G T = G f (product) G f (reactant) = kj/mol Ka (550 O C) = exp(8840/(8.314*( )) = 3.369

Transcription

1 (7.) An equimolar mixture of H 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 H, CO, and HO. The reaction is H O + CO = H + CO. G f data at 550 C are given. G T = G f (product) G f (reactant) = kj/mol Ka (550 O C) = exp(8840/(8.34*( )) = HO CO H CO Total ( )(0.333 )* Ka = (0.333 )*(0.333 )* = = 0.76; Ans. y = 0.57, y = 0.57, y 3 = 0.509, y 4 = 0.76 (7.) One method for the production of hydrogen cyanide is by the gas-phase nitrogenation of acetylene according to the reaction below. For a stoichiometric feed at 300C, calculate the product composition at and 00 bar given G T = kj/mole. N + C H = HCN Solution: Two mole basis. Assume ideal gas. # of moles (N ) = # of moles (C H ) =.0 N CH HCN Total Ka = exp(-30080/(8.34*( )) = (* ) * Ka = ( ) * = 0.008, = (Ans. y = , y = , y 3 = 0.046) J.R. Elliott, C.T. Lira, 00-04, all rights reserved. (/7/4) 6

2 (.4.3) Butadiene can be prepared by the gas-phase catalytic dehydrogenation of -Butene: C4H8 = C4H6 + H. 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 -butene to,3-butadiene at a reactor pressure of bar from a feed consisting of mol of steam per mole of -butene. (b) If the initial mixture consists of 50 mol% steam and 50mol% -butene, how will the required temperature be affected? G f 600K 700K 800K 900K C4H C4H Solution: C4H8 - C4H6 H HO Total = bar * 0 3 * Ka - 3 * Ka Noting that lnka = -G tot /RT, we can -6 identify the temperature by fitting a -7 y = -3996x trendline to the given data. -8 ln = -3.95, -9 substitute in the equation of straight line, -0 x = -( )/3996. / T x = = /T, T = 765. K = 49 O C ln Ka C4H8 - C4H6 H HO Total * Ka 0.8 O Similarly, T K C We need higher T. * J.R. Elliott, C.T. Lira, 00-04, all rights reserved. (/7/4) 7

3 (7.4) The standard Gibbs energy change for ethylene oxide at 98K for the reaction is kj/mole. This large negative value of G T indicates that equilibrium is far to the right at 98K but what about 550K? Heat capacity expressions are given as C = a + b T. Solution: The heat of reaction must be looked up. Referring to Apx E.6 for ethylene and the DIR handbook for ethylene oxide, H 98 = = -05. kj/mol. Following Eqs. 4.8 and 4.30, = J + ( /) 98 + ( /) 98 / = J.55 * *98 / J =-0037 J/mol /(8.34*98) = -0037/(8.34*98) ln II * * *8.34 G ln * * *8.34 G = kj/mol. If T = 550 K, increasing T will give an adverse effect on equilibrium, but the reaction is still very strongly favored and the impact will be indistinguishable. The water gas shift is to be carried out at a specified temperature and pressure employing a feed containing only CO and HO. Show that the maximum equilibrium mole fraction of H in the product results when the feed contains CO and HO in their stoichiometric proportions. Assume ideal gas behavior. Solution: CO H O CO H CO z z- HO CO H Total +z +z Ka z Kaz 0 0 Ka[ z ( z) ] (-Ka) +Ka (+z) - Ka z = 0 = {-Ka(+z) + [Ka (+z) +4z(-Ka)Ka] / }/[(-Ka)] y H = /(+z) = {-Ka(+z) + [Ka (+z) +4z(-Ka)Ka] / }/[(+z)(-ka)] y H = /(+z) = {-Ka+ [Ka +4z(-Ka)Ka/(+z) ] / }/[(-Ka)] To find maximum, take derivative and set equal to zero. dy/dz = 0.5[Ka +4z(-Ka)Ka/(+z) ] -/ [4(-Ka)Ka/(+z) 8z(-Ka)Ka/(+z) 3 ] = 0 = z/(+z) z =. QED. (7.6) Assuming ideal gas behavior, estimate the equilibrium composition at 400K and bar of a reactive mixture containing the three isomers of pentane. Formation data are given at 400K. Solution: This is best solved by the Gibbs minimization method, adapting Example 7.3 and GibbsMin from the Rxns.xls workbook, we obtain the following. Gf(J/mole) Gf400/RT feed ni log(ni) yi ni(gi/rt+lnyi) nentane ientane neoentane Tot Out In C-bal 5 5 Hbal J.R. Elliott, C.T. Lira, 00-04, all rights reserved. (/7/4) 8

