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

ID: 3398610 • Letter: C

Question

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

Concentration (M):[XY] [X] + [Y]

initial: 0.500 0.100 0.100

change: -x +x +x

equilibrium: 0.500 0.100 0.100

The change in concentration, x, is negative for the reactants because they are consumed and positive for the products because they are produced.

Based on a Kc value of 0.250 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?

Express the molar concentrations numerically.

Lets construct an ICE table for this

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

Kc = [0.1+2.1x][0.1+2.1x}/0.5-1.5x

Given value of Kc = 0.250

0.25 = [0.1+2.1x][0.1+2.1x]/[0.5-1.5x]

0.25 (0.5-1.5x) = 0.01+0.21x+0.21x+4.41x^2

0.125-0.375x = 0.01+0.42x+4.41x^2

0=0.115+0.795x+4.41x^2

solving this quadratic equation:

x = 0.043

[XY] = 0.5-0.5(0.043)-0.043 = 0.435 M

[X] = 0.1+0.1(0.043)+0.043 = 0.1473M

[Y] = 0.1473M

XY-------> X+ Y Initial 0.5-0.5x-x 0.1+0.1x+x 0.1+0.1x+x Change -x +x +x Euilibrium 0.5+0.5x-2x=0.5-1.5x 0.1+0.1x+2x=0.1+2.1x 0.1+2.1x

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