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 concept of degree of dissociation
The degree of dissociation (\( \alpha \)) is defined as the fraction of the initial amount of a substance that has dissociated at equilibrium. It is expressed mathematically as:
\[ \alpha = \frac{\text{Dissociated moles}}{\text{Initial moles}} \]

### Step 2: Identify initial moles and dissociated moles
In this case, we are given:
- Initial moles of A = \( a \)
- Moles of A that get dissociated at equilibrium = \( x \)

### Step 3: Substitute the values into the formula
Using the formula for degree of dissociation, we can substitute the values we identified:
\[ \alpha = \frac{x}{a} \]

### Step 4: Conclusion
Thus, the degree of dissociation (\( \alpha \)) for the reaction \( A \rightleftharpoons nB \) is:
\[ \alpha = \frac{x}{a} \]