Based on a kc value of 0.200 and the given data table, what are the equilibrium concentrations of xy, x, and y, respectively? express the molar concentrations numerically.
missing in your question : Concentration by (M):           Xy:       y:      X initial     0.2      0.3     0.3 change    +X      -X      -X equilibrim (0.2+x)  (0.3-x)   (0.3-x) according to Kc formula: when Kc = 0.2 Kc = [XY]/[X]*[Y] 0.2 = (0.2+x) / (0.3-x)*(0.3-x) 0.2=(0.2+x) / (0.3-x)^2 by solving this equation 0.2*(0.3-x)^2 = 0.2+x 0.2* (0.09-0.6x+x^2)= 0.2 +x 0.0018 - 0.12 X +0.2X^2 = 0.2 + X 0.2X^2 -1.12 X -0.1982 = 0 ∴X= 0.17 ∴[XY] = 0.2 + 0.17 = 0.37 m ∴[X] = 0.3 - 0.17 =0.13 m ∴[y] = 0.3 - 0.17 = 0.13 m  Â
