Consider the following chemical equilibrium of the gas phase reaction at a constant temperature: A(g) <= B(g) + C(g) If…

Step 1 — Derive $K_p$ expression

| | A | B | C | Total | |---|---|---|---|---| | Eq moles | $1-\alpha$ | $\alpha$ | $\alpha$ | $1+\alpha$ |

$P_A = \frac{(1-\alpha)p}{1+\alpha}$; $P_B = P_C = \frac{\alpha p}{1+\alpha}$

$K_p = \frac{P_B \cdot P_C}{P_A} = \frac{\alpha^2 p^2/(1+\alpha)^2}{(1-\alpha)p/(1+\alpha)} = \frac{\alpha^2 p}{1-\alpha^2}$

Step 2 — Effect of pressure

$K_p = \frac{\alpha^2 p}{1-\alpha^2}$ = constant at constant T.

If p increases → $\alpha$ must decrease to keep $K_p$ constant. ✓ (Le Chatelier: more pressure → backward reaction → fewer moles)

Step 3 — Check other options

(a) $p \gg K_p$: $\alpha^2 \approx K_p/p \ll 1$ → $\alpha \ll 1$ (not ≈ 1) ✗

(c) $K_p \gg p$: $\alpha^2 \approx K_p/p \gg 1$ → $\alpha \approx 1$ (not ≪ 1) ✗

Answer: When p increases, $\alpha$ decreases → Option (b)

Key Points to Remember:

• $K_p = \frac{\alpha^2 p}{1-\alpha^2}$ for $\ce{A <=> B + C}$
• Increasing p → $\alpha$ decreases (Le Chatelier)
• $p \gg K_p$ → $\alpha \ll 1$; $K_p \gg p$ → $\alpha \approx 1$