Consider the fraction The equation representing correct relationship between the degree of dissociation of (g) with its equilibrium constant is _______ . Assume to be very small. [2025]

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Solution

Q.
Consider the fraction

$X_{2} Y (g) = X_{2} (g) + \frac{1}{2} Y_{2} (g)$

The equation representing correct relationship between the degree of dissociation $(x)$ of $X_{2} Y$ (g) with its equilibrium constant $Kp$ is _______ .

Assume $x$ to be very small. [2025]

1 $x = \sqrt[3]{\frac{2 K p^{2}}{p}}$

2 $x = \sqrt[3]{\frac{K p}{2 p}}$

3 $x = \sqrt[3]{\frac{K p}{p}}$

4 $x = \sqrt[3]{\frac{2 K p}{p}}$

Ans.
(1)

$\begin{matrix} X_{2} Y (g) & ⇌ & X_{2} (g) & + & \frac{1}{2} Y_{2} (g) \\ t=0 (moles) & a & 0 & 0 \\ t =∞ (moles) & a (1 - x) & a x & \frac{a x}{2} \\ t =∞ (mole fraction) & \frac{a (1 - x)}{a (1 + \frac{x}{2})} & \frac{a x}{a (1 + \frac{x}{2})} & \frac{a x}{2 a (1 + \frac{x}{2})} \\ = (\frac{1 - x}{1 + \frac{x}{2}}) & = (\frac{x}{1 + \frac{x}{2}}) & \frac{x}{2 (1 + \frac{x}{2})} \\ \approx 1 & \approx x & \approx \frac{x}{2} \\ Partial pressures & 1 \times P & x \times P & \frac{x \times P}{2} \end{matrix} \begin{matrix} Total = a (1 + \frac{x}{2}) \end{matrix}$

$K_{p} = \frac{P_{X_{2}} \times {(P_{Y_{2}})}^{1 / 2}}{P_{X_{2} Y}} = \frac{x P \times {(\frac{x P}{2})}^{1 / 2}}{P}$

$K_{p} = \frac{x^{3 / 2} P^{1 / 2}}{2^{1 / 2}}$

$K_{p}^{2} = \frac{x^{3} P}{2}$

$x = \sqrt[3]{\frac{2 K_{p}^{2}}{P}}$

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