Question

Problem 15.52

For the equilibrium
Br2(g)+Cl2(g)⇌2BrCl(g)
at 400 K, Kc = 7.0.

Part A

If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of Br2?

Express your answer to two significant figures and include the appropriate units.

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Part B

If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of Cl2?

Express your answer to two significant figures and include the appropriate units.

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Part C

If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of BrCl?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Br2 + Cl2 <> 2BrCl

K = [BrCl]^2 / [Br2][Cl2]

K = 7

a)

If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of Br2?

initiallly:

[Br2] = 0.25/3 = 0.0833

[Cl2] = 0.55/3 = 0.183

[BrCl2] = 0

then; in equilibrium

[Br2] = 0.0833-x

[Cl2] = 0.183-x

[BrCl2] = 0 +2x

substitute in K

K = [BrCl]^2 / [Br2][Cl2]

7 = (2x)^2/ (0.0833-x)(0.183-x)

(0.0833*0.183 -(0.0833+0.183)x + x^2) = 4/7x^2

0.0152439 - 0.2663x + x^2= 0.5714285x^2

0.42857x^2  - 0.2663x + 0.0152439 = 0

x = 0.0637

substitute

[Br2] = 0.0833-x = 0.0833-0.0637 = 0.0196 M

[Br2] = 0.0196 M

[Cl2] = 0.183-x = 0.183-0.0637

[Cl2] = 0.1193 M

[BrCl2] = 0 +2x = 2*0.0637  = 0.1274

[BrCl2] =0.1274 M