Phosphorous pentachloride when heated in a sealed tube at `700 K` it undergoes decomposition as
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g), K_(p)=38` atm
Vapour density of the mixture is `74.25`.
If pressure is increased then the equilibrium will

Options

A. Be unaffected
B. Shift in backward direction
C. Shift in forward direction
D. Cannot be predicted

Explanation

To solve the problem, we will analyze the decomposition of phosphorus pentachloride (PCl5) and apply Le Chatelier's Principle to determine the effect of increased pressure on the equilibrium.

### Step-by-Step Solution:

1. **Write the Reaction:**
The decomposition of phosphorus pentachloride can be represented as:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

2. **Identify Moles of Gases:**
- On the left side (reactants), we have 1 mole of PCl5.
- On the right side (products), we have 1 mole of PCl3 and 1 mole of Cl2, totaling 2 moles of gases.
- Therefore, the total number of moles of gas on the reactants side is 1, and on the products side is 2.

3. **Apply Le Chatelier's Principle:**
Le Chatelier's Principle states that if an external change is applied to a system at equilibrium, the system will adjust to counteract that change and restore a new equilibrium.
- In this case, an increase in pressure will favor the side of the reaction that has fewer moles of gas.

4. **Determine the Effect of Increased Pressure:**
- Since the reactants side (PCl5) has 1 mole of gas and the products side (PCl3 + Cl2) has 2 moles of gas, increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, which is the reactants side (PCl5).

5. **Conclusion:**
Therefore, if pressure is increased, the equilibrium will shift to the left (backward direction) towards the formation of PCl5.

### Final Answer:
The equilibrium will shift in the backward direction towards PCl5.

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