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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.

Respuesta :

superman1987
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
  


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