Part B Consider mixture B, which will cause the reaction to proceed forward: Con
ID: 822955 • Letter: P
Question
Part B
Consider mixture B, which will cause the reaction to proceed forward:
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
Based on a Kc value of 0.200 the given data table, what are the equilibrium concentrations of XY, X, and Y respectively?
Express the molar concentrations numerically
Part C
Consider mixture C, which will cause the net reaction to proceed in reverse
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
Based on a Kc value of 0.20 and the data table given, what are the equilibrium concentrations of XY, X, and Y respectively?
Express the molar concentrations numerically
Explanation / Answer
(B) XY <=> X + Y
Kc = [X][Y]/[XY]
= (0.100 + x)(0.100 + x)/(0.500 - x) = 0.200
x^2 + 0.2x + 0.01 = 0.1 - 0.2x
x^2 + 0.4x - 0.09 = 0
The positive root of the quadratic equation is:
x = 0.1606
[XY] = 0.500 - x = 0.3394 M = 0.339 M
[X] = 0.100 + x = 0.2606 M = 0.261 M
[Y] = 0.100 + x = 0.2606 M = 0.261 M
(C) XY <=> X + Y
Kc = [X][Y]/[XY]
= (0.300 - x)(0.300 - x)/(0.200 + x) = 0.200
x^2 - 0.6x + 0.09 = 0.04 + 0.2x
x^2 - 0.8x + 0.05 = 0
The positive roots of the quadratic equation are:
x = 0.0683 and x = 0.7317 (not possible since x < 0.300)
[XY] = 0.200 + x = 0.2683 M = 0.268 M
[X] = 0.300 - x = 0.2317 M = 0.232 M
[Y] = 0.300 - x = 0.2317 M = 0.232 M