Thermal decomposition of gaseous `X_(2)` to gaseous `X` at `298K` takes place according to the following equation:
`X(g)hArr2X(g)`
The standard reaction Gibbs energy `Delta_(r)G^(@)`, of this reaction is positive. At the start of the reaction, there is one mole of `X_(2)` and no `X`. As the reaction proceeds, the number of moles of `X` formed is given by `beta`. Thus `beta_("equilibrium")` is the number of moles of `X` formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally.
[Given, `R=0.083L` bar `K^(-1) mol^(-1)`)
The equilibrium constant `K_(p)` for this reaction at `298K`, in terms of `beta_("equilibrium")` is

Options

A. `(8(beta^(2))_("equilibrium"))/(2-beta_("equilibrium"))`
B. `(8(beta^(2))_("equilibrium"))/(4-beta_("equilibrium")^(2))`
C. `(4(beta^(2))_(equilibrium))/(2-beta_("equilibrium"))`
D. `(4(beta^(2))_("equilibrium"))/(4-beta^(2)_("equilibrium"))`

Explanation

To solve the problem step by step, we need to analyze the reaction and the conditions given.

### Step 1: Write the reaction and initial conditions
The reaction is given as:
\[ X_2(g) \rightleftharpoons 2X(g) \]

At the start of the reaction:
- Moles of \( X_2 = 1 \)
- Moles of \( X = 0 \)

### Step 2: Define the change in moles
Let \( \beta \) be the number of moles of \( X \) formed at equilibrium. Therefore, at equilibrium:
- Moles of \( X_2 = 1 - \frac{\beta}{2} \) (since 1 mole of \( X_2 \) produces 2 moles of \( X \))
- Moles of \( X = \beta \)

### Step 3: Calculate total moles at equilibrium
The total number of moles at equilibrium is:
\[ n_{\text{total}} = \left(1 - \frac{\beta}{2}\right) + \beta = 1 + \frac{\beta}{2} \]

### Step 4: Calculate mole fractions
The mole fraction of \( X \) and \( X_2 \) can be calculated as follows:
- Mole fraction of \( X \):
\[ Y_X = \frac{\beta}{n_{\text{total}}} = \frac{\beta}{1 + \frac{\beta}{2}} \]

- Mole fraction of \( X_2 \):
\[ Y_{X_2} = \frac{1 - \frac{\beta}{2}}{n_{\text{total}}} = \frac{1 - \frac{\beta}{2}}{1 + \frac{\beta}{2}} \]

### Step 5: Calculate partial pressures
Using the total pressure \( P = 2 \, \text{bar} \):
- Partial pressure of \( X \):
\[ P_X = Y_X \cdot P = \frac{\beta}{1 + \frac{\beta}{2}} \cdot 2 \]

- Partial pressure of \( X_2 \):
\[ P_{X_2} = Y_{X_2} \cdot P = \frac{1 - \frac{\beta}{2}}{1 + \frac{\beta}{2}} \cdot 2 \]

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_X)^2}{P_{X_2}} \]

Substituting the expressions for \( P_X \) and \( P_{X_2} \):
\[ K_p = \frac{\left(\frac{2\beta}{1 + \frac{\beta}{2}}\right)^2}{\frac{2(1 - \frac{\beta}{2})}{1 + \frac{\beta}{2}}} \]

### Step 7: Simplify the expression
1. The numerator becomes:
\[ \left(\frac{2\beta}{1 + \frac{\beta}{2}}\right)^2 = \frac{4\beta^2}{\left(1 + \frac{\beta}{2}\right)^2} \]

2. The denominator simplifies to:
\[ \frac{2(1 - \frac{\beta}{2})}{1 + \frac{\beta}{2}} \]

Thus, we have:
\[ K_p = \frac{4\beta^2}{\left(1 + \frac{\beta}{2}\right)^2} \cdot \frac{1 + \frac{\beta}{2}}{2(1 - \frac{\beta}{2})} \]

3. Simplifying further:
\[ K_p = \frac{4\beta^2}{2(1 - \frac{\beta}{2}) \cdot (1 + \frac{\beta}{2})} = \frac{2\beta^2}{(1 - \frac{\beta}{2})(1 + \frac{\beta}{2})} \]

4. Recognizing that \( (1 - \frac{\beta}{2})(1 + \frac{\beta}{2}) = 1 - \left(\frac{\beta}{2}\right)^2 = 1 - \frac{\beta^2}{4} \), we can write:
\[ K_p = \frac{2\beta^2}{1 - \frac{\beta^2}{4}} \]

### Final Expression for \( K_p \)
Thus, the equilibrium constant \( K_p \) in terms of \( \beta \) is:
\[ K_p = \frac{8\beta^2}{4 - \beta^2} \]