For the reaction $A (g) ⇌ 2 B (g)$ , the equilibrium partial pressures of gases A and B are 4 atm and 2atm respectively. If equilibrium mixture is slowly and isothermally compressed to 12 atm, maintaining equilibrium, then the new equilibrium pressure (in atm) of B(g) is

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

4 atm

3 atm

8 atm

6 atm

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$A (g) ⇄ 2 B (g); K_{P} = \frac{2^{2}}{4} = 1$

At new equilibrium, let partial pressue of A(g) = x atm, and partial pressure of B(g) = y atm
Then total pressure = (x+y) = 12 atm and $K_{P} = \frac{y^{2}}{x} = 1$
Solving for x and y, we get partial pressure of B(g) = 3 atm

Watch 3-min video & get full concept clarity

course

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

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

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!