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 0.260 and the given data table, what are the equilibrium concentrations of  XY, X, and Y, respectively?

Express the molar concentrations numerically.

[XY], [X], [Y] =		M

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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 0.260 and the data table given, what are the equilibrium concentrations of  XY, X, and Y, respectively?

Express the molar concentrations numerically.

[XY], [X],  [Y] =		M

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Answer 1

                      XY --------- X +Y

Initial             0.50        0.10 0.10

Final            (0.50-x)    (0.10+x) (0.10+x)

Concentration of X = Concentration of Y = 0.100 + 0.1858 = 0.2858M

Concentration of XY = 0.500 - 0.1858 = 0.3142

2)

                      XY --------- X +Y

Initial             0.20        0.30 0.30

Final            (0.20+x)    (0.30-x) (0.30-x)

Concentration of X = Concentration of Y = 0.300 - 0.046 = 0.254M

Concentration of XY = 0.200 + 0.046 = 0.246M