Chap14prac - Chapter 14 Practice Problem Solutions (P14.1)...

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Chapter 14 Practice Problem Solutions To accompany Introductory Chemical Engineering Thermodynamics J.R. Elliott, C.T. Lira, 2001, all rights reserved. (02/11/02) 16 (P14.1) An equimolar mixture of H2 and CO can be obtained by the reaction of steam with coal. Compute the equilibrium compositions at 550 C based on an equimolar feed of H2, CO, and H2O. The reaction is H 2 O + CO = H 2 + CO 2 . G f ° data at 550 C are given. G T = Σ∆ G f (product) ± Σ∆ G f (reactant) = -8.84 kJ/mol Ka (550 O C) = exp(8840/(8.314*(550+273.15)) = 3.369 ± Compound In Out H2O 0.333 0.333−ξ CO 0.333 0.333−ξ H2 0.333 0.333+ξ CO2 0 ξ Total 1 1 Ka = 2 2 ( )(0.333 )* (0.333 )*(0.333 P P ξξ + −− = 3.369 ξ = 0.176; Ans. y 1 = 0.157, y 2 = 0.157, y 3 = 0.509, y 4 = 0.176 (P14.2) One method for the production of hydrogen cyanide is by the gas-phase nitrogenation of acetylene according to the reaction below. ²or a stoichiometric feed at 300 ° C, calculate the product composition at 1 and 200 bar given G T ° = 30.08 kJ/mole. N 2 + C 2 H 2 = 2HCN Solution: Two mole basis. Assume ideal gas. # of moles (N 2 ) = # of moles (C 2 H 2 ) = 1.0 Compound In Out N2 1 1−ξ C2H2 1 1−ξ HCN 0 Total 2 2 Ka = exp(-30080/(8.314*(300+273.15)) = 0.0018 Ka = 22 (2* ) * (1 ) * P P ξ = 0.0018 , ξ = 0.0208 (Ans. y 1 = 0.4896, y 2 = 0.4896, y 3 = 0.0416)
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Chapter 14 Practice Problem Solutions To accompany Introductory Chemical Engineering Thermodynamics J.R. Elliott, C.T. Lira, 2001, all rights reserved. (02/11/02) 17 (P.14.3) Butadiene can be prepared by the gas-phase catalytic dehydrogenation of 1-Butene: C4H8 = C4H6 + H2. In order to suppress side reactions, the butene is diluted with steam before it passes into the reactor. (a) Estimate the temperature at which the reactor must be operated in order to convert 30% of the 1-butene to 1,3-butadiene at a reactor pressure of 2 bar from a feed consisting of 12 mol of steam per mole of 1-butene. (b) If the initial mixture consists of 50 mol% steam and 50mol% 1-butene, how will the required temperature be affected? G f ° 600K 700K 800K 900K C4H6 195.73 211.71 227.94 244.35 C4H8 150.92 178.78 206.89 235.35 Solution: Compound In Out C4H8 1 1- ξ C4H6 0 ξ H2 0 ξ H2O 12 12 Total 13 13+ξ P = 2 bar () ( ) 2 2 2 * 13 * 13 * 1 1 13 0.01933 P P Ka P Ka ξ ξξ   +  == +− + = Noting that ln Ka = - G tot /RT , we can identify the temperature by fitting a trendline to the given data. ln 0.01933 = -3.95, substitute in the equation of straight line, x = -(-14.34 & 3.95 )/13996. x = 0.001306 = 1/ T , T = 765.22 K = 492 O C y = -13996x + 14.34 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 0.001 0.0012 0.0014 0.0016 0.0018 1 / T ln Ka Compound In Out C4H8 1 1- ξ C4H6 0 ξ H2 0 ξ H2O 1 1 Total 2 2+ξ ( ) C K T Similarly P Ka O 6 . 573 65 . 846 , 1118 . 0 1 * 2 * 2 = = = + = We need higher T .
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Chapter 14 Practice Problem Solutions To accompany Introductory Chemical Engineering Thermodynamics J.R. Elliott, C.T. Lira, 2001, all rights reserved. (02/11/02) 18 (P14.4) The standard Gibbs energy change for ethylene oxide at 298K for the reaction is &79.79 kJ/mole.
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Chap14prac - Chapter 14 Practice Problem Solutions (P14.1)...

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