The values of Kp1and KP2 for the reactions X ⇋ Y +&

Discussion Forum : Chemical Equilibrium

Question -

The values of $K_{p} 1$ and $K_{P} 2$ for the reactions
$X ⇋ Y + Z a n d A ⇋ 2 B$ Are in the ratio of $9 : 1.$ If the degree of dissociation of $X a n d A$ be equal, then total pressure at equilibrium $(1) a n d (2)$ are in the ration

Options:

A . 3:1

B . 1:9

C . 36:1

D . 1:1

Answer: Option C
:
C
Suppose total pressure at equilibrium for reactions 1 and 2 are

P_{1} a n d P_{2}

respectively

\begin{matrix} X ⇋ & Y + & Z I n i t i a l m o l e s & 1 & 0 & 0 A t e q u i l i b r i u m & 1 - x & x & x \end{matrix}

T o t a l m o l e s = 1 + α

Partial pressure of

X = p_{x} = \frac{(1 - x) X P 1}{1 + x}

P_{y} <= \frac{(x) X P 1}{1 + x} = P_{z}

(K_{P 1} = P_{Y} P_{Z} / P_{x} = x^{2} P_{1} / (1 - x^{2}) \approx x^{2} P_{1}

\begin{matrix} A ⇋ & 2 B h l i n e I n i t i a l m o l e s & 1 & 0 A t e q u i l i b r i u m & 1 - x & 2 x \end{matrix}

T o t a l m o l e s = 1 + α

P_{A} = \frac{(1 - x) X P 2}{1 + x} P_{B} = \frac{(2 x) X P 2}{1 + x}

K_{P 2} = P_{B}^{2} / P_{A} = 4 x^{2} P_{2} / (1 - x^{2}) \approx 4 x^{2} P_{2}

K_{P} 1 / K_{P} 2 = x^{2} P_{1} / 4 x^{2} P_{2} = 9 / 1

P_{1} / P_{2} = 36 / 1

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