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

Question

Concentration (M)initial:change:equilibrium:[XY]0.500x0.500xnet[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 0.170 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively? and Part C Consider mixture C, which will cause the net reaction to proceed in reverse. Concentration (M)initial:change:equilibrium:[XY]0.200+x0.200+xnet[X]0.300x0.300x+[Y]0.300x0.300x The change in concentration, x, is positive for the reactants because they are produced and negative for the products because they are consumed. Based on a Kc value of 0.170 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Explanation / Answer

Concentration (M): XY             X                 Y

initial:                                                     0.500       0.100          0.100

change:                                                     -x             +x                +x

equilibrium:                                            0.500-x       0.100+x     0.100+x

Part B

Kc value is 0.170 and reactants are consumed and products are produced

Reaction will be X + Y <-> XY

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

Let us apply values from at equilibrium given above.

0.170 = (0.500 – x) / (0.1 + x) (0.1 + x)

0.170 * (0.1 + x) (0.1 + x) = (0.500 – x)

0.170 * (0.01 + 0.2x + x2) = (0.500 – x)

0.0017 + 0.034x + 0.170x2 = 0.500 – x

0.170x2 + 1.0034x – 0.4983 = 0

Solving this equation, for x

x = 0.460

Hence equilibrium concentrations are:

[XY] = 0.500 – 0.460 = 0.040

[X] = 0.100 + 0.460 = 0.560

[Y] = 0.100 + 0.460 = 0.560

Part C

Kc value is 0.170 and reaction is reversed.

Reaction will be XY <-> X + Y

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

Let us apply values from at equilibrium given above.

0.170 = (0.1 + x) (0.1 + x) / (0.500 – x)

0.170 * (0.500 – x) = 0.01 + 0.2x + x2

0.085 – 0.17x = 0.01 + 0.2x + x2

x2 + 0.37x – 0.075 = 0

Solving this equation, for x

We get two values of x, but negative value is not feasible as it will give concentrations in negative.

x = 0.145

Hence equilibrium concentrations are:

[XY] = 0.500 – 0.145 = 0.645

[X] = 0.100 + 0.145 = 0.245

[Y] = 0.100 + 0.145 = 0.245

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