Question

The equilibrium constant Kc for the reaction below is 0.00427 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentrations are [Br2] = 0.0844 M and [Br] = 0.0763 M, calculate the concentrations of these species at equilibrium.

Answer 1

Given balanced decomposition reaction,

Br₂(g) ⇌ 2Br(g)

Kc = [Br]²/[Br₂] = 0.00427 ------- (given) ------------ (1)

Initially, [Br₂] = 0.0844 M and [Br] = 0.0763 M

Let at equilibrium 'X' M of Br2 decomposes then we write ICE table as,

Br₂(g) ⇌ 2Br(g)

Initially 0.0844 0.0763

Change -X +2X

At equilibrium (0.0844-X) (0.0763+2X)

Using equilibrium concentrations in eq. (1) we get,

(0.0763 + 2X)² / (0.0844 - X) = 0.00427

(0.0763 + 2X)² = 0.00427 x (0.0844 - X)

0.00582 + 0.3052X + 4X² = 0.00036 - 0.00427X

4X² + 0.3052X +0.00427X = 0.00036 - 0.00582

4X² + 0.3095X = -0.00546

On dividing through out by 4,

X² + 0.07737 X = -0.001365

Let us solve this quadratic equation for X using perfect square methode.

Add, Third term = (1/2 x 0.07737)² = (0.03868)² = 0.001496 then

X² + 0.07737 X + 0.001496 = -0.001365 + 0.001496

(X + 0.03868)² = 0.000131

(X + 0.03868) = +/- 0.01145

X + 0.03868 = + 0.01145 or X + 0.03868 = -0.01145

X = -0.02723 or X = -0.05013 (Rejected. See reason at bottom of page)

i) With X = -0.02723, Equilibrium concentrations are

[Br₂] = 0.0844 - (-0.02723) = 0.1116 M

[Br-] = 0.0763 + 2 (-0.02723) = 0.2184 M

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ii) With X = -0.05013

[Br₂] = 0.0844 - (-0.05013) = 0.1345 M

[Br-] = 0.0763 + 2 (-0.05013) = -0.02766 M

Concentration of Br- is coming out -ve with X = -0.05013 which is meaningless so we reject X = -0.05013.