`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 find the equilibrium constant \( K_p \) for the reaction

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

given that nitrosyl bromide (NOBr) is 40% dissociated at a total pressure of 0.30 atm, we can follow these steps:

### Step 1: Determine Initial Moles
Assume we start with 2 moles of NOBr. Therefore, the initial amounts are:
- \([NOBr] = 2\)
- \([NO] = 0\)
- \([Br_2] = 0\)

### Step 2: Calculate Moles at Equilibrium
Since 40% of NOBr dissociates, we can calculate the amount that dissociates:
- Amount dissociated = \(0.4 \times 2 = 0.8\) moles

At equilibrium, the amounts will be:
- \([NOBr] = 2 - 0.8 = 1.2\) moles
- \([NO] = 0 + 0.4 \times 2 = 0.8\) moles (2 moles of NO are produced for every 2 moles of NOBr that dissociate)
- \([Br_2] = 0 + 0.4 = 0.4\) moles (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:
\[ \text{Total moles} = [NOBr] + [NO] + [Br_2] = 1.2 + 0.8 + 0.4 = 2.4 \text{ moles} \]

### Step 4: Calculate Mole Fractions
Now we can calculate the mole fractions of each component:
- Mole fraction of NOBr:
\[ \chi_{NOBr} = \frac{1.2}{2.4} = 0.5 \]
- Mole fraction of NO:
\[ \chi_{NO} = \frac{0.8}{2.4} = \frac{1}{3} \approx 0.333 \]
- Mole fraction of Br2:
\[ \chi_{Br_2} = \frac{0.4}{2.4} = \frac{1}{6} \approx 0.167 \]

### Step 5: Calculate Partial Pressures
Using the total pressure of 0.30 atm, we can find the partial pressures:
- \( P_{NOBr} = \chi_{NOBr} \times P_{total} = 0.5 \times 0.30 = 0.15 \text{ atm} \)
- \( P_{NO} = \chi_{NO} \times P_{total} = \frac{1}{3} \times 0.30 = 0.10 \text{ atm} \)
- \( P_{Br_2} = \chi_{Br_2} \times P_{total} = \frac{1}{6} \times 0.30 = 0.05 \text{ atm} \)

### Step 6: Write the Expression for \( K_p \)
The expression for \( K_p \) for the reaction \( 2NO + Br_2 \rightleftharpoons 2NOBr \) is:
\[ K_p = \frac{(P_{NOBr})^2}{(P_{NO})^2 \cdot (P_{Br_2})} \]

### Step 7: Substitute the Values into the Expression
Substituting the values we found:
\[ K_p = \frac{(0.15)^2}{(0.10)^2 \cdot (0.05)} = \frac{0.0225}{0.01 \cdot 0.05} = \frac{0.0225}{0.0005} = 45 \]

### Final Answer
Thus, the value of \( K_p \) for the reaction \( 2NO + Br_2 \rightleftharpoons 2NOBr \) is approximately **45**.

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