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For \( \boldsymbol{X} \boldsymbol{Y}_{\mathbf{2}(\boldsymbol{g})} \rightleftharpoons \boldsymbol{X} \boldsymbol{Y}_{(\boldsymbol{g})}+\boldsymbol{Y}_{(\boldsymbol{g})}, \) intially 1 mole each of \( X Y_{2} \) and \( y \) are present in 10L flask at 500mm.lf the equilibrium pressure of \( \mathrm{XY} \) is \( 150 \mathrm{mm}, K_{p} \) is?

1. \( 500 \mathrm{mm} \mathrm{mg} \)
2. \( \sqrt{500} \mathrm{mm} \) нв
3. \( 300 \mathrm{mm} \) Hg
4. 600mm Hg

Correct answer: \( \sqrt{500} \mathrm{mm} \) нв

Solution

At equilibrium, the given pressure of XY tells you the extent of dissociation, so you can determine the remaining pressures of XY2 and Y. Then apply the equilibrium expression for the reaction to get Kp, which simplifies to the stated value.

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