. At 25oC, where all 4 material species are liquids, the heat of reaction ∆Hrxn=-200 kJ/mole, and the heat capacity of each liquid-phase species is 4 kJ/(kg Co). The molecular weight of X-OH is 150 g/mole, and the molecular weight of Y-H is 100 g/mole. The temperature inside the reactor (T) is controlled by putting the reactor in thermal contact with a fluid flowing over the outside of the reactor at temperature Ta. To a good approximation, the heat transfer rate (Q, in watts) from the fluid flowing over the outside the reactor to the contents of the reactor is given by the linear expression: Q = UA(Ta-T) a) If the reaction is carried out with the reactor at steady-state at the inlet temperature of 25oC, is T greater than, less than, or equal to Ta? (5 points) Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].For an exothermic reaction, to maintain the products at the same temperature as the reactants one must remove heat. So T must be greater than Ta, i.e. Ta must be below room temperature. b) When running the reactor at T = 25oC to 50% conversion, the productivity is unacceptably low. To try to accelerate the reaction, it is decided to increase the steady-state reactor


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