Question

Figure 1 represents the flow diagram of the production of ammonia from hydrogen and nitrogen (Haber process), according to the following reaction at 400°C and atmospheric pressure: N2(g) + 3H2(g) + 2NH3(9) TN).6 TH).6 7 Yis 1,6 = 9.7 6 5TH = 0 kmol h-1 3 TH(a). T.1 ANN (9) TIN2(g) 4 = 1,623.4 = 0 kmol h-1 Figure 1. Flow diagram for question 1. The process is designed for a target production of 200 kmol h of pure ammonia. The hydrogen and nitrogen leaving the reactor are separated in a distillation column and half of the flow rate (and volumetric rate) of this mixture is purged out of the system whilst the other half is recycled into the reactor. If the fresh feed contains 600 kmol h of hydrogen and 400 kmol h - of nitrogen, calculate: (i) Fractional excess of the non-limiting reactant at the fresh feed [3 marks] (ii) Composition of the purge stream in terms of molar rates (nii,orkmol h 2) [7 marks) (iii) Overall and single step fractional conversions of nitrogen and hydrogen. (10 marks] (iv) Estimate the heat absorbed or released at the reactor. (5 marks]

Answer 1

rinal - 400km/h ㅋ n N2,6 n H₂,6 5 MH2, 1=600 / 2 3. 4 n NH3(4) krudh Basis 200 kom din of pure NH3 as product N2 + 3H2 2 NH3 lut 2 aq molar flow rate of of Ny & Ha in stream 2 reep 300 komora H₂ A 100 kmrch M is NH3 200 km olla reachd so Amount y N2 remains in stream 3 = N2, 3 = x-loo H2 - H2,3 - Y - 300 H₂ is so from fresh sud Limiting reagent and Ny is in excass 3 mole H₂ require I mole Nz 600 600 = 200 kmol N2 x 3 excess fractional - n Nail na, rendred 400 - 200 - 1 200 ring, required It is That 5 two equal flow rate stream splet în to given recycle Total molar 4 purge. flow rate in stream r 12,5 + " H2,5 x-100 ty-300 x + y - 400 orla

Stream G = x+y - 400 2 ㅋ x+y -400 2 Molar flow rate of N₂ in 6 & -loo 2 N2 7 a-loo 2 H₂ 6 Y_300 2 H2 7 - y-300 2 at ft P Overall molar balance ni tnt = n2 + x +4-400 aty 2 > 1600 = x+y N₂ balance. razil + N27 n N2 12 400 t x-loo x - 2 700 = R a = 700 komnih

rito, 6 = 300kmillh. from Y = 900 loma/h 300 Na, 6 = x-100 2 V N2, 6 = 300 kmnih nh2.6 4-300 2 300 2 for single step step conversion ring, a rN212 n N2,3 700 - 600 700 XS Na = 0.142 Xs H2 n H2, 2 n H2 , 2 n H2,3 900 600 P w/- 900 XS H₂ = 0.3 3 3 XON rinal - [riNa, 6 + H24 ] 400 - 300 to ni Nail 400 Xona = 0.25 Xo H2 n Hal [[H2, 6+ in Har4 I 600 300+02 ק 600 nitril XOH2 0.5