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Step-by-Step Detailed Solution
Step 1: Identify the Reaction and Given Data
The reaction under consideration is:
$ \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 $.
We have the following initial data:
Moles of Ar (inert) introduced: 4.0 mol
Moles of PCl5 introduced: 5.0 mol
Volume of the container, $V = 100\,\text{L}$
Temperature, $T = 610\,\text{K}$
Total equilibrium pressure, $P_{\text{total}} = 6.0\,\text{atm}$
Gas constant, $R = 0.082\,\text{L atm K}^{-1}\text{mol}^{-1}$
Step 2: Define the Extent of Reaction
Let $x$ be the number of moles of PCl5 that decompose at equilibrium. Therefore:
• Moles of PCl5 left at equilibrium = $5.0 - x$
• Moles of PCl3 formed at equilibrium = $x$
• Moles of Cl2 formed at equilibrium = $x$
• Moles of Ar remain 4.0 (since it is inert).
Hence, total moles at equilibrium = $(5.0 - x) + x + x + 4.0 = 9.0 + x$.
Step 3: Apply the Ideal Gas Law to Find $x$
The total pressure $P_{\text{total}}$ is given by the ideal gas law expression:
$ P_{\text{total}} = \dfrac{n_{\text{total}} \, R \, T}{V} ,
\quad \text{where } n_{\text{total}} = 9.0 + x.$
Substituting the values:
$ 6.0 = \dfrac{(9.0 + x) \times 0.082 \times 610}{100} .
$
Simplifying the right-hand side:
$6.0 = \dfrac{(9.0 + x) \times (0.082 \times 610)}{100}
= \dfrac{(9.0 + x) \times 50.02}{100}
= 0.5002 \,(9.0 + x).$
Hence,
$ 9.0 + x = \dfrac{6.0}{0.5002} \approx 12.0 \quad \Rightarrow \quad x \approx 3.0. $
Step 4: Calculate Equilibrium Moles
With $x \approx 3.0$:
• Moles of PCl5 at equilibrium = $5.0 - 3.0 = 2.0$
• Moles of PCl3 at equilibrium = $3.0$
• Moles of Cl2 at equilibrium = $3.0$
• Moles of Ar = $4.0$
Total moles at equilibrium = $2.0 + 3.0 + 3.0 + 4.0 = 12.0$.
Step 5: Determine Partial Pressures
The total pressure is 6.0 atm, and total moles are 12. Each gas’s partial pressure is determined by its mole fraction times the total pressure:
$ P_i = \left(\dfrac{n_i}{n_{\text{total}}}\right) \times P_{\text{total}}. $
So:
• $P_{\text{PCl}_5} = \dfrac{2.0}{12.0} \times 6.0 = 1.0\,\text{atm}$
• $P_{\text{PCl}_3} = \dfrac{3.0}{12.0} \times 6.0 = 1.5\,\text{atm}$
• $P_{\text{Cl}_2} = \dfrac{3.0}{12.0} \times 6.0 = 1.5\,\text{atm}$
• $P_{\text{Ar}} = \dfrac{4.0}{12.0} \times 6.0 = 2.0\,\text{atm}$
Step 6: Calculate $K_p$
For the decomposition reaction
$ \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2, $
$ K_p = \dfrac{P_{\text{PCl}_3} \times P_{\text{Cl}_2}}{P_{\text{PCl}_5}}. $
Substituting the partial pressures:
$ K_p = \dfrac{(1.5)\times(1.5)}{1.0} = 2.25. $
Final Answer
The value of $K_p$ for the reaction is $2.25.$
