At 35 degree C, K = 1.6 Times 10^-5 for the reaction 2NOCl(g) 2NO(g) + Cl_2(g) C

Question

At 35 degree C, K = 1.6 Times 10^-5 for the reaction 2NOCl(g) 2NO(g) + Cl_2(g) Calculate the concentrations of all species at equilibrium for each of the following original mixtures. 2.0 mol pure NOCl in a 2.0-L flask 1.0 mol NOCl and 1.0 mol NO in a 1.0-L flask 2.0 mol NOCl and 1.0 mol Cl_2 in a 1.0-L flask At a particular temperature, K = 4.0 Times 10^-7 for the reaction N_2O_4(g) 2NO_2(g) In an experiment, 1.0 mol N_2O_4 is placed in a 10.0-L vessel. Calculate the concentrations of N_2O_4 and NO_2 when this reaction reaches equilibrium. At a particular temperature, K = 2.0 Times 10^-6 for the reaction 2CO_2(g) 2CO(g) + O_2(g) If 2.0 mol CO_2 is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species. Lexan is a plastic, used to make compact discs, eyeglass lenses, and bullet-proof glass. One of the compounds used to make Lexan is phosgene (COCl_2), an extremely poisonous gas. Phosgene decomposes by the reaction

Explanation / Answer

57.

K = [NO]^2[Cl2]/[NOCl]^2

a. [NOCl] = 2 mol/1 L = 2 M

Let x be the change at equilibrium

1.6 x 10^-5 = 4x^3/(2 - 2x)^2

4x^3 - 6.4 x 10^-5x^2 + 1.28 x 10^-4x - 6.4 x 10^-5 = 0

x = 0.025 M

Equilibrium concentration of,

[NOCl] = 2 - 2 x 0.025 = 1.5 M

[NO] = 2 x 0.025 = 0.05 M

[Cl2] = 0.025 M

b. [NOCl] = 1 mol/1 L = 1 M

[NO] = 1 mol/1 L = 1 M

let x be the change at equilbrium

1.6 x 10^-5 = (1 + 2x)^2(x)/(1 - 2x)^2

3.96x^3 + 4x^2 + 0.96x - 1.6 x 10^-5 = 0

x = 1.6 x 10^-5 M

Equilibrium concentration of,

[NOCl] = 2 - 2 x 1.6 x 10^-5 = 1.96 M

[NO] = 1 - 2 x 1.6 x 10^-5 = 0.96 M

[Cl2] = 1.6 x 10^-5 M

c. [NOCl] = 2 mol/1 L = 2 M

[Cl2] = 1 mol/1 L = 1 M

let x be the change at equilibrium

1.6 x 10^-5 = (2x)^2(1+x)/(2 - 2x)^2

4x^3 + 4x^2 + 1.28 x 10^-4x - 6.4 x 10^-5 = 0

x = 0.004 M

Equilibrium concentration of,

[NOCl] = 2 - 2 x 0.004 = 1.992 M

[NO] = 2 x 0.004 = 0.008 M

[Cl2] = 1 - 0.004 = 0.996 M

59. K = [CO]^2[O2]/[CO2]^2

[CO2] = 2 mol/5 L = 0.4 M

let x be the change at equilibrium

2 x 10^-6 = (2x)^2(x)/(0.4 - 2x)^2

4x^3 - 8 x 10^-6x^3 + 3.2 x 10^-6x - 3.2 x 10^-7 = 0

x = 0.004 M

Equilibrium concentration of,

[CO2] = 0.4 - 2 x 0.004 = 0.392 M

[CO] = 2 x 0.004 = 0.008 M

[O2] = 0.004 M

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