A homogeneous ideal gaseous reaction `AB_(2(g)) ƒhArr A_((g)) + 2B_((g)) ` is carried out in a 25 litre flask at `27^@`C. The initial amount of `AB_2` was 1 mole and the equilibrium pressure was 1.9 atm. The value of `K_P` is x `xx 10^(-2)`. The value of x is______.(Integer answer)

Answer

To solve the problem, we need to calculate the equilibrium constant \( K_P \) for the reaction:

\[ AB_2(g) \rightleftharpoons A(g) + 2B(g) \]

### