For the reaction : `AhArr nB`, if 'a' mole of A is taken initially and 'x' moles of A get dissociated at equililbrium. What is the value of degree of dissociation?

Options

A. `((x)/(a^(n)))`
B. `ax`
C. `((a)/(x))^(n)`
D. `((x)/(a))`

Answer

To find the degree of dissociation for the reaction \( A \rightleftharpoons nB \), we can follow these steps:

### Step 1: Understand the Reaction
The reaction indicates that one mole of A dissociates to form n moles of B.

### Step 2: Initial Moles of A
We start with 'a' moles of A. This is our initial amount before any dissociation occurs.

### Step 3: Moles Dissociated at Equilibrium
Let 'x' be the number of moles of A that dissociate at equilibrium.

### Step 4: Moles of A at Equilibrium
At equilibrium, the number of moles of A remaining will be:
\[ \text{Moles of A at equilibrium} = a - x \]

### Step 5: Moles of B at Equilibrium
According to the stoichiometry of the reaction, for every mole of A that dissociates, n moles of B are formed. Therefore, the number of moles of B formed will be:
\[ \text{Moles of B at equilibrium} = n \cdot x \]

### Step 6: Define Degree of Dissociation
The degree of dissociation (\( \alpha \)) is defined as the fraction of the initial amount of reactant that has dissociated. It can be expressed mathematically as:
\[ \alpha = \frac{\text{Moles of A dissociated}}{\text{Initial moles of A}} = \frac{x}{a} \]

### Step 7: Final Expression
Thus, the degree of dissociation (\( \alpha \)) for the reaction is:
\[ \alpha = \frac{x}{a} \]

### Conclusion
The degree of dissociation is given by the ratio of the moles of A that dissociate (x) to the initial moles of A (a).

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