Calculating equilibrium concentrations when the net reaction proceeds in reverse

Question

Calculating equilibrium concentrations when the net reaction proceeds in reverse

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

Part C

Part complete

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.

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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+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.

Part C

Part complete

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.

View Available Hint(s)

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

Explanation / Answer

             XY    <------>   X   +   Y

initially   0.2              0.3     0.3

Change       +x               -x      -x

At equilib (0.2+x)          (0.3-x) (0.3-x)

since X and Y are more concentration initially so the reaction shifts to backward direction.

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

0.26 = ((0.3-x)*(0.3-x))/(0.2+x)

X = 0.04672

Equilibrium concentrations of [XY] = 0.2+0.04672 = 0.2467

[X] = (0.3-x) = (0.3-0.04672) = 0.02533

[Y] = (0.3-x) = (0.3-0.04672) = 0.02533

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