Consider mixture B, which will cause the net reaction to proceed forward net onc

Question

Consider mixture B, which will cause the net reaction to proceed forward net oncentration M initial change: equilibrium 0.500 0.100 0.100 0.500- 0.100+r 0.100+r The change in concentration, , is negative for the reactants because they are consumed and positive for the products because they are produced Part B Based on a Kc value of 0.170 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively? Express the molar concentrations numerically Submit Hints My Answers Give Up Review Part

Explanation / Answer

If Kc = 0.17

and

Kc = [X][Y]/[XY]

0.17 = (0.1+x)(0.1+x)/(0.5-x)

solve for x

0.17(0.5-x) = (0.1^2 + 2*0.1x +x^2)

0.085 - 0.17x = 0.01 +0.2x + x^2

x^2 +(0.2+0.17)x + 0.01-0.085 = 0

x^2 +0.34x + -0.075 = 0

x = 0.15233

then

[X]= 0.1 +x = 0.1 +0.15233 = 1.15233

[Y]= 0.1 +x = 0.1 + 0.15233 = 1.15233

[XY] = 0.5 - x = 0.5 - 0.15233= 0.34767

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