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Derivation of Kp and Kc

The dissociat...

Question

The dissociation equilibrium of a gas ${A B}_{2}$ can be represented as:

$2 A B_{2} (g) ⇌ 2 A B (g) + B_{2} (g)$

The degree of dissociation is $x$ and is small compared to $1$ . The expression relating the degree of dissociation $x$ with equilibrium constant $K_{p}$ and total pressure $p$ is:

(2 K_{p} / p)

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{(2 K_{p} / p)}^{1 / 3}

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{(2 K_{p} / p)}^{1 / 2}

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(K_{p} / p)

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Solution

The correct option is A ${(2 K_{p} / p)}^{1 / 3}$
Let $P$ be the initial pressure of $A B_{2}$
The equilibrium pressures of $A B_{2},$ $A B$ and $B_{2}$ are $P (1 - x),$ $x P$ and $0.5 x P$ respectively.
The equilibrium constant expression is
$K_{p} = \frac{P_{A B}^{2} P_{B_{2}}}{P_{A B_{2}}^{2}}$
$K_{p} = \frac{[x P]^{2} \times 0.5 x P}{[P (1 - x)]^{2}}$
$K_{p} = P \frac{0.5 x^{3}}{[(1 - x)]^{2}}$
Since the degree of dissociation is small compared to $1, 1 - x$ can be approximated to 1.
$K_{p} = P \times 0.5 x^{3}$ .... (1)
The total pressure is

$P (1 - x) + P x + 0.5 P x = P (1 + 0.5 x) = p$

$P = \frac{p}{1 + 0.5 x}$ .....(2)
Substitute (2) in (1),
$K_{p} = \frac{p}{1 + 0.5 x} \times 0.5 x^{3}$
Since the degree of dissociation is small compared to $1, 1 + 0.5 x$ can be approximated to 1.
$K_{p} = p \times 0.5 x^{3}$
$x = {(2 K_{p} / p)}^{1 / 3}$
Hence, the expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure p is $x = {(2 K_{p} / p)}^{1 / 3}$

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