Government Engineering College, Valsad. Department of Chemical Engineering Subject : Chemical Engg. Thermodynamics-2 (2150503) GTU TOPIC: Multireaction equilibria Prepared By: Enrolment no.   Name of member 140193105015 Patel Rakeshkumar G. 140193105016 Patil Tushar D. 140193105017 Halpati Pratikkumar Guided By: Prof. G.D.Vegad

Multireaction equilibria When the equilibrium state in the reacting system depend on two or more independent chemical reactions. The equilibrium composition can be found by the direct extension of the methods developed for single reactions. ………….(1) (j=1,2,…..r)

For a gas phase reaction equation 1 takes the form, ………(2) If the equilibrium mixture is an ideal gas, ………(3)

Example 1 A feedstock of pure n-butane is cracked at 750 K and 1.2 bar to produce olefins. The rxns are: C4H10C2H4 + C2H6 (I) C4H10C3H6 + CH4 (II) The equilibrium constants are KI = 3.856 and KII = 268.4. If there xns reach equilibrium, what is the product composition?

solution eII =(KII/KI)1/2 eI nI = 1; nII = 1; n0 = 1 yC4H10 = (1-eI-eII)/(1+eI+eII) yC2H4=yC2H6= eI/(1+eI+eII) yC3H6=yCH4= eII/(1+eI+eII) Assuming ideal gas mixture (low pressure): eII =(KII/KI)1/2 eI

Using the expressions of yi in terms of eI and eII, we get Solving: eII = 0.8914; eI = 0.1068 And the product compositions are: yC4H10 = 0.001; yC2H4 = yC2H6 = 0.0534; yC3H6=yCH4=0.4461

Example 2 Synthesis gas may be produced by the catalytic reforming of methane with steam: CH4 (g) + H2O(g)  CO(g) +3H2(g) CO(g) + H2O(g)  CO2 (g) +H2(g) Assume equilibrium is obtained at 1 bar and 1,300 K for both reactions. a) Would it be better to carry out the reaction at pressures above 1 bar?

solution At 298 K from table C for the 1st rxn DH298K = 205813 J/mol; DG 298 k= 141863 J/mol, and calculate DG1300 K = -1.031 x 105 J/mol, K1 = 13845 For the 2nd rxn, see Problem 13.32, DH298K = -41166 J/mol; DG 298k= -28618 J/mol, and calculate DG1300 K = 5.892 x 103 J/mol, K2 = 0.5798 Since K1 >> K2, 1st rxn is the primary rxn. Since n = 2 (>0), rxn shifts to left when P increases

Solution (cont) (b) Would it be better to carry out the rxn at T < 1,300 K? No, because the primary rxn is endothermic, so, it shifts left when T decreases (c) Estimate the molar ratio of hydrogen to CO in the synthesis gas if the feed consists of an equimolar mixture of steam and methane CH4 (g) + H2O(g)  CO(g) +3H2(g) CO(g) + H2O(g)  CO2 (g) +H2(g) If the feed is equimolar, no more water for the 2nd rxn. Ratio H2/CO???

Solution (cont.) (d) Repeat (c) for a steam to methane mole ratio of 2 in the feed. Rxn 2 may proceed. However rxn 1 proceeds to completion and provides feed to rxn 2. CH4 (g) + H2O(g)  CO(g) +3H2(g) CO(g) + H2O(g)  CO2 (g) +H2(g) 1 mole CH4 and 2 mol H2O initially, Therefore for rxn 2 there is 1 mole H2O, 1 mol CO and 3 mol H2, no = 5 yCO=yH2O =(1-e)/5, yCO2 =e/5, yH2=(3+e)/5 Solve for e, get yH2/yCO

Solution (cont.) (e) how could the feed comp. be altered to yield a lower ratio of H2 to CO in the synthesis gas than that found in (c). CH4 (g) + H2O(g)  CO(g) +3H2(g) CO(g) + H2O(g)  CO2 (g) +H2(g) Add CO2 to the feed. Some H2 reacts with CO2 and generates more CO, lowering the H2/CO ratio.

Solution (cont) Is there any danger that C will deposit by the rxn 2CO C+CO2 under the conditions of part (c)? In part d? If so, how could the feed be altered to prevent C deposition? 2CO C+CO2 Calculate DG at 1300 K = 5.674x104 J/mol K = 5.255x10-3 Deposition of C depends on the ratio yCO2/(yCO)2 If in the reactor this ratio is higher than that given by K, there is no carbon deposition. If yCO2 0, there is danger of C deposition.

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