A gaseous substance $A B_{2} (g)$ convert to $A B (g)$ presence of solid $A (s)$ as : $A B_{2} (g) + A (s) ⇌ 2 A B (g)$
The initial pressure and equilibrium pressure are 0.7 and 0.95 bar. Now the equilibrium mixture is expanded reversibly and isothermally till the gas pressure falls to 0.4 bar. The volume percentage of $A B_{2} (g)$ at the final equilibrium is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

22.5

32.5

67.5

answer is C.

(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

$A B_{2 (g)} + A_{(s)} ⇌ 2 A B_{(g)}$

Initial 0.7 0
0.7-x 2x
Final y (0.4 - y)

$K_{p} = \frac{{(2 x)}^{2}}{0.7 - x} = \frac{1}{4} \times \frac{1}{0.45}$ , $P_{t o t} = 0.95 = 0.7 + x$

$∴ x = 0.25$

$\frac{{(0.4 - y)}^{2}}{y} = \frac{5}{9}$ , $y = 0.13$
At second equilibrium $V_{A B_{2}} = \frac{0.13}{0.4} \times 100 = 32.5 %$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE