Question

2. Liquid N203 is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO2. At 25°C, a value of Kp = 1.91 has been established for this decomposition. If 0.236 moles of N2O3 are placed in a 1.52 L vessel at 25°C. calculate the equilibrium pressures of N2O3(), NO(g), and NO2(2).

Answer 1

Answer! 2. let, the reaction be, Nios (g) = NO(g) + NOZG) let. Initial pressure of N203= MRI whese, n=0.236 V=1152 T = 298.15 R=0.08206 Lathmallka P- 0.236) (0.08206) (292-15 ) 1:52 P= 5.77402 1.52 [P= 3.80 atson Now, l₂O3 - - Not NO2 Initial 3.80 change - Equilibri 3.80- kp. (Pro) LPNO) (PN203)

kp: (x(2) (3.80-x) 1. X2 1.91 atm . 3.80-2 (1.91) (3-80-x) = x2 7.28 -1.919 >*? x?+1.919 –7.26 = 0 By solving the above equation, -1.90 Therefore, Equilibrium pastical pressures! Here, all partical pressures are 1.90 atm PNO = x= 1.90 utu PNO2 = x= 1.90 atm PN2O3 = 3.80- 3 .20-.90 - 1.90 at o