What is the equilibrium concentration of each at the substances in equilibrium. When the initial concentration at ICl was 0.70?
$I_{C l} \to I_{2} + C l_{2}$
Kc = 0.14

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Solution

$2 I C l \to I_{2} + c h$
Initially $0.70 00$
At equilibrium $0.70 - 2 x x x$
$K_{C} = \frac{[I_{2}] [C l_{2}]}{[I C l_{2}]^{2}}$
$K_{C} = \frac{x \times x}{(0.7 - 2 x)}$
$0.14 = \frac{x^{2}}{0.7 - 2 x}$
$0.098 - 0.28 x = x^{3}$
$x^{2} + 0.28 x - 0.098 = 0$
$= \frac{- 0.28 + \sqrt{(0.28)^{2}} - 4 \times 0.098}{2}$
$= \frac{- 0.28 + \sqrt{0.0784} - 0.392}{2}$
$= \frac{- 0.28 + 0.56}{2} = 0.14$
$[I C l] = 0.70 - 0.14 \times 2 = 0.42$
$[I_{2}] = 0.14 [C l_{2}] = 0.14$

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