JEE Main · 2025 · Shift-IhardCEQ-062

At temperature T, compound AB2(g) dissociates as: AB2(g) <= AB(g) + 12 B2(g) having degree of dissociation x (small…

Chemical Equilibrium · Class 11 · JEE Main Previous Year Question

Question

At temperature T, compound $\ce{AB_{2(g)}}$ dissociates as:

$\ce{AB_{2(g)} <=> AB_{(g)} + \frac{1}{2} B_{2(g)}}$

having degree of dissociation $x$ (small compared to unity). The correct expression for $x$ in terms of $K_p$ and $p$ is:

Options

a
$\left(\dfrac{2K_p}{p}\right)^{1/3}$
b
$\left(\dfrac{2K_p}{p}\right)^{1/4}$
c
$\left(\dfrac{2K_p^2}{p}\right)^{1/3}$
✓
d
$\sqrt{K_p}$

Correct Answerc

$\left(\dfrac{2K_p^2}{p}\right)^{1/3}$

Detailed Solution

Step 1 — ICE table (with $x \ll 1$ )

| | $\ce{AB2}$ | $\ce{AB}$ | $\ce{\frac{1}{2}B2}$ | |---|---|---|---| | Eq moles | $1-x \approx 1$ | $x$ | $x/2$ | | Total | $\approx 1$ | | |

Step 2 — Partial pressures

$P_{\ce{AB2}} \approx p$ ; $P_{\ce{AB}} \approx xp$ ; $P_{\ce{B2}} \approx \frac{xp}{2}$

Step 3 — $K_p$ expression

$K_p = \frac{P_{\ce{AB}} \cdot P_{\ce{B2}}^{1/2}}{P_{\ce{AB2}}} = \frac{(xp) \cdot \left(\frac{xp}{2}\right)^{1/2}}{p} = \frac{x^{3/2} p^{1/2}}{\sqrt{2}}$

Step 4 — Solve for x

$K_p^2 = \frac{x^3 p}{2}$ → $x^3 = \frac{2K_p^2}{p}$ → $x = \left(\frac{2K_p^2}{p}\right)^{1/3}$ → Option (c)

Key Points to Remember:

Use $x \ll 1$ : total moles $\approx 1$ , $P_{\ce{AB2}} \approx p$
$K_p = \frac{x^{3/2} p^{1/2}}{\sqrt{2}}$ → square both sides → $x = \left(\frac{2K_p^2}{p}\right)^{1/3}$

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