Question

0.100-2 O in the so Solving for a, asing the kmown (HI) at equilibriam 011M -0100 -2-0078 M concentrations ane -011 ad we are given [H In a so find equilit focuses or Therefore the equilibrium HI0.078 M Substituting into the reaction quotient [HJ11] Sample Thus x - (0.011x0.011) K =-0.078 Problem to regul of H2O equilib PlanV the eq to wr (0.25 set u conc Sol Check Rounding gives 0.010.08 0.02. Because the initial HIl of slightly at equilibrium to 0.078 M, relatively little product formed; so we expect K. 0 and 100 atm of O2. At equilibrium, po,?0.506 atr. Calc? NOg 01ti,?2NOfg. 2NO4g), was stidied at 184'C with initial pressures Using the Equilibrium Constant to Find Quantities type of problem that involves finding equilibrium quantities also has several variCa tions. Sample Problem 17. 7 is one variation, in which we know K and some of the tions and must find another equilibriun

please solve for highlighted section. show as many steps as possible I'm just learning this

Answer 1

2N_0 + O_2 rightarrow 2NO_2 P_NO = 1 atm, PO_2 = 1 atm at initial ta equilibrium [PO_2]_eq = 0.506 atm Given [PO_2]_Sigma Q = 0.506 = 1 - x Therefore x = 1 - 0.506 = 0.499 atm K_p = (P_NO_2)_epsilon a/(P_NO)_epsilon a (P_O_2)_epsilon a = 2x/(1 - x)(1 - 2x) = 2(0.494)/(1 - 0.494)(1 - 2(0.494)) = 0.988/(0.506)(0.012) = 162.7 Therefore K_p = 162.7