Question

22. The initial pressure of N204 = 0.14 atm and the initial pressure of NO2 = 0.55 atm. If the Kp for the reaction is 4.32, what is the equilibrium pressure of the NO2 gas? N2O4 (g) 2 NO2 (g) a) 0.0949 atm b) 0.0451 atm c) 0.658 atm d) 0.640 atm e) 0.731 atm x = - -b + vb2 - 4ac 2a

Answer 1

Given that - N2O4 (9) Initial Pressione = 0.14 atm 2 NO (9) 0.55 atm Pressure at equilibrium => 0.14- O. 55 tau N204 Ko = (b no.)" (0.55 tax)? 4.32 - PN2O4 (0.14 - 4) = (055+2x)} = 4.32 (0:14-2) - a 40²+2.20+ 0.3025 = 0.6048 - 4.324 4212 +2.24 +4.324 + 0.3025 -0.6048 = 0

4x2 + 6.528 -0.3023 = 0 a=4 na - bet 56²-uac b=6.52 ga co -0.30.23 -6.52 + 56:52)2-4(4)(-0.3023 - 2x4 - -6.52 +542.5104 + 4.8368 U = = -6.52 + 6.88 = 0,045, -1.675 Taking positiva value of n = 0.045 Pressure of No, at equilibrium = 0.55 +2u = 0.55+2 (0.045) = 0.640 atm Hence (d) 0.640 is correct answer