Calculating equilibrium concentrations when the net reaction proceeds forward Co

ID: 1036630 • Letter: C

Question

Calculating equilibrium concentrations when the net reaction proceeds forward

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

Concentration (M)initial:change:equilibrium:[XY]0.500?x0.500?xnet??[X]0.100+x0.100+x+[Y]0.100+x0.100+x

The change in concentration, x, 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 0.140 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?

Express the molar concentrations numerically. ?????

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Calculating equilibrium concentrations when the net reaction proceeds in reverse

Consider mixture C, which will cause the net reaction to proceed in reverse.

Concentration (M)initial:change:equilibrium:[XY]0.200+x0.200+x?net?[X]0.300?x0.300?x+[Y]0.300?x0.300?x

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

Part C

Based on a Kc value of 0.140 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Express the molar concentrations numerically. ?????

XY <-----> X + Y

initially 0.5 0.1 0.1

change -x +x +x

equilibrium (0.5-x) (0.1+x) (0.1+X)

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

0.14 = (0.1+x)*(0.1+x)/(0.5-x)

x = 0.128

equilibrium concentrations [XY] = 0.5-0.128 = 0.372

[X] = [Y] = 0.1+x = 0.1+0.128 = 0.228

XY <-----> X + Y

initially 0.2 0.3 0.3

change +x -x -x

equilibrium (0.2+x) (0.3-x) (0.3-X)

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

0.14 = (0.3-x)*(0.3-x)/(0.2+x)

x = 0.0963

equilibrium concentrations [XY] = 0.2+0.0963 = 0.2963

[X] = [Y] = 0.3-x = 0.3-0.0963 = 0.2037

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