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

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22.5

32.5

67.5

answer is C.

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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 $V_{A B_{2}} = \frac{0.13}{0.4} \times 100 = 32.5 %$

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