If for the reaction, $2 I C l \to I_{2} + C l_{2}; K_{c} = 0.14$ and initial concentration of $I C L$ is $0.6 M$ then equilibrium concentration of $I_{2}$ is

0.37 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.128 M

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.224 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.784 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0.128 M$
For given reaction,

$K_{c} = 0.14 = \frac{[I_{2}] [C l_{2}]}{[I C l]^{2}} = \frac{x^{2}}{(0.6 - 2 x)^{2}}$

$0.37 = \frac{x}{0.6 - 2 x}$
$0.222 - 0.74 x = x$
$\Rightarrow 1.74 x = 0.222$
$\Rightarrow x = 0.128 M$

Suggest Corrections

Similar questions

I_{C l} \to I_{2} + C l_{2}

(K_{c})

2 I C l (s) ⇌ I_{2} (s) + C l_{2} (g)

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2 ICl(g) ⇌ I₂(g) + Cl₂(g) ; K_C = 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?

C l_{2} + 2 I^{-} \to I_{2} + 2 C l^{-}

I^{-}

0.20 m o l l i t^{- 1}

20

0.18 m o l l i t^{- 1}

I_{2}

m o l l i t^{- 1} m i n^{- 1}

Join BYJU'S Learning Program