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Characteristics of Equilibrium Constant

XY2 dissociat...

Question

$X Y_{2}$ dissociates as: $X Y_{2} (g) ⇌ X Y (g) + Y (g)$
Initial pressure of $X Y_{2}$ is $600 m m$ $H g$ . The total pressure at equilibrium is $800 m m$ $H g$ . Assuming volume of system to remain constant, the value of $K_{p}$ is:

50

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100

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200

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400

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Solution

The correct option is B $100$
$X Y_{2}$ dissociates as:

$X Y_{2} (g) ⇌ X Y (g) + Y (g)$

Initial pressure of $X Y_{2}$ is $600 m m$ $H g$ . The total pressure at equilibrium is $800 m m$ $H g$ .

Let $x m m$ $H g$ of $X Y_{2}$ dissociates to reach equilibrium. $x m m$ $H g$ of $X Y$ and $x m m$ $H g$ of $Y$ will be formed at equilibrium. $600 - x m m$ $H g$ of $X Y_{2}$ will be present at equilibrium.

Total pressure $= (600 - x) + x + x = 600 + x m m$ $H g$

But total pressure $= 800 m m$ $H g$

$600 + x = 800$ or $x = 200$
Thus, the equilibrium pressures are

$P_{X Y_{2}} = 600 - x = 600 - 200 = 400 m m$ $H g$

$P_{X Y} = x = 200 m m$ $H g$

$P_{Y} = x = 200 m m$ $H g$

Assuming volume of system to remain constant, the value of $K_{p} = \frac{P_{X Y} P_{Y}}{P_{X Y_{2}}}$

$K_{p} = \frac{200 \times 200}{400}$

$K_{p} = 100 m m$ $H g$

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