1. Set up the equilibrium

For $\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)+\mathrm{C}(g)$ let initial moles of $A$ be $1$.

If the degree of dissociation is $\alpha$, then at equilibrium:

• moles of $A = 1-\alpha$
• moles of $B = \alpha$
• moles of $C = \alpha$

So total moles at equilibrium are $1-\alpha+\alpha+\alpha=1+\alpha$

2. Write partial pressures

If total pressure is $p$, then $p_A=\frac{1-\alpha}{1+\alpha}p, \quad p_B=\frac{\alpha}{1+\alpha}p, \quad p_C=\frac{\alpha}{1+\alpha}p$

3. Expression for $K_p$

For the reaction, $K_p=\frac{p_B p_C}{p_A}$

Substituting the partial pressures: $K_p=\frac{\left(\frac{\alpha p}{1+\alpha}\right)\left(\frac{\alpha p}{1+\alpha}\right)}{\frac{(1-\alpha)p}{1+\alpha}}$

Simplify: $K_p=\frac{\alpha^2 p}{1-\alpha^2}$

Thus, $K_p(1-\alpha^2)=\alpha^2 p$ $K_p=\alpha^2(p+K_p)$ $\alpha^2=\frac{K_p}{p+K_p}$

Hence, $\alpha=\sqrt{\frac{K_p}{p+K_p}}$

4. Analyze the options

Option A

If $K_p \gg p$, then $\alpha=\sqrt{\frac{K_p}{p+K_p}}\approx \sqrt{\frac{K_p}{K_p}}=1$ So $\alpha$ is close to unity, not much less than unity.

❌ A is false

Option B

As $p$ increases, from $\alpha=\sqrt{\frac{K_p}{p+K_p}}$ we see the denominator increases, so $\alpha$ decreases.

❌ B is false

Option C

If $p \gg K_p$, then $\alpha\approx \sqrt{\frac{K_p}{p}} \ll 1$ So $\alpha$ is much less than $1$, not approximately $1$.

❌ C is false

Option D

When $p$ increases, $\alpha$ decreases.

✅ D is true

5. Final answer

The correct option is: $\boxed{\text{D}}$

6. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.