Question

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of $ICl$ was $0.78 \ M$?
$2ICl (g) ⇋ I_2 (g) + Cl_2 (g); \ K_c = 0.14$

Answer 1

The given reaction is:
$2ICl(g) ⇋ I_2(g) + Cl_2(g)$
Initial conc. $0.78 \ M$ 0 0
At equilibrium $(0.78 - 2x) M$ $x M$ $x M$
Now, we can write,
$\frac {[I_2] [Cl_2]}{[ICl]^2}= K_c$

⇒ $\frac {x × x }{(0.78 - 2x)^2} = 0.14$

⇒ $\frac {x^2 }{(0.78 - 2x)^2} = 0.14$

⇒ $\frac {x}{0.78 - 2x}= 0.374$

⇒ $x = 0.292 - 0.748 x$
⇒ $1.748 x = 0.292$
⇒ $x = 0.167$
Hence, at equilibrium,
$[H_2]=[I_2] = 0.167 \ M$
$[HI] = (0.78 - 2 \times 0.167) M$
$= 0.446\ M$