What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?
$2 I C l_{(g)} \leftrightarrow l_{2 (g)} + C l_{2 (g)}; K_{c} = 0.14$

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Solution

The given reaction is:

$\begin{matrix} 2 I C l_{(g)} & \leftrightarrow & I_{2 (g)} & C l_{2 (g)} I o n i c c o n e . & 0.78 M & 0 & 0 A t e q u i l i b r i u m & (0.78 - 2 x) M & x M & x M \end{matrix}$
Now we can wirte, $\frac{[I_{2}] [C l_{2}]}{[I C l]^{2}} = K_{c}$
$\Rightarrow \frac{x \times x}{(0.78 - 2 x)^{2}} = 0.14 \Rightarrow \frac{x^{2}}{(0.78 - 2 x)^{2}} = 0.14 \Rightarrow \frac{x}{0.78 - 2 x} = 0.374 \Rightarrow x = 0.292 - 0.748 x \Rightarrow 1.748 x = 0.292 \Rightarrow x = 0.167$
Hence, at equilibrium
$[H_{2}] = [I_{2}] = 0.167 M [H I] = (0.78 - 2 \times 0.167) M = 0.446 M$

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What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M? 2ICl(g)↔l2(g)+Cl2(g); Kc=0.14

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?
$2 I C l_{(g)} \leftrightarrow l_{2 (g)} + C l_{2 (g)}; K_{c} = 0.14$