`2NOBr(g)hArr2NO(g)+Br2(g).` If nitrosyl bromide (NOBr) `40%` dissociated at certain temp. and a total pressure of `0.30` atm `K_(p)` for the reaction `2NO(g)+Br_(2)(g)hArr2NOBr(g)` is

Options

A. `45`
B. `25`
C. `0.022`
D. `0.25`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the Reaction and Initial Conditions
The reaction given is:
\[ 2 \text{NOBr} (g) \rightleftharpoons 2 \text{NO} (g) + \text{Br}_2 (g) \]

At the start (t = 0), we assume we have 1 mole of NOBr:
- Moles of NOBr = 1 mole
- Moles of NO = 0
- Moles of Br2 = 0

### Step 2: Determine the Change in Moles Due to Dissociation
Given that 40% of NOBr dissociates:
- Amount dissociated = \( 0.4 \times 1 \) mole = 0.4 moles
- Moles of NOBr at equilibrium = \( 1 - 0.4 = 0.6 \) moles
- Moles of NO produced = \( 2 \times 0.4 = 0.8 \) moles (since 2 moles of NO are produced for every 2 moles of NOBr that dissociate)
- Moles of Br2 produced = \( 0.4 \) moles (since 1 mole of Br2 is produced for every 2 moles of NOBr that dissociate)

### Step 3: Calculate Total Moles at Equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = \text{Moles of NOBr} + \text{Moles of NO} + \text{Moles of Br2} = 0.6 + 0.8 + 0.4 = 1.8 \text{ moles} \]

### Step 4: Calculate Partial Pressures
Given the total pressure at equilibrium is 0.30 atm, we can calculate the partial pressures using mole fractions.

- Mole fraction of NOBr:
\[ \text{Mole fraction of NOBr} = \frac{0.6}{1.8} = \frac{1}{3} \]
- Mole fraction of NO:
\[ \text{Mole fraction of NO} = \frac{0.8}{1.8} = \frac{4}{9} \]
- Mole fraction of Br2:
\[ \text{Mole fraction of Br2} = \frac{0.4}{1.8} = \frac{2}{9} \]

Now, we can calculate the partial pressures:
- Partial pressure of NOBr:
\[ P_{\text{NOBr}} = \text{Mole fraction of NOBr} \times \text{Total Pressure} = \frac{1}{3} \times 0.30 = 0.10 \text{ atm} \]
- Partial pressure of NO:
\[ P_{\text{NO}} = \text{Mole fraction of NO} \times \text{Total Pressure} = \frac{4}{9} \times 0.30 = \frac{1.2}{9} \approx 0.1333 \text{ atm} \]
- Partial pressure of Br2:
\[ P_{\text{Br2}} = \text{Mole fraction of Br2} \times \text{Total Pressure} = \frac{2}{9} \times 0.30 = \frac{0.6}{9} \approx 0.0667 \text{ atm} \]

### Step 5: Write the Expression for Kp
The equilibrium constant \( K_p \) for the reaction \( 2 \text{NO} + \text{Br}_2 \rightleftharpoons 2 \text{NOBr} \) is given by:
\[ K_p = \frac{(P_{\text{NO}})^2 \cdot (P_{\text{Br2}})}{(P_{\text{NOBr}})^2} \]

### Step 6: Substitute the Values into the Kp Expression
Substituting the partial pressures:
\[ K_p = \frac{(0.1333)^2 \cdot (0.0667)}{(0.10)^2} \]
Calculating:
\[ K_p = \frac{(0.01778) \cdot (0.0667)}{0.01} = \frac{0.001185}{0.01} = 0.1185 \]

### Step 7: Find Kp for the Reverse Reaction
Since we need \( K_p \) for the reverse reaction:
\[ K_p' = \frac{1}{K_p} = \frac{1}{0.1185} \approx 8.44 \]

### Final Answer
Thus, the value of \( K_p \) for the reaction \( 2 \text{NO} + \text{Br}_2 \rightleftharpoons 2 \text{NOBr} \) is approximately **8.44**.

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