Part B.) Concentration (M) [XY] [X] + [Y] I 0.500 0.100 0.100 C -x +x +x E 0.500
ID: 897742 • Letter: P
Question
Part B.)
Concentration (M) [XY] [X] + [Y]
I 0.500 0.100 0.100
C -x +x +x
E 0.500 – x 0.100 + x 0.100 + x
Express the molar concentrations numerically.
Based on a Kc value of 0.260 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?
Part C.)
Concentration (M) [XY] [X] + [Y]
I 0.200 0.300 0.300
C +x - x - x
E 0.200 + x 0.300 - x 0.300 - x
Express the molar concentrations numerically.
Based on a Kc value of 0.260 and the data table given, what are the equilibrium concentrations of XY, X, and Y respectively?
Explanation / Answer
Part(B):
Kc = [XY]/[X][Y]
0.260 = (0.500-x)/(0.100+x)^2
=> x = 0.43
[XY] = 0.500 - 0.43 = 0.07 M
[X] = 0.100 + 0.43 = 0.53 M
[Y] = 0.100 + 0.43 = 0.53 M
Part(C):
Kc = [X][Y]/[XY]
0.260 = (0.300-x)(0.300-x)/(0.200+x)
0.260 = ((0.300-x)(0.300-x))/(0.200+x)
=> x = 0.05
[XY] = 0.200 + 0.05 = 0.250 M
[X] = 0.300 - 0.05 = 0.250 M
[Y] = = 0.300 - 0.05 = 0.250 M