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Characteristics of Equilibrium Constant

AB2 dissociat...

Question

$A B_{2}$ dissociates as $A B_{2} (g) ⇌ A B (g) + B (g)$
When the initial pressure of AB is $500$ mm, the total pressure becomes $700$ mm when the equilibrium is attained. Calculte $K_{p}$ for the reaction assuming volume remains constant.

133.3

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125

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750

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375

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Solution

The correct option is A $133.3$
$A B_{2} (g) ⇌ A B (g) + B (g) 500 m m 00 500 - x x x 500 - 200200200$
Hence,
$500 - x + x + x = 700$
$x = 200$
$K_{p} = \frac{P_{A B . P_{b}}}{P_{A B_{2}}}$
$= \frac{200 \times 200}{300} = 133.3 m m$

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