Equillibrium Hydrazine Decomposition If we allowed "infinite" time for the reaction, hydrazine products would reach an equilibrium with little ammonia left(depending on pressure). The product, N2, H2, NH3, must then satisfy =K2(T); Kp=1.089×10°e(atm2) (Ps in atm) and the pressure is To conserve moles of H and N, starting from(arbitrarily )3 moles of N 2H4, we must H:3nk+2n2=3×4=12 dividing +2Pn2=2 nm+2n=3×2=6 (ammonia mole fraction) 2 Solv P PN, 16. 522, Space Propulsion Prof. Manuel martinez-Sanchez Page 8 of 1216.522, Space Propulsion Lecture 5 Prof. Manuel Martinez-Sanchez Page 8 of 12 Equilibrium Hydrazine Decomposition If we allowed “infinite” time for the reaction, hydrazine products would reach an equilibrium with little ammonia left (depending on pressure). The product, N2, H2, NH3, must then satisfy ( ) 3 2 2 NH 1 3 p 2 2 N H P =K T P P ; ( ) 6289 -6 T -1 K 1.089×10 e atm p  (P’s in atm) and the pressure is NH N H 322 P=P +P +P To conserve moles of H and N, starting from (arbitrarily) 3 moles of N2H4, we must have ×   ×  3 2 3 2 3 2 3 2 NH H NH H NH N NH N H : 3n + 2n = 3 4 = 12 3P + 2P dividing, = 2 N : n + 2n = 3 2 = 6 P + 2P Define NH3 P y = P (ammonia mole fraction). Then 2 2 H N 3y + 2P P = 2 y + 2P P and H N 2 2 P P y+ + =1 P P Solving, H2 P 2 5 = 1- y P3 4       N2 P 1 1 = 1- y P3 2      