Question

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Instructions:Select the UNE correct answer from the given options.
For XY2(8) = XY() + Y(g) initial pressure of XY, is 800mm and equilibrium pressure is 960mm. Degree
of dissociation of XY2(g) is:
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Answer 1

Final Answer : K_p = 100 mm Hg

Steps:

1) We have,

Since, Volume is constant .

XY_2(g) \rightarrow XY(g) + Y(g) \\ \\ \\t=0 => 600. \rightarrow 0 \quad + 0 \\ \\ t=t_{eq. } => 600-x \rightarrow x.\quad + x

2) Then at Equilibrium

P_T = P_{XY_2}+ P_{XY} + P_Y \\ \\ => P_T = (600-x) + x + x \\ \\ => 800 = 600+ x \\ \\ => x = 200 mm \:Hg

3) Now,

K_p = \frac{x*x}{(600-x) } \\ \\ K_p = \frac{200*200}{400} = 100 mm\: Hg

Hence, Value of K_p = 100 \: mm Hg

Answer 2

Answer:

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