Consider thr reaction where `K_(p)=0.497 `at 500K
`PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)`
If the htree gasses are mixed in a right container so that the partial pressure of each gas in initially 1 atm ,then which is correct observation ?

Options

A. More `PCl_(5)` will be produced.
B. More `PCl_(3)` will be produced.
C. Equilibrium will be established when `50%` reaction is complete
D. None of the above

Answer

To solve the problem step by step, we will analyze the given reaction and the equilibrium constant, and then determine the direction of the reaction based on the initial conditions.

### Step 1: Write the Reaction and Given Data
The reaction given is:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

We are provided with:
- \( K_p = 0.497 \) at \( 500 \, K \)
- Initial partial pressures of each gas: \( P_{\text{PCl}_5} = 1 \, \text{atm}, P_{\text{PCl}_3} = 1 \, \text{atm}, P_{\text{Cl}_2} = 1 \, \text{atm} \)

### Step 2: Calculate the Reaction Quotient \( Q_p \)
The reaction quotient \( Q_p \) is calculated using the same expression as \( K_p \), but it is evaluated at the initial conditions:
\[ Q_p = \frac{P_{\text{PCl}_3} \cdot P_{\text{Cl}_2}}{P_{\text{PCl}_5}} \]
Substituting the initial pressures:
\[ Q_p = \frac{(1 \, \text{atm}) \cdot (1 \, \text{atm})}{1 \, \text{atm}} = 1 \]

### Step 3: Compare \( Q_p \) with \( K_p \)
Now we compare \( Q_p \) with \( K_p \):
- \( Q_p = 1 \)
- \( K_p = 0.497 \)

Since \( Q_p > K_p \), we conclude that the reaction will shift to the left (backward direction) to reach equilibrium.

### Step 4: Determine the Direction of the Reaction
When \( Q_p > K_p \), the equilibrium shifts toward the reactants. Therefore, more \( \text{PCl}_5 \) will be produced as the reaction proceeds in the backward direction.

### Conclusion
The correct observation is that the equilibrium will shift to the left, resulting in the production of more \( \text{PCl}_5 \).

### Final Answer
The correct observation is that more \( \text{PCl}_5 \) will be produced.

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