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Problem 1 The combined hydrogenation of ethylene and propylene in a system consisting of 1 mol H2, 2 mol CHA, and 2 mol Cale, is performed at 50 bar and at a temperature range of 500-1000 K to produce ethane and propane. Only four reactions have favorable equilibrium conversions at these conditions: CaHa + C3Hs - CaHe + CaH6 (1) C3H6 + H2 = CH: CaHA + 2CHA = 202H6 (3) CHA + CHA = C3Hs (4) Assume that these reactions reach equilibrium, calculate the equilibrium product composition for the given temperature range. Notes:" The equilibrium constant Keg for reaction j of a multiple-reaction system is given by the expression Kegj Po) = (5) where y is the equilibrium mole fraction of the ? specie, P is the pressure of the system and P" = 1 bar is the standard pressure. The equilibrium mole fraction y, is related to the extent of the j" reaction, 6,, by the ex- pression #. = (6) no + E, VSwhere my is the stoichiometric number of component i in the je reaction and total stoichio- metric number ;, is defined as vj = >vij (7) Use Eq. 5 to set-up the four non-linear equations to be able to solve for the equilibrium product composition. The equilibrium constant is computed by the following empirical formula, In Keqj = Aj In T + B; T+ C, T' + Ej + T + Fi (8) The coefficients of Eq. 8 are given below reaction A B x 10 C x 10 D E 10 F 1 0.131 -0.624 0.1233 1.426171 0.2202 2 -3.673 2.8285 -0.318167 -4150 14.127914 7.6652 3 -2.566 2.947 8.028224 -0.5437 4 -1.913 2.655 -0.378 0 9.588485 -4.9589 Additional Information: a. Guess value: $1 = 1.5, 6 = 1.25, 43 = -0.5, 64 = 0.5 b. For consistency, number each specie, i, as follows: (1) He. (2) CHA. (3) CH, (4) CHs, (5) CaH6. (6) C3Hs