The equilibrium constant, Kp for the reaction A⇔2B is released to degree of dissociation α of A and total pressure P as

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The equilibrium constant, $K_{p}$ for the reaction $A \Leftrightarrow 2 B$ is released to degree of dissociation $α$ of A and total pressure P as

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{4 α^{2} P}{1 - α}$

$\frac{4 α^{2} P}{1 - α^{2}}$

$\frac{4 α^{2} P^{2}}{1 - α^{2}}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

$A \Leftrightarrow 2 B 1 0 (1 - α) 2 α Total moles = 1 - α + 2 α = 1 + α K_{p} = \frac{{P_{B}}^{2}}{P_{A}} \frac{{(\frac{2 α}{1 + α} P)}^{2}}{(\frac{1 - α}{1 + α} P)} = \frac{4 α^{2}}{1 - α^{2}} P$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th