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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

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4 atm

3 atm

8 atm

6 atm

answer is A.

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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

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