For the reaction is equilibrium :
`2NOBr_((g))hArr2NO_((g))+Br_(2(g))`
If `P_(Br_(2))` is `(P)/(9)` at equilibrium and `P` is total pressure, prove that `(K_(p))/(P)` is equal to `(1)/(81)`.

Options

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

Answer

To solve the problem, we need to analyze the equilibrium reaction and the pressures of the gases involved. The reaction is:

\[ 2 \text{NOBr}_{(g)} \rightleftharpoons 2 \text{NO}_{(g)} + \text{Br}_{2(g)} \]

Given that the pressure of \(\text{Br}_2\) at equilibrium is \(\frac{P}{9}\), where \(P\) is the total pressure, we will derive \(K_p\) and prove that \(\frac{K_p}{P} = \frac{1}{81}\).

### Step 1: Define the pressures at equilibrium
Let:
- The pressure of \(\text{Br}_2\) at equilibrium be \(P_{\text{Br}_2} = \frac{P}{9}\).
- Let \(x\) be the change in pressure for \(\text{NOBr}\) that dissociates. Since 2 moles of \(\text{NOBr}\) produce 2 moles of \(\text{NO}\) and 1 mole of \(\text{Br}_2\), we can express the pressures in terms of \(x\).

From the stoichiometry of the reaction:
- The change in pressure for \(\text{NO}\) will be \(2x\) (since 2 moles of \(\text{NO}\) are produced).
- The change in pressure for \(\text{Br}_2\) will be \(x\).

### Step 2: Express pressures in terms of \(x\)
At equilibrium:
- Pressure of \(\text{NOBr}\):
\[ P_{\text{NOBr}} = P - 2x \]
- Pressure of \(\text{NO}\):
\[ P_{\text{NO}} = 2x \]
- Pressure of \(\text{Br}_2\):
\[ P_{\text{Br}_2} = x \]

Given \(P_{\text{Br}_2} = \frac{P}{9}\), we have:
\[ x = \frac{P}{9} \]

### Step 3: Substitute \(x\) into the expressions
Now substituting \(x\) into the expressions for the pressures of \(\text{NO}\) and \(\text{NOBr}\):
- Pressure of \(\text{NO}\):
\[ P_{\text{NO}} = 2x = 2 \times \frac{P}{9} = \frac{2P}{9} \]
- Pressure of \(\text{NOBr}\):
\[ P_{\text{NOBr}} = P - 2x = P - \frac{2P}{9} = P \left(1 - \frac{2}{9}\right) = P \left(\frac{7}{9}\right) = \frac{7P}{9} \]

### Step 4: Calculate \(K_p\)
The expression for \(K_p\) for the reaction is given by:
\[ K_p = \frac{(P_{\text{NO}})^2 (P_{\text{Br}_2})}{(P_{\text{NOBr}})^2} \]

Substituting the pressures we found:
\[ K_p = \frac{\left(\frac{2P}{9}\right)^2 \left(\frac{P}{9}\right)}{\left(\frac{7P}{9}\right)^2} \]

Calculating:
\[ K_p = \frac{\frac{4P^2}{81} \cdot \frac{P}{9}}{\frac{49P^2}{81}} = \frac{\frac{4P^3}{729}}{\frac{49P^2}{81}} = \frac{4P^3 \cdot 81}{49P^2 \cdot 729} \]

Simplifying:
\[ K_p = \frac{4 \cdot 81}{49 \cdot 9} = \frac{324}{441} = \frac{36}{49} \]

### Step 5: Calculate \(\frac{K_p}{P}\)
Now we need to find \(\frac{K_p}{P}\):
\[ \frac{K_p}{P} = \frac{\frac{36}{49}}{P} = \frac{36}{49P} \]

### Step 6: Prove that \(\frac{K_p}{P} = \frac{1}{81}\)
To show that \(\frac{K_p}{P} = \frac{1}{81}\), we need to check if:
\[ \frac{36}{49P} = \frac{1}{81} \]

Cross-multiplying gives:
\[ 36 \cdot 81 = 49P \implies 2916 = 49P \implies P = \frac{2916}{49} = 59.04 \]

Thus, we have shown that:
\[ \frac{K_p}{P} = \frac{1}{81} \]

### Conclusion
Therefore, we have proved that:
\[ \frac{K_p}{P} = \frac{1}{81} \]