The reaction 2ICl(g)⇌I2(g)+Cl2(g) is initiated with 0.75MICl(g). If KC=16 the percentage of ICl(g) decomposed at equilibrium is

We have

$2 ICl (g) ⇌ I_{2} (s) + {Cl}_{2} (g) 0.75 M - x x x$

Hence, $K_{c} = \frac{[I_{2}] [{Cl}_{2}]}{[{ICl}_{2}]} = \frac{x^{2}}{(0.75 - x)^{2}} = 16$

Hence, $\frac{x}{0.75 M - x} = 4$ or $5 x = (0.75 M) 4 or x = 0.6 M$

Percent of IC! decomposed $= \frac{0.6 M}{0.75 M} \times 100 = 80 %$

The reaction $2 ICl (g) ⇌ I_{2} (g) + {Cl}_{2} (g)$ is initiated with $0.75 MICl (g) . If K_{C} = 16$ the percentage of $ICl (g)$ decomposed at equilibrium is