Given the following informations.
(i) ${NH}_{3} (aq) + H_{2} O (l) ⇌ {NH}_{4}^{+} (aq) + {OH}^{-} (aq) K_{c l}$
(ii) ${CO}_{2} (g) + H_{2} O (l) ⇌ H_{2} {CO}_{3} (aq) K c_{2}$
(iii) $H_{2} {CO}_{3} (aq) ⇌ 2 H^{+} (aq) + {CO}_{3}^{2 -} (aq) K_{c 3}$
(iv) $H_{2} O (l) ⇌ H^{+} (aq) + {OH}^{-} (aq)$ $K_{c 4}$
The expression of $K_{c}$ for $2 {NH}_{3} (aq) + {CO}_{2} (g) + H_{2} O (l) ⇌ 2 {NH}_{4}^{+} (aq) + {CO}_{3}^{2 -} (aq)$ will be equal to

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$K_{c 1} K_{c 2} K_{c 3} / K_{c 4}$

$K_{c 1} K_{c 2}^{2} K_{c 3} / K_{c 4}^{2}$

$K_{c 1}^{2} K_{c 2} K_{c 3} / K_{c 4}^{2}$

$K_{c 1} K_{c 2} / K_{c 3} K_{c 4}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The given expression is obtained as 2Eq.(i) + Eq.(ii) + Eq.(iii)- 2Eq.(iv)

Hence $K_{c} = K_{c 1}^{2} K_{c 2} K_{c 3} / K_{c 4}^{2}$

The constant of equilibrium Kc is a constant that indicates how far a reaction will advance at a certain temperature.

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test