For `2NOBr(g) iff 2NO(g) + Br_2(g)` at equilibrium, `P_(Br_2) = p/q` and p is the total pressure, the ratio `(k_p)/p` will be

Options

A. `1/3`
B. `1/9`
C. `1/27`
D. `1/81`

Explanation

To solve the problem, we need to find the ratio \( \frac{K_p}{p} \) for the equilibrium reaction:

\[ 2 \text{NOBr}(g) \iff 2 \text{NO}(g) + \text{Br}_2(g) \]

Given that the partial pressure of \( \text{Br}_2 \) is \( \frac{p}{q} \) and \( p \) is the total pressure.

### Step 1: Define the pressures at equilibrium
Let:
- \( P_{\text{Br}_2} = \frac{p}{q} \) (given)
- Let \( P_{\text{NO}} = x \)

Since the stoichiometry of the reaction shows that for every 2 moles of \( \text{NOBr} \) that dissociate, 2 moles of \( \text{NO} \) and 1 mole of \( \text{Br}_2 \) are produced, we can express the pressures in terms of \( P_{\text{Br}_2} \).

### Step 2: Express \( P_{\text{NO}} \) in terms of \( P_{\text{Br}_2} \)
From the stoichiometry, we can derive:
- For every 1 mole of \( \text{Br}_2 \), there are 2 moles of \( \text{NO} \).
- Therefore, if \( P_{\text{Br}_2} = \frac{p}{q} \), then \( P_{\text{NO}} = 2 \times P_{\text{Br}_2} = 2 \times \frac{p}{q} = \frac{2p}{q} \).

### Step 3: Calculate the total pressure \( p \)
The total pressure \( p \) at equilibrium can be expressed as:
\[ p = P_{\text{NOBr}} + P_{\text{NO}} + P_{\text{Br}_2} \]
Assuming \( P_{\text{NOBr}} \) is \( y \), we have:
\[ p = y + \frac{2p}{q} + \frac{p}{q} \]
This simplifies to:
\[ p = y + \frac{3p}{q} \]
Rearranging gives:
\[ y = p - \frac{3p}{q} = p \left(1 - \frac{3}{q}\right) \]

### Step 4: Calculate \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{\text{NO}})^2 \cdot (P_{\text{Br}_2})}{(P_{\text{NOBr}})^2} \]
Substituting the values we have:
\[ K_p = \frac{\left(\frac{2p}{q}\right)^2 \cdot \left(\frac{p}{q}\right)}{(y)^2} \]
Substituting \( y \) from the previous step:
\[ K_p = \frac{\left(\frac{2p}{q}\right)^2 \cdot \left(\frac{p}{q}\right)}{\left(p \left(1 - \frac{3}{q}\right)\right)^2} \]

### Step 5: Simplify \( K_p \)
After simplification, we will find that:
\[ K_p = \frac{4p^3}{q^3 \cdot p^2 \left(1 - \frac{3}{q}\right)^2} = \frac{4p}{q^3 \left(1 - \frac{3}{q}\right)^2} \]

### Step 6: Find the ratio \( \frac{K_p}{p} \)
Now, we can find the ratio:
\[ \frac{K_p}{p} = \frac{4}{q^3 \left(1 - \frac{3}{q}\right)^2} \]

### Conclusion
Thus, the final ratio \( \frac{K_p}{p} \) is:
\[ \frac{K_p}{p} = \frac{4}{q^3 \left(1 - \frac{3}{q}\right)^2} \]