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2008Exam1sol

2008Exam1sol - CE 561 Exam 1 November 7 2008 This exam...

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CE 561, Exam 1, November 7, 2008 This exam consists of 3 questions, each with multiple parts. You should be careful not to get stuck on one part. If you do not know how to do a problem, move on and return to it if you have time at the end. You may use a calculator and three letter-size sheets (2-sided) of notes to aid you on this exam. You may not exchange notes with or otherwise consult your fellow students. If you talk to your fellow students during the exam, I will assume that you are cheating, you will be asked to leave, and you will fail the exam. You will have 2 hours to complete the exam. Please use a separate blue book for each exam problem. Carefully explain any assumptions you make, label what part of what problem you are working on, and define the symbols that you use. The point value of each part is indicated – budget your effort accordingly. There are 100 points total. 1. Consider the following two reactions: (1) A B with rate r 1 = k 1 C A (2) B C with rate r 2 = k 2 C B k 3 C C (a) Write these reactions in matrix form. (5 points) In matrix form, these can be written as: 11 0 0 11 0 01 1 A B C     −=   If the forward and reverse directions of the second reaction are treated as separate reactions, or as 0 01 1 A B C   =  If the second reaction is treated as a single reaction (b) Write the rate equations for the concentrations of the three species in matrix form. (5 points) The rate equations can be written terms of the vector of concentrations and a matrix of rate constants as: 1 1 23 00 0 AA BB CC Ck C d C k kkC dt C k kC   =  (c) Explain how you would solve these equations to obtain expressions for the concentrations of A, B, and C as functions of time, for initial conditions of C A ( t = 0) = C Ao , C B ( t = 0) = C C ( t = 0) = 0. Take your solution as far as you can in the time available. (10 points) If M is the matrix of rate constants from part (b), and C o is the vector of concentrations at t = 0, then we know that the solution can be written as ( ) exp( ) o C t Mt C = or -1 ( ) ( exp( ) ) o Ct T t T C = Λ where Λ is the diagonal matrix of the eigenvalues of M , and T is the matrix whose columns contain the corresponding eigenvectors of M . To write the solution in this form, we find the

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eigenvalues of M by solving ( ) det 0 MI λ −= for . For each solution, i , we find the corresponding eigenvector x i by solving i ii Mx x = . The eigenvectors and eigenvalues are then placed in the matrices T and Λ , respectively. We then take the inverse of T , then multiply the matrices to get C ( t ) according to the equation given above. ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) 1 1 2 3 1 2 3 23 2 1 2 3 1 2 3 00 det det 0 0 0 k M I k k k k k k kk k k k k k k λλ −−   − ± ± −− −− −− − ±  + + + − = = So, the three eigenvalues are λ 1 = - k 1 , λ 2 = -( k 2 + k 3 ), and λ 3 = 0. Corresponding eigenvectors are obtained by substituting the eigenvalues into i x = . For λ 1 = - k 1 this gives ( ) ( ) ( ) ( ) ( ) 1,1 1 2 1 3 1,2 2 3 1 1,3 1 1,1 1 2 3 1,3 2 1,2 1 3 2 31 2 21 3 2 2 1 3 1 12 13 1 1 1 13 1 0 0 0 0 0 x k k x k kkx kx
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2008Exam1sol - CE 561 Exam 1 November 7 2008 This exam...

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