For the equilibrium `2NOBr(g) hArr2NO + Br_(2)(g)` , calculate the ratio `(K_(p))/(P)`, where P is the total pressure and `P_(Br_(2)) = (P)/(9)` at a certain temperature

Options

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

Explanation

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

\[ 2NOBr(g) \rightleftharpoons 2NO(g) + Br_2(g) \]

Given that the partial pressure of \( Br_2 \) is \( \frac{P}{9} \) at a certain temperature, we can proceed with the following steps:

### Step 1: Define the pressures in terms of total pressure \( P \)

From the information given, we know:
- The partial pressure of \( Br_2 \) is \( P_{Br_2} = \frac{P}{9} \).
- Let the partial pressure of \( NO \) be \( P_{NO} \) and the partial pressure of \( NOBr \) be \( P_{NOBr} \).

### Step 2: Calculate the partial pressures of \( NO \) and \( NOBr \)

Since the stoichiometry of the reaction shows that 2 moles of \( NOBr \) produce 2 moles of \( NO \) and 1 mole of \( Br_2 \), we can express the partial pressure of \( NO \) in terms of \( P \).

Let’s denote the partial pressure of \( NO \) as \( P_{NO} \). Since 2 moles of \( NO \) are produced for every 2 moles of \( NOBr \), we can say:

\[ P_{NO} = 2 \times P_{Br_2} = 2 \times \frac{P}{9} = \frac{2P}{9} \]

### Step 3: Calculate the total pressure \( P \)

The total pressure \( P \) can be expressed as the sum of the partial pressures of all components at equilibrium:

\[ P = P_{NOBr} + P_{NO} + P_{Br_2} \]

We know \( P_{Br_2} = \frac{P}{9} \) and \( P_{NO} = \frac{2P}{9} \). Thus, we can find \( P_{NOBr} \):

\[ P_{NOBr} = P - P_{NO} - P_{Br_2} = P - \frac{2P}{9} - \frac{P}{9} = P - \frac{3P}{9} = P - \frac{P}{3} = \frac{2P}{3} \]

### Step 4: Write the expression for \( K_p \)

The equilibrium constant \( K_p \) is given by the expression:

\[ K_p = \frac{(P_{NO})^2 (P_{Br_2})}{(P_{NOBr})^2} \]

Substituting the values we found:

\[ K_p = \frac{\left(\frac{2P}{9}\right)^2 \left(\frac{P}{9}\right)}{\left(\frac{2P}{3}\right)^2} \]

### Step 5: Simplify the expression for \( K_p \)

Calculating the numerator:

\[ \left(\frac{2P}{9}\right)^2 = \frac{4P^2}{81} \]
\[ K_p = \frac{\frac{4P^2}{81} \cdot \frac{P}{9}}{\left(\frac{2P}{3}\right)^2} = \frac{\frac{4P^3}{729}}{\frac{4P^2}{9}} = \frac{4P^3}{729} \cdot \frac{9}{4P^2} = \frac{9P}{729} = \frac{P}{81} \]

### Step 6: Calculate the ratio \( \frac{K_p}{P} \)

Now we can find the ratio:

\[ \frac{K_p}{P} = \frac{\frac{P}{81}}{P} = \frac{1}{81} \]

Thus, the final answer is:

\[ \frac{K_p}{P} = \frac{1}{81} \]