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${AB}_{2}$ dissociates as : ${AB}_{2 (g)} ⇌ {AB}_{(g)} + B_{(g)}$ When the intial pressure of ${AB}_{2}$ is $500 mm$ $Hg$ , the total equilibrium pressure is $700 mm$ Hg. Calculate equilibrium constant for the reaction, assuming that the volume of the system remains unchanged.

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$214.6 mmHg$

$100 mmHg$

$133.3 mmHg$

$200 mmHg$

answer is B.

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Detailed Solution

${AB}_{2} (g) ⇌ AB (g) + B (g)$

$500 mm 0 0$

500-x x x

Hence,

$500 - x + x + x = 700$

$x = 200$

$K_{p} = \frac{P_{A B} x P_{B}}{P_{A B_{2}}}$

$P_{A B} = x = 200 P_{B} = x = 200 P_{A B_{2}} = x = 200$

$K_{p} = \frac{200 \times 200}{300} = 133.3 mm$

Hence the correct option is (B) $133.3 mmHg$

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