In the dissociation of `2HI hArr H_(2)+I_(2)`, the degree of dissociation will be affected by

Options

A. Increase of temperature
B. Addition of an inert gas
C. Addition of `H_(2)` and `I_(2)`
D. Increase of pressure

Answer

To solve the problem regarding the dissociation of \(2HI \rightleftharpoons H_2 + I_2\) and how the degree of dissociation (\(\alpha\)) is affected, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Reaction**:
The reaction given is:
\[ 2HI \rightleftharpoons H_2 + I_2 \]
This means that 2 moles of hydrogen iodide (HI) dissociate to form 1 mole of hydrogen gas (H2) and 1 mole of iodine gas (I2).

2. **Setting Up Initial Concentrations**:
Let’s assume we start with \(C\) moles of HI. At the beginning (time \(t=0\)):
- Concentration of HI = \(C\)
- Concentration of H2 = 0
- Concentration of I2 = 0

3. **Change in Concentrations at Equilibrium**:
If \(\alpha\) is the degree of dissociation, then at equilibrium:
- Concentration of HI = \(C - C\alpha = C(1 - \alpha)\)
- Concentration of H2 = \(\frac{C\alpha}{2}\) (since 2 moles of HI produce 1 mole of H2)
- Concentration of I2 = \(\frac{C\alpha}{2}\) (similarly for I2)

4. **Equilibrium Constant Expression**:
The equilibrium constant \(K_c\) for the reaction can be expressed as:
\[ K_c = \frac{[H_2][I_2]}{[HI]^2} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{C\alpha}{2}\right)\left(\frac{C\alpha}{2}\right)}{(C(1 - \alpha))^2} \]
Simplifying this gives:
\[ K_c = \frac{\frac{C^2\alpha^2}{4}}{C^2(1 - \alpha)^2} = \frac{\alpha^2}{4(1 - \alpha)^2} \]

5. **Factors Affecting Degree of Dissociation**:
From the expression derived, we can see that \(\alpha\) depends on \(K_c\). The equilibrium constant \(K_c\) is affected by temperature. Therefore, the degree of dissociation will be affected by:
- **Temperature**: As temperature increases, \(K_c\) changes, which in turn affects \(\alpha\).

6. **Conclusion**:
The degree of dissociation (\(\alpha\)) in the reaction \(2HI \rightleftharpoons H_2 + I_2\) is primarily affected by temperature.