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

Question

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

  net

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

initial:   0.500 .100 .10

change: -x +x +x

equilibrium: 0.500x   0.100+x   0.100+x

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

Question #1 Based on a Kc value of 0.190 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?

Express the molar concentrations numerically.

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

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

net

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

initial:   0.200 .300 .300

change: +x -x -x

equilibrium: 0.200+x   0.300-x   0.300-x

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

QUESTION #2   

Based on a Kc value of 0.190 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Express the molar concentrations numerically.

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

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

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

net

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

initial:   0.200 .300 .300

change: +x -x -x

equilibrium: 0.200+x   0.300-x   0.300-x

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

QUESTION #2   

Based on a Kc value of 0.190 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Express the molar concentrations numerically.

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

Explanation / Answer

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

initial:   0.500 .100 .10

change: -x +x +x

equilibrium: 0.500x   0.100+x   0.100+x

   Kc   = [X[[Y]/[XY]

0.19 = (0.1+x )*(0.1+x)/0.5-x

0.19*(0.5-x)   = (0.1+x )*(0.1+x)

   x    = 0.1557

[X]   = 0.1+x   = 0.1 +0.1557 = 0.2557M

[Y]   = 0.1+x   = 0.1 +0.1557 = 0.2557M

[XY] = 0.5-x   = 0.5-0.1557 = 0.3443M

2.

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

initial:   0.200 .300 .300

change: +x -x -x

equilibrium: 0.200+x   0.300-x   0.300-x

         Kc   = [X[[Y]/[XY]

         0.19 = (0.3-x)(0.3-x)/0.2+x

        0.19*(0.2+X) = (0.3-x)*(0.3-x)

      0.038 +0.19X   = (0.3-x)^2

     0.038+0.19x     = 0.09 +x2 -0.6x

       x2 -0.79x+0.052 =0

       X = 0.072

[x]   = 0.3-x = 0.3-0.072 = 0.228M

[Y]   = 0.3-x = 0.3-0.072 = 0.228m

[XY]   = 0.2+X = 0.2+0.072 = 0.272M

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