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Part B

Consider mixture B, which will cause the reaction to proceedforward:


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

I 0.500 0.100 0.100

C -x +x +x

E 0.500-x 0.100+x 0.100+ x

Based on a Kc value of 0.200 the given data table, what are theequilibrium concentrations of XY, X, and Y respectively?


Express the molar concentrations numerically


Part C

Consider mixture C, which will cause the net reaction to proceedin reverse


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

I 0.200 0.300 0.300

C +x -x -x

E 0.200+ x 0.300 -x 0.300 - x

Based on a Kc value of 0.20 and the data table given, what arethe equilibrium concentrations of XY, X, and Y respectively?


Express the molar concentrations numerically




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Patrina SchowalterLv2
28 Sep 2019

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Related questions

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

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

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

Express the molar concentrations numerically.

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

Concentration (M) [XY] ⇌ [X] + [Y]

initial: 0.200 0.300 0.300

Change: +x -x -x

equilibrium:0.200+x 0.300-x 0.300-x

Part C

Based on a Kc value of 0.180 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Express the molar concentrations numerically.

XY(aq

XY(aq)?X(aq)+Y(aq)

Qc=[X][Y]/[XY]

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

Concentration (M)initial:change:equilibrium:

[XY (I)]0.500(C)?x(E)0.500?x

net??

[X](I)0.100(C)+x(E)0.100+x

+[Y](I)0.100(C)+x(E)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.

Part B

Based on a Kc value of 0.200 and the given data table, what are the equilibrium concentrations of XY , X , and Y , respectively?

Express the molar concentrations numerically.

SubmitHintsMy AnswersGive UpReview Part

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](I)0.200(C)+x(E)0.200+x

?net?

[X](I)0.300(C)?x(E)0.300?x+

[Y](I)0.300(C)?x(E)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.

Part C

Based on a Kc value of 0.200 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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