Test1Sol - CE 561 Exam 1 This exam consists of 4 questions...

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CE 561, Exam 1, October 30, 2009 This exam consists of 4 questions, some 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 reactions: (1) C 3 H 8 C 3 H 6 + H 2 with rate r 1 = k 1 [C 3 H 8 ] (2) C 3 H 8 → C 2 H 4 + CH 4 with rate r 2 = k 2 [C 3 H 8 ] (3) C 2 H 6 → C 2 H 4 + H 2 with rate r 3 = k 3 [C 2 H 6 ] (4) C 2 H 4 → C 2 H 2 + H 2 with rate r 4 = k 4 [C 2 H 4 ] (a) Write these reactions in matrix form. (5 points) In matrix form, these can be written as: 38 36 2 24 4 26 22 1110000 1001100 0 0 01 1 0 10 0 01 10 0 1 CH H CH   =  (b) Write the rate equations for the concentrations of the seven species in matrix form (for isothermal reaction at constant volume). (5 points) The rate equations can be written terms of the vector of concentrations and a matrix of rate constants as: [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 12 1 1 43 2 2 44 3 4 0 00000 0 00 0 0 0 0 0 0 0 000 0 0 0 0 kk k k HH d k dt k CH CH k k −− =
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(c) Describe briefly how you would solve these equations using matrix methods (you do not have to actually solve them) (5 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 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. Note that we would not necessarily have to do this for the full 7 by 7 matrix. Only C 3 H 8 , C 2 H 6 , and C 2 H 4 appear as reactants. Thus, we could solve for these three species (using a 3 by 3 matrix). (d) Describe a numerical method that could be used to integrate the rate equations. Outline the algorithm used in this method and state the advantages and disadvantages of the method (5 points).
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Test1Sol - CE 561 Exam 1 This exam consists of 4 questions...

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