What is the effect of increasing pressure on the dissociation of `PCl_5`, according to the equation `PCl_(5(g)) hArrPCl_(3(g)) + Cl_(2(g)) - x` cal

Options

A. Dissociation decreases
B. Dissociation increases
C. Dissociation does not change
D. None of these

Explanation

To determine the effect of increasing pressure on the dissociation of \( PCl_5 \) according to the reaction:

\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

we can follow these steps:

### Step 1: Identify the Reaction and Its Components
The reaction involves one mole of \( PCl_5 \) dissociating into one mole of \( PCl_3 \) and one mole of \( Cl_2 \).

### Step 2: Calculate the Change in Moles of Gas (Δn)
To understand the effect of pressure, we need to calculate the change in the number of moles of gas (Δn):
- Reactants: 1 mole of \( PCl_5 \)
- Products: 1 mole of \( PCl_3 \) + 1 mole of \( Cl_2 \) = 2 moles

Thus, the change in moles of gas is:
\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 1 = 1 \]

### Step 3: Apply Le Chatelier's Principle
According to Le Chatelier's Principle, if the pressure of a system at equilibrium is increased, the equilibrium will shift in the direction that reduces the number of moles of gas.

Since we calculated that \( \Delta n = 1 \) (which is greater than 0), increasing the pressure will shift the equilibrium to the left (towards the reactants).

### Step 4: Conclusion on the Effect of Increasing Pressure
As a result of the shift to the left, the dissociation of \( PCl_5 \) will decrease. Therefore, increasing the pressure will favor the formation of \( PCl_5 \) and reduce the amount of \( PCl_3 \) and \( Cl_2 \) in the system.

### Final Answer
Increasing the pressure on the dissociation of \( PCl_5 \) will decrease its dissociation.

---