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Characteristics of Equilibrium Constant

AB dissociate...

Question

$A B$ dissociates as $2 A B (g) ⇌ 2 A (g) + B_{2} (g)$ . When the initial pressure of $A B$ is $500 m m$ , the total pressure becomes $625 m m$ when the equilibrium is attained. Calculate $K_{P}$ for the reaction assuming volume remains constant.

500

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125

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750

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375

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Solution

The correct option is B $125$

Considering the equation:

$2 A ⟶ 2 A + B 2$

Initial= $500.00$

Final= $500 - . P$ $\frac{P}{2}$

Total pressure= $625 = 500 - P + P +$ $\frac{P}{2}$

P= $250 m m H g$

$P_{A B} = 250$

$P_{A} = 250$

$P_{B_{2}} = 125$

$∴ K_{p} = 125 m m$

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