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 solve the problem, we need to determine the equilibrium constant \( K \) for the reaction:

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

### Step-by-Step Solution:

1. **Initial Conditions**:
At time \( t = 0 \), we have:
- \( [CO_2] = 1 \) mole
- \( [CO] = 0 \) moles
- \( [O_2] = 0 \) moles

2. **Degree of Dissociation**:
Let \( \alpha \) be the degree of dissociation of \( CO_2 \). The moles of \( CO_2 \) that dissociate will be \( \alpha \) moles. Therefore, at equilibrium:
- Moles of \( CO_2 \) remaining = \( 1 - \alpha \)
- Moles of \( CO \) formed = \( \alpha \)
- Moles of \( O_2 \) formed = \( \frac{1}{2} \alpha \)

3. **Total Moles at Equilibrium**:
The total moles at equilibrium will be:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \frac{1}{2} \alpha = 1 + \frac{1}{2} \alpha \]

4. **Equilibrium Concentrations**:
Since we assume the volume is 1 L, the equilibrium concentrations are:
- \( [CO_2] = 1 - \alpha \)
- \( [CO] = \alpha \)
- \( [O_2] = \frac{1}{2} \alpha \)

5. **Expression for the Equilibrium Constant \( K \)**:
The equilibrium constant \( K \) is given by:
\[ K = \frac{[CO] \cdot [O_2]^{1/2}}{[CO_2]} \]
Substituting the equilibrium concentrations:
\[ K = \frac{\alpha \cdot \left(\frac{1}{2} \alpha\right)^{1/2}}{1 - \alpha} \]

6. **Simplifying the Expression**:
Simplifying the expression:
\[ K = \frac{\alpha \cdot \frac{1}{\sqrt{2}} \cdot \sqrt{\alpha}}{1 - \alpha} = \frac{\alpha^{3/2}}{\sqrt{2}(1 - \alpha)} \]

7. **Assuming \( \alpha \) is Very Small**:
Given that \( \alpha \) is very small (\( \alpha \ll 1 \)), we can approximate \( 1 - \alpha \approx 1 \):
\[ K \approx \frac{\alpha^{3/2}}{\sqrt{2}} \]

### Final Result:
Thus, the equilibrium constant \( K \) is approximately:
\[ K \approx \frac{\alpha^{3/2}}{\sqrt{2}} \]