Question

< Problem Set #5 (Ch 15) Exercise 15.83 Ammonia can be synthesized according to the following reaction: N2 (9) + 3 H2 (9) = 2 NH3 (9) Kp = 5.3 x 10-5 at 725 K A 200.0-L reaction container initially contains 1.27 kg of N, and 0.310 kg of H2 at 725 K

Answer 1

mass of NT, - 18.48 X 17.029/mol = 314.5m 2 here, 45:36 mole of N2 react with 2x45.36 = 136111 mole of H₂ 4 Ng Slimiting agent - mote of the formed - 45.36 X2 that of My = 90.72 mol of why Theoretical yield go.72 x 17.02 I = 1544.05gin Hence, % yield - Actual yield xoo Theoretical 814.5 non 1544 2014

- 1:27X103 200.0L * Answer -we have to calculate number of moles moles of Na- mass Molar Mass = 4536 mole 28 almal Number of moles of 4 -0.310x logen = 153.8mol we know, Py=nRT : Pressure of N-HXRXI.45.36 x 0.08.2) 17254 = 13 satm so .. pressure of th - 153.8x0.0821 x 725k 200.0t - 45 8 atm now, ICE Table build Nerant 3 Hergs Initial 1) 3.5 458 chang (6) -x -3x t +2x Equilibrium E) (3.5-2) (4582) 2x Kp = Pitt a (2x) 2 PR Pro 12=2.75 we get this so partial pressure with=22 =282.75 = 5. Sam no ofruotes interv =5.5X 200 RT 0.0821 X 725 2NH3 g) (53X105) 58-73) pre 5.5x 200 18h &mole

- 1:27X103 200.0L * Answer -we have to calculate number of moles moles of Na- mass Molar Mass = 4536 mole 28 almal Number of moles of 4 -0.310x logen = 153.8mol we know, Py=nRT : Pressure of N-HXRXI.45.36 x 0.08.2) 17254 = 13 satm so .. pressure of th - 153.8x0.0821 x 725k 200.0t - 45 8 atm now, ICE Table build Nerant 3 Hergs Initial 1) 3.5 458 chang (6) -x -3x t +2x Equilibrium E) (3.5-2) (4582) 2x Kp = Pitt a (2x) 2 PR Pro 12=2.75 we get this so partial pressure with=22 =282.75 = 5. Sam no ofruotes interv =5.5X 200 RT 0.0821 X 725 2NH3 g) (53X105) 58-73) pre 5.5x 200 18h &mole