Question

Suppose a 250. mL flask is filled with 2.0 mol of NO and 0.30 mol of NO . The following reaction becomes possible: NO(g) + NO(g) - 2NO() The equilibrium constant K for this reaction is 0.662 at the temperature of the flask. Calculate the equilibrium molarity of NO2. Round your answer to two decimal places. x ?

Answer 1

Write the reaction as follows: NO;() + NO(g) → 2N0 (8) Equilibrium constant, K = 0.662 Volume of the flask = 250 mL Initi al moles of NO = 2.0 mol Initi al moles of NO, = 0.30 mol - Calculate the initial concentrations of the components as follows: Inol- Initial moles of NO 2.0 mol 1000 mL = 8.0 mol/L = 8.0M Volume of the flask 250 mL 1L Initial moles of NO, 0.30mol 1000 mL [NO]= = 1.2 mol/L = 1.2M Volume of the flask 250 mL 1L ----- -------------- ------ ------ -------------- Draw the ICE table as follows: NO(g) + NO(g) + 2NO,(8) I(M): 0 8.0 1.2 C(M): +x +x -2x E(M): +X 8.0+x 1.2-2x Equilibrium constant is defined as the ratio of product of concentrations of products raised to the powers of their stoichiometric coefficients to the product of concentrations of reactants raised to the powers of their stoichiometric coefficients. (NO KE ... (1) (NO3 ][NO]

Substitute the corresponding values in equation (1) to calculate x. 0.662 = (1.2–2x) 02 (8.0+x) 0.662 ( 8.0x+x) = 1.44+4x2 - 4.8x 5.296x+0.662xº = 1.44 +47° -4.8x 3.338x² -10.096x +1.44 = 0 The above equation is in the form of quadratic equation, axº + bx+c = 0 solution is x=-bvb2-4ac X= I za 10.096+ (-10.096) -4(3.338) (1.44) x= 2(3.338) 10.096 82.7 6.676 10.096+9.094 6.676 10.096+9.094 10.096-9.094 6.676 6.676 19.191.002 ** 6.676 and * 6.676 x = 2.87 and x = 0.15 - and x Since 1.2-2x must be positive, we take x= 0.15 ----- ------------------ -- ------ At equilibrium, molarity of NO, is (NO2) = 1.2-2x = 1.2-2(0.15) = 1.2-0.30 = 0.90 M Therefore, the required solution is 0.90 M