Consider the equilibrium `CO_2(g) hArrCO(g)+1/2O_2(g)` The equilibrium constant K is given by (when `a lt lt lt 1` )

Options

A. `K = (alpha^(3//2))/sqrt2`
B. `K=alpha^3/2`
C. `K=(alpha^3//2)/2`
D. `K=(alpha^(3//2))/sqrt3`

Answer

To find the equilibrium constant \( K \) for the reaction

\[ CO_2(g) \rightleftharpoons CO(g) + \frac{1}{2} O_2(g) \]

given that \( \alpha \) (the degree of dissociation) is very small (i.e., \( \alpha \ll 1 \)), we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_p \)

The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the products and reactants:

\[ K_p = \frac{(P_{CO}) (P_{O_2})^{1/2}}{P_{CO_2}} \]

### Step 2: Set up the initial conditions

Assume we start with 1 mole of \( CO_2 \) and no \( CO \) or \( O_2 \):

- Initial moles:
- \( CO_2 = 1 \)
- \( CO = 0 \)
- \( O_2 = 0 \)

### Step 3: Define the change in moles at equilibrium

Let \( \alpha \) be the degree of dissociation of \( CO_2 \). At equilibrium, the changes in moles will be:

- Moles of \( CO_2 \) at equilibrium: \( 1 - \alpha \)
- Moles of \( CO \) at equilibrium: \( \alpha \)
- Moles of \( O_2 \) at equilibrium: \( \frac{1}{2} \alpha \)

### Step 4: Calculate the total moles at equilibrium

The total number of moles at equilibrium will be:

\[ \text{Total moles} = (1 - \alpha) + \alpha + \frac{1}{2} \alpha = 1 + \frac{1}{2} \alpha \]

### Step 5: Calculate the partial pressures

The partial pressures can be expressed in terms of the total pressure \( P \):

- \( P_{CO_2} = P \cdot \frac{1 - \alpha}{1 + \frac{1}{2} \alpha} \)
- \( P_{CO} = P \cdot \frac{\alpha}{1 + \frac{1}{2} \alpha} \)
- \( P_{O_2} = P \cdot \frac{\frac{1}{2} \alpha}{1 + \frac{1}{2} \alpha} \)

### Step 6: Substitute the partial pressures into the \( K_p \) expression

Substituting these into the expression for \( K_p \):

\[ K_p = \frac{\left(P \cdot \frac{\alpha}{1 + \frac{1}{2} \alpha}\right) \left(P \cdot \frac{\frac{1}{2} \alpha}{1 + \frac{1}{2} \alpha}\right)^{1/2}}{P \cdot \frac{1 - \alpha}{1 + \frac{1}{2} \alpha}} \]

### Step 7: Simplify the expression

This simplifies to:

\[ K_p = \frac{\left(\frac{\alpha}{1 + \frac{1}{2} \alpha}\right) \left(\frac{\frac{1}{2} \alpha}{1 + \frac{1}{2} \alpha}\right)^{1/2}}{\frac{1 - \alpha}{1 + \frac{1}{2} \alpha}} \]

### Step 8: Approximate for small \( \alpha \)

Since \( \alpha \) is very small, we can approximate:

- \( 1 - \alpha \approx 1 \)
- \( 1 + \frac{1}{2} \alpha \approx 1 \)

Thus, the expression simplifies to:

\[ K_p \approx \frac{\alpha \cdot \left(\frac{1}{2} \alpha\right)^{1/2}}{1} = \frac{\alpha \cdot \frac{1}{\sqrt{2}} \cdot \sqrt{\alpha}}{1} = \frac{\alpha^{3/2}}{\sqrt{2}} \]

### Final Result

Thus, the equilibrium constant \( K_p \) is approximately:

\[ K_p \approx \frac{\alpha^{3/2}}{\sqrt{2}} \]