4 (7.7) One method for the manufacture of synthesis gas depends on the vapor-phase catalytic reaction of methane with steam according to the equation below. The water-gas shift reaction also is important. Bases on stoichiometric feed of methane and steam, compute the eq composition at 600K, 300K and, 00 bars. CH 4 H O CO 3H rxn() H O CO H CO rxn() CH4 HO CO H CO Total + G600K kj / mole... rxn() G kj / mole... rxn() Ka Ka K 7300J / mole Ka600K, exp * Ka600K, exp * ( )(3 ) * 4 ( ) * ( ) 3 ( )(3 ) * ( ) ( )* 4 ( ) * 4 Note: high pressure tends to disfavor rxn (). Rxn is negligible at 600K, and rxn () requires CO to run or - will be less than zero. So both reactions are zero. At 300 K, the situation is quite different. G300K Ka = 63 G300K Ka = 0.54 Solving by method of Example 4.9, = 0.97 and = 0.05 at bar. At 00 bar, = 0.45 and = 0.49 at 00 bar E Two simultaneous reactions: CH4 + HO =CO + 3H H+CO = CO + HO (Details of equations described in text) (bars) 00 T(K) 300 K a 63 K a y 0.89 y y y y ntot.905 Objective Functions err err J.R. Elliott, C.T. Lira, 00-04, all rights reserved. (/7/4) 9

5 (7.8) Is there any danger that solid carbon will form at 550C and bar by the reaction: CO CS CO H K kj / mole G K kj / mole IN Out 98 CO C s (gas) CO Tot - G * kJ / mole 98K, T 98 H 98K, T * kJ / mole Increasing T, adverse affect on equilibrium G98K 0060 Ka98K exp exp.08e RT 8.34* 98.5 Using Shortcut Van t Hoff Eq. 4.3 Ka H 98 Ka 7450 ln ln Ka98 R T T98.08 E ln Ka ln.08e Ka exp(4.06) yco * * Ka yco * = ratio of carbon solid to feed is = There is danger. Note: this exemplifies a very important and undesirable side reaction in many catalytic reactions know as coking. The carbon tends to clog the catalyst pores and substantially reduce its effectiveness. Because of this problem, fluidized catalytic crackers were developed (aka. Cat crackers). The solid catalyst particles are fluidized by the upflow of gaseous reactants. As they ultimately settle at the bottom, they are removed and recirculated through an oxidation zone that burns off the coke then recycles the catalyst to the top of the bed. This is a good example of how thermodynamics impacts reactor design. (7.9) Calculate the equilibrium percent conversion of ethylene oxide to ethylene glycol at 98K and bar if the initial molar ratio of ethylene oxide to water is 3. In Out(z i ) EtO 3 3- Water (-) Glycol 0 Tot 4- K w = sat / = 0.045; K EtO =.76; K Gly = 8.6E-4; y i = z i K i /[ K i +L/F*(- K i )] ygly 784 Ka exp 3.5 yeto * yw * 98.5*8.34 and y i = are constraints, and L/F are unknown. Guess, = 0.99, L/F = /3 (all glycol in liquid, all EtO in vapor). J.R. Elliott, C.T. Lira, 00-04, all rights reserved. (/7/4) 0

6 (7.9) Sample solution of one reaction with vle: (Details of input equations described in text by Elliott and Lira) (bar) T(K) Ka psat(bar) K-ratios zfeed yi xi EtO Water E Glycol E sum(yi-xi) L/F 0.77 ErrKa 5.583E-07 As it turns out, the ethylene oxide is not so volatile after all and dissolves a fair amount in the liquid. The guess about the extent of conversion being high was good though. A more clever engineering approach would be to assume complete conversion and solve the simple flash. Then back out the exact conversion assuming L/F does not change. (7.0) Acetic acid vapor dimerizes according to A = A. Assume that no higher-order associations occur. Supposing that a value for Ka is available, and that the monomers and dimers behave as an ideal gas, derive an expression for y A in terms of and Ka. Then develop an expression for V/n 0 RT in terms of y A, where n0 is the superficial number of moles neglecting dimerization. Hint: write n 0 /n T in terms of y A where n T = n +n. Solution: n0 n n ya ( ya ) ya n n T T ya 4Ka Ka y A ya Ka ya ya Ka V V nt Ideal gas nrt T nrt 0 n0 ya Note: as Ka, V/n 0 RT ½ because the monomer is converted to dimer. Note also that V/n 0 RT is what we normally refer to as the compressibility factor, Z. This is an interesting result with regard to equations of state and phase equilibria. Since Ka is simply a function of temperature [ie. exp(-g/rt)], it says that we can compute Z given a pressure and temperature. This is analogous to the pressure explicit virial equation (Section 7.4), but the form of the pressure dependence is more complex. Exploring this perspective, generalizing to densitydependent equations, and adapting to multimer-forming species and mixtures is the subject of Chapter 9. Most of the physical insight contained in Chapter 9 is contained in this simple practice problem. J.R. Elliott, C.T. Lira, 00-04, all rights reserved. (/7/4)