Question

5.67 A mixture of one mole of nitrogen and one mole of hy- drogen is equilibrated over a catalyst for the ammonia reaction at 500 K and 1 bar. (a) What are the equilibrium mole fractions of N2, H2, and NH3? (b) The experiment is repeated with 1.2 mol of nitrogen and 1 mol of hydrogen. What is the equilibrium partial pressure of ammonia? (c) How do you explain this result in terms of Le Châtelier's principle?

Answer 1

It is based on Le-Chatelier's Principle.

- Solution - The given values are - Heteroffiffiffi Temperatury = 500K Pusswu (P) z 1 bau reaction of The “fastmation of ammonia from Hz and Nz 18 as- N2cg) + 3H2(g) _2143 (8) Part ca) - Nacg) + 3H₂(g) = 2NH₂ (9) at initial Here males of Heeg) is limited because if Nacy) gives 1 males then He must give z male but H₂ has only one male reacted 1 3 2X1 At equilibrium 1-1/3 1-3 27 1 Henu at equilibrium total males Z +0+2 Male fraction of Ne z male of Ne - total mole8 . mole of H₂ Male fraction of H2 z / 22 total males

Male fraction of NHg z mos • миз total mole Part (6) N2 + 3 Hyg) - 2NH₃ Co. at initial 1.2 at equilitacin 12-12 1-1 } 20.867 0 2 2 0:67 Irrrrrrrrrrrrr Total mole at equilibaciune z 0.867+0+0.67 z 1.54 Mole facc Hon of whaz muito 2 0.435 Partial Pressure of NH3 z mole fraction of NH3 X total pousguru z 0.435 X 1 Partial Prusowe of NM, 20.435 bar (6) Part- According to Le Chatelier's Principle is a 898tem at equilibrium is disturbed, the system will move in such way as to counteract the westwiband a A8 Hence in Part to the males of the the system will Nz is incuased than Part a move towards formation of NH3-