In a reaction `PCI_(5)hArrPCI_(3) + CI_(2)` degree of dissociation is `30%` . If initial moles of `PCI_(3)` is one then total moles at equilibrium is

Options

A. `1.3`
B. `0.7`
C. `1.6`
D. `1.0`

Explanation

To solve the problem step-by-step, we will analyze the dissociation of phosphorus pentachloride (PCl5) into phosphorus trichloride (PCl3) and chlorine (Cl2) at equilibrium.

### Step 1: Write the balanced chemical equation
The reaction can be represented as:
\[ \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \]

### Step 2: Define the initial conditions
We are given that the initial moles of PCl5 is 1. Therefore, we can set up an initial table for the moles of each species:

| Species | Initial Moles |
|-----------|----------------|
| PCl5 | 1 |
| PCl3 | 0 |
| Cl2 | 0 |

### Step 3: Define the degree of dissociation
The degree of dissociation (α) is given as 30%, which can be expressed as:
\[ \alpha = \frac{30}{100} = 0.3 \]

### Step 4: Calculate the change in moles at equilibrium
Since PCl5 dissociates to form PCl3 and Cl2, the change in moles can be represented as follows:
- For every mole of PCl5 that dissociates, 1 mole of PCl3 and 1 mole of Cl2 are produced.
- Therefore, if α = 0.3, then:
- Moles of PCl5 that dissociate = \( \alpha \times \text{initial moles of PCl5} = 0.3 \times 1 = 0.3 \)

Now we can calculate the moles at equilibrium:
- Moles of PCl5 at equilibrium = Initial moles - Moles dissociated = \( 1 - 0.3 = 0.7 \)
- Moles of PCl3 at equilibrium = Moles produced = \( 0 + 0.3 = 0.3 \)
- Moles of Cl2 at equilibrium = Moles produced = \( 0 + 0.3 = 0.3 \)

### Step 5: Calculate total moles at equilibrium
Now, we can sum the moles of all species at equilibrium:
\[ \text{Total moles at equilibrium} = \text{Moles of PCl5} + \text{Moles of PCl3} + \text{Moles of Cl2} \]
\[ = 0.7 + 0.3 + 0.3 = 1.3 \]

### Final Answer
The total moles at equilibrium is **1.3 moles**.

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