Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g) = Br2 (g) + Cl2 (g) for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 mbar what is its partial pressure in the mixture at equilibrium

Question

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g) = Br2 (g) + Cl2 (g) for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 mbar what is its partial pressure in the mixture at equilibrium

Vikas TU · Answer

2BrCl (g) = Br2 (g) + Cl2 (g) t=0 3.3 * 10^-3 0 0 t=t 3.3 * 10^-3 – x x/2 x/2 Kc = x^2/4(3.3 * 10^-3 – x)^2 = 32 x/(3.3 * 10^-3 – x) = 2 x 4 x root(2) x = 37.3 – 11.3x 12.3x = 37.3 x = 3.035 mBar. At eqm. initial partial pressure would be hence, x/2 for each Br2 and Cl2 that is = > 1.517 mBar.

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Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g) = Br2 (g) + Cl2 (g) for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 mbar what is its partial pressure in the mixture at equilibrium

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g) = Br2 (g) + Cl2 (g) for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 mbar what is its partial pressure in the mixture at equilibrium

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Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g) = Br2 (g) + Cl2 (g) for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 mbar what is its partial pressure in the mixture at equilibrium

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g) = Br2 (g) + Cl2 (g) for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 mbar what is its partial pressure in the mixture at equilibrium

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