X _ 2 ( ~g )+ Y _ 2 ( ~g ) 2 Z ( g ) X _ 2 ( ~g ) and Y _ 2 ( ~g ) are added to a 1 L flask and it is found that the system attains the above equilibrium at T ( K ) with the number of moles of X _ 2 ( ~g ), Y _ 2 ( ~g ) and Z ( g ) being 3,3 and 9 mol respectively (equilibrium moles). Under this con

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X _ 2 ( ~g )+ Y _ 2 ( ~g ) 2 Z ( g ) X _ 2 ( ~g ) and Y _ 2 ( ~g ) are added to a 1 L flask and it is found that the system attains the above equilibrium at T ( K ) with the number of moles of X _ 2 ( ~g ), Y _ 2 ( ~g ) and Z ( g ) being 3,3 and 9 mol respectively (equilibrium moles). Under this condition of equilibrium, 10 mol of Z ( g ) is added to the flask and the temperature is maintained at T ( K ). Then the number of moles of Z ( g ) in the flask when the new equilibrium is established is \_\_\_\_. (Nearest integer)

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Correct answer: A. A

The equilibrium constant for the reaction X_2(g) + Y_2(g) 2Z(g) is calculated from the initial equilibrium state: K_c = [Z]^2 [X_2][Y_2] = 9^2 3 3 = 9 mol/L. When 10 mol of Z(g) is added, the new concentrations are: [X_2] = 3, [Y_2] = 3, [Z] = 19 mol. The reaction quotient is Q = 19^2 3 3 = 40.11. Since Q > K_c, the equilibrium shifts left (toward reactants). Let x mol of Z decompose. At the new equilibrium: [X_2] = 3 + x 2 , [Y_2] = 3 + x 2 , [Z] = 19 - x. Applying the equilibrium expression: (19-x)^2 (3+ x 2 )^2 = 9. Taking the square root: 19-x 3+ x 2 = 3. Solving: 19 - x = 9 + 3x 2 , which gives 10 = 5x 2 , so x = 4. The moles of Z at new equilibrium = 19 - 4 = 15 mol.

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