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

Step-by-step solution

The equilibrium constant for the reaction X₂(g) + Y₂(g) 2Z(g) is calculated from the initial equilibrium state: K_c = [Z]^2 [X₂][Y₂] = 9^2 3 3 = 9 mol/L. When 10 mol of Z(g) is added, the new concentrations are: [X₂] = 3 , [Y₂] = 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: [X₂] = 3 + x 2 , [Y₂] = 3 + x 2 , [Z] = 19 - x . Applying the equilibrium expression: (19-x)^2 (3+ x 2 )^2 = 9