Suppose a 250 flask is filled with 0.90 mol of N 2 and 0.10 mol of NO . The following reaction becomes possible : N 2 (g)+O 2 (g) 2NO * (g) The equilibrum constant for this reaction is 0.572 at the temperature of the Calculate the equilibrium molanty of NO. Round your answer to two decimal places. 9.6.10Suppose a 250. mL flask is filled with 0.90 mol of N, and 0.10 mol of NO. The following reaction becomes possible:N,(g) +0,(e) - 2NO(g)The equilibrium constant K for this reaction is 0.572 at the temperature of the flask.Calculate the equilibrium molarity of NO. Round your answer to two decimal places.

Given : Moles of N2 taken = 0.90 mol. Moles of NO taken = 0.10 mol. And volume of flask = 250 mL = 0.250 L (since 1 L = 1000 mL) The equilibrium reaction taking place is given as, => N2 (g) + O2 (g) ------> 2 NO (g) K = 0.572 Since moles of gas = molarity X volume of flask in L => 0.10 = Initial molarity of NO X 0.250 => Initial molarity of NO = 0.4 M And 0.90 = Initial molarity of N2 X 0.250 => Initial molarity of N2 = 3.6 M Assuming y molarity of O2 is formed to reach equilibrium. Hence molarity of N2 formed = molarity of O2 formed = y And molarity of NO reacted = molarity of O2 formed X 2 = 2y Hence the equilibrium molarities of gases are, => [N2 ] = Initial + formed = 3.6 + y => [O2 ] = Initial + formed = 0 + y = y And [NO] = Initial - reacted = 0.4 - 2y The equilibrium constant expression for the reaction is given by, => K = [NO]2[N2] [O2]Hence substituting the values we get,=> 0.572 = (0.4 - 2y)2(3.6 + y) X y=> y = 0.0457 M = 4.57 X 10-2 M approx. Hence the equilibrium molarity of NO = 0.4 - 2y = 0.4 - 2 X 4.57 X 10-2 = 0.31 M approx.