Under constant temperature conditions, the equilibrium constant (kc) value was measured to be 0.85 for the following chemical reaction: 2 icl (g) ⇋ i2 (g) + cl2 (g). determine the equilibrium concentration of icl if 0.57 mol of i2 and 0.57 mol of cl2 are mixed in a 3.0-l flask with no icl present initially.(a) 0.067 m(b) 0.40 m(c) 0.20 m(d) 0.10 m(e) 0.13 m

E) 0,13 M.

Explanation:

For the equilibrium reaction:

2 ICl (g) ⇋ I₂ (g) + Cl₂(g)

The equilibrium constant, kc, is:

(1)

The equilibrium moles are:

[I₂] = 0,57 mol - x

[Cl₂] = 0,57 mol -x

[ICl] = 2x

-Where x are the moles that react-

Replacing in (1):

2,4x² + 1,14x - 0,3249 = 0

Solving:

X =-0,675 -No chemical sense. There are not negative concentrations-

X = 0,200 -Real answer-

Thus, moles of ICl are 2×0,200 = 0,400 moles

As the volume is 3 L

The concentration of ICl in equilibrium is 0,400mol/3L = 0,13 M

I hope it helps!

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