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18 Nov 2019

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

Express the molar concentrations numerically.

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29 Aug 2019

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Related questions

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 0.170 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+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. Based on a Kc value of 0.170 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

(1) The following reaction was carried out in a 3.50 L reaction vessel at 1100 K:

C(s)+H2O(g)⇌CO(g)+H2(g)

If during the course of the reaction, the vessel is found to contain 5.25 mol of C, 12.0 mol of H2O, 4.00 mol of CO, and 7.40 mol of H2, what is the reaction quotient Q?

(2) 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.

Based on a Kc value of 0.250 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?

(3) 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.

Based on a Kc value of 0.250 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

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 0.200 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

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

Express the molar concentrations numerically.

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

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