Fugacity: It is derived from Latin, expressed as fleetness or escaping ...

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Effect of variables on equilibrium : Effect of temperature: G RT lnK since ΔG depends only on temperature as the 0 298 298 pressure is fixed at 1.0 atmosphere, K value varies with temperature. It is not affected by Pressure, Concentration, etc,. Variation of K with T is given by ln K d H 2 dT RT Van't Hoff equation. For endothermic reactions ΔH 0 is + ve as T increases K also increases. This means that the equilibrium conversion is more at higher pressure. For exothermic reactions ΔH 0 is –ve. Therefore K increases with T. Hence the equilibrium conversion decreases as T increases. Eg. SO2 + 1/2 O2 ↔ SO3 Effect of pressure: Consider a reaction of ideal gases, then K = KyP Δn or Ky = K / P Δn When Δn > 0. An increase in pressure decreases Ky. Hence equilibrium product yield is less at high pressures When Δn< 0. An increase in pressure increases Ky and equilibrium product yield pressure is required C2H4 + H2O C2H5OH Δn = 1 – (1+1) = -1, Δn < 0 N2 + 3H2 2NH3 Δn = 2 – 4 = – 2 Δn < 0 Here high When Δn = 0 Here pressure has no effect on reaction. For ideal gases the effect of pressure depends on the variation of γ and Φ with pressure Effect of Inert: Presence of inert has the opposite effect of increase of pressure. Therefore when Δn > 0, addition of inert increases Ky and equilibrium conversion And Δn < 0, addition of inert decreases equilibrium conversion. Effect of excess reactants: Presence of excess reactants increases equilibrium conversion of the limiting reactant Presence of products in feed: decreases the equilibrium conversion of reactants. Eg. CH3COOH + C2H5OH CH3COOC2H5 + H2O If water is added by 1.0 mole to the feed, equilibrium conversion of CH3COOH reduces from 30% to 15%
HCN is produced by the reaction N2 (g)+ C2H2 (g) 2HCN (g). The reactants are taken in stoichiometric ratio at 1.0 atmosphere and 300 0 C. At this temperature ΔG 0 = - 30100 k J / k mol. Calculate equilibrium composition of product stream and maximum conversion of C2H2 30100 8.314(573) 0 3 Δn=2– (1+1) = 0 we have G RT ln K or lnK = or K = 1.8029x10 Component Moles in feed Moles reacted Moles present Mole fraction N2 1 X 1 – X (1 – X) / 2 C2H2 1 X 1 – X (1 – X) / 2 HCN -- 2X 2X K K K K P K K P P n n 0 y Assume 1 and 1 y X K K Solving for X 2 2 3 1.8029(10) y HCN 1 1 yN y 2 C2H2 1 X 1 X , X = 0.0207 Equilibrium Composition: Equilibrium Conversi on of C H 2 2 1 X 1 0.0207 N2 100 100 48.965% 2 2 1 X 1 0.0207 C2H2 100 100 48.965% 2 2 HCN ()1 X 00 2.007% 2 2 Equilibrium conversion is max imum for reversible reaction Moles of C H reacted X 0.0207 2 2 Max. Conversion of C2H2 100 100 100 2.07% Moles of C2H2 in feed 1 1 (2X) / X = Workout the problem, when the pressure is changed to 203 bar. Fugacity co efficients of N2,,C2H2 and HCN are 1.1, 0.928, and 0.54 Assume KΦ= 1 X
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Fugacity: It is derived from Latin,
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II method Using compressibility fac
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50 ln f ln 9. 86 10 1 f=26.7
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