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Common Ion Effect

2 mol of SO 2...

Question

2 mol of $S O_{2}$ and I mol of $O_{2}$ are heated in a closed vessel to reach the equilibrium:
$2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)}$ . The equilibrium mixture exerted a pressure of 5 atm and required $\frac{1}{3}$ mol of $K_{2} C r_{2} O_{7}$ in acidic medium. $K_{P}$ for the reaction is

A

2.0

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B

0.5

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C

1.0

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D

none of these

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Solution

The correct option is C
1.0

$S O_{2} \to S O_{3}$ , Eq.mass=molar mass/2
$C r_{2} O_{7}^{2} \to 2 C r^{3 +}$ , Eq. mass = molar mass/6
$\frac{1}{3}$ mol of $K_{2} C r_{2} O_{7} = \frac{6}{3} = 2$ equiv.of $K_{2} C r_{2} O_{7}$
= 2 equiv. of $S O_{2} = 1$ mol of $S O_{2}$
$\begin{matrix} 2 S O_{2} & + & O_{2 (g)} & ⇌ & 2 S O_{3 (g)} 2 m o l & 1 m o l & 0 m o l 1 m o l & 0.5 m o l & 1 m o l \end{matrix}$

$P_{S O_{2}} = \frac{1}{2.5} \times 5 = 2 a t m, P_{O_{2}} = \frac{0.5 \times 5}{2.5} = 1 a t m, P_{S O_{3}} = \frac{1}{2} .5 \times 5 = 2 a t m K_{p} = \frac{P_{S O_{3}^{2}}}{P_{S O_{3}^{2}} \times P_{o_{2}}} = \frac{2^{2}}{2^{2} \times 1} = 1.0$

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Similar questions

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