At a certain temperature, the equilibrium constant, K_c, for the reaction Cl_2(f

Question

At a certain temperature, the equilibrium constant, K_c, for the reaction Cl_2(f) 2Cl(g) is 5.92 times 10^4. A. If 2.77 g of Cl_2 are placed in a 2.50 L flask at this temperature, what are the equilibrium concentrations of Cl_2 and Cl? B. Following the establishment of equilibrium in part A, the volume of the flask is suddenly increased to 4.00 L while the temperature is held constant. What are the new equilibrium concentrations of Cl_2 and Cl? C. Following the establishment of equilibrium in part A, the volume of the flask is instead suddenly decreased to 1.00 L while the temperature is held constant. What are the new equilibrium concentrations of Cl_2 and Cl?

Explanation / Answer

Given

the reaction is

Cl2 (g) <-----> 2 Cl (g) Kc = 5.92 * 10-3

Given

2.77 g of Cl2

Molar mass of Cl2 = 70.9 g/mol

No. of moles of Cl2 = 2.77 g / 70.9 g/mol = 0.039 moles

Volume = 2.5 L

Concentration [Cl2] = 0.039 moles / 2.5 L = 0.01563 mol/L

Cl2 Cl

Intial 0.01563 -

Converted - x 2x

Equilibrium 0.01563-x 2x

Kc = [Cl]2 / [Cl2] = (2x)2 / (0.01563 - x) = 5.92 * 10-3

4x2 = 5.92 * 10-3 ( 0.01563 - x)

x = 4.13 * 10-3 = 0.00413

[Cl2] = 0.01563 - 0.00413 = 0.0115 M Answer

[Cl] = 2x = 2 * 0.00413 = 0.00816 M Answer

b)

Given

Volume = 4 L

[Cl2] = 0.039 moles/ 4 L = 0.00975 mol/L

Cl2 Cl

Intial 0.00975 -

Converted - x 2x

Equilibrium 0.00975-x 2x

Kc = [Cl]2 / [Cl2] = (2x)2 / (0.00975 - x) = 5.92 * 10-3

4x2 = 5.92 * 10-3 ( 0.00975 - x)

x = 3.13 * 10-3 = 0.00313

[Cl2] = 0.00975 - 0.00313 = 0.00662 M Answer

[Cl] = 2x = 2 * 0.00313 = 0.00616 M Answer

C)

Given

Volume = 1 L

[Cl2] = 0.039 moles/ 1 L = 0.039 mol/L

Cl2 Cl

Intial 0.039 -

Converted - x 2x

Equilibrium 0.039-x 2x

Kc = [Cl]2 / [Cl2] = (2x)2 / (0.039 - x) = 5.92 * 10-3

4x2 = 5.92 * 10-3 ( 0.039 - x)

x = 6.89 * 10-3 = 0.00689

[Cl2] = 0.039 - 0.00689 = 0.0321 M Answer

[Cl] = 2x = 2 * 0.00689 = 0.01378 M Answer

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