SATHEE: Chemical and Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 49

####49. The equilibrium constant $K_{p}$ of the reaction,

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$

is $900 a t m$ at $800 K$ . A mixture containing $S O_{3}$ and $O_{2}$ having initial pressure of 1 and $2 a t m$ respectively is heated at constant volume to equilibrate. Calculate the partial pressure of each gas at $800 K$ .

$(1989, 3 M)$

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Solution:

$\begin{array}{lccc} 2 S O_{2} (g) + O_{2} (g) & ⇌ S O_{3} (g) Initial p_{i} : & 0 & 2 & 1 Equilibrium p_{i} : 2 p & 2 + p & 1 - 2 p \end{array}$

$K_{p} = 900 = \frac{(1 - 2 p)^{2}}{(2 + p) (2 p)^{2}} [Ignoring p in comparison to 2]$

$p = \frac{1}{87} atm$

Partial pressure of $S O_{2} = 2 p = \frac{2}{87}$ atm

$Partial pressure of O_{2} = 2 + p = 2 + \frac{1}{87} = \frac{175}{87} a t m$

Partial pressure of $S O_{3} = 1 - 2 p = 1 - 2 (\frac{1}{87}) = \frac{85}{87} a t m$

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Chemical and Ionic Equilibrium - Result Question 49

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