For the equilibrium below at 400 K, K c = 7.0. If (initially) 0.22 mol of Br 2 a

Question

For the equilibrium below at 400 K, Kc = 7.0.

If (initially) 0.22 mol of Br2 and 0.22 mol Cl2 are introduced into a 1.0 L container at 400. K, what will be the equilibrium concentration of BrCl? Assume the initial product concentration is zero.

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For the equilibrium below at 400 K, Kc = 7.0. Br2(g) + Cl2(g) 2 BrCl(g) If (initially) 0.22 mol of Br2 and 0.22 mol Cl2 are introduced into a 1.0 L container at 400. K, what will be the equilibrium concentration of BrCl? Assume the initial product concentration is zero.

Explanation / Answer

initial concentration Br2 = initial concentration Cl2 = 0.30/ 1.0 = 0.22 M

at equilibrium : [Br2]= [Cl2] = 0.22 -x
[BrCl]= 2x

7.0 = (2x)^2/ (0.22-x)^2

square root

2.6 = 2x / 0.22-x

0.572 - 2.6 x = 2x

0.572 = 4.6 x
x = 0.124

[Br2]= [Cl2] = 0.30 - 0.124 =0.2876 M
[BrCl]= 2x = 0.124 x 2 = 0.248 M

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