The equilibrium constant for the reaction 2SO2 + O2 ⇋

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 . 0.97atm

B . 0.012atm

C . 2.01atm

D . 0.024atm

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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