1. Write the equilibrium reaction

$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$

Argon is inert, so it does not participate in equilibrium.

2. Let degree of dissociation of $PCl_5$ be $x$ moles

Initially:

• $PCl_5 = 5$
• $PCl_3 = 0$
• $Cl_2 = 0$
• Ar $= 4$

At equilibrium:

• $PCl_5 = 5 - x$
• $PCl_3 = x$
• $Cl_2 = x$
• Ar $= 4$

So total moles at equilibrium are

$n_{\text{total}} = (5-x) + x + x + 4 = 9 + x$

3. Use total pressure relation

Given:

• $P_{\text{total}} = 6.0\,\text{atm}$
• $V = 100\,\text{L}$
• $T = 610\,\text{K}$
• $R = 0.082\,\text{L atm mol}^{-1}\text{K}^{-1}$

Using ideal gas equation:

$P_{\text{total}}V = n_{\text{total}}RT$

$6.0 \times 100 = (9+x)(0.082)(610)$

Now,

$0.082 \times 610 = 50.02$

So,

$600 = (9+x)(50.02)$

$9+x \approx \frac{600}{50.02} \approx 12$

Hence,

$x = 3$

4. Equilibrium moles

Thus,

• $PCl_5 = 5-3 = 2$
• $PCl_3 = 3$
• $Cl_2 = 3$
• Ar $= 4$

Total moles $= 12$

5. Find partial pressures

Partial pressure of each gas:

$p_i = \frac{n_i}{n_{\text{total}}} P_{\text{total}}$

Therefore,

$p_{PCl_5} = \frac{2}{12} \times 6 = 1\,\text{atm}$

$p_{PCl_3} = \frac{3}{12} \times 6 = 1.5\,\text{atm}$

$p_{Cl_2} = \frac{3}{12} \times 6 = 1.5\,\text{atm}$

6. Calculate $K_p$

For the reaction,

$K_p = \frac{p_{PCl_3} \cdot p_{Cl_2}}{p_{PCl_5}}$

Substitute values:

$K_p = \frac{(1.5)(1.5)}{1} = 2.25$

7. Match with options

$K_p = 2.25$

So the correct option is A.

8. Comparison with stored answer

Stored correct answer: A

Derived answer: A

They agree.