The equilibrium constant for the reaction 2SO2 + O2 ⇋&#x

Discussion Forum : Chemical Equilibrium

Question -

The equilibrium constant for the reaction $2 S O_{2} + O_{2} ⇋ 2 S O_{3} i s 900 a t m^{- 1} a t 800 K .$ A mixture containing $S O_{3} a n d O_{2}$ having initial pressures of $1 a t m a n d 2 a t m$ is heated to equilibrate. The partial pressure of $O_{2}$ at equilibrium will be

Options:

A .

0.97 a t m

B .

0.012 a t m

C .

2.01 a t m

D .

0.024 a t m

Answer: Option C
:
C
Correct option is c). Considering the backward reaction we have,

\begin{matrix} 2 S O_{3} ⇋ & 2 S O_{2} + & O_{2} & K_{P} = 1 / 900 a t m I n i t i a l p r e s s u r e & 1 a t m & 0 & 2 a t m P r e s s u r e a t e q u i l i b r i u m & 1 - x & x & 2 + x / 2 \end{matrix}

Let partial pressure of $S O_{2} = p_{1}$ and partial pressure of $O_{2} = p_{2}$ and partial pressure of $S O_{3} = p_{3}$
$K <_{p} = (p_{1})^{2} \times (p_{2}) / (p_{3})^{2} = x^{2} (2 + x / 2) / (1 - x)^{2} = 1 / 900$
$A s K_{p}$ for this reaction is very small, $x << 1.$
Taking $2 + x / 2$ as x and $(1 - x)$ as x and solving for x, we get
$X = 0.0236$
Hence, at equilibrium, partial pressure of $S O_{3} = 1 - X = 0.9764 a t m,$
Partial pressure of $S O_{2} = x = 0.0236 a t m$
And Partial pressure of $O_{2} = 2 + x / 2 = 2.0118 a t m$

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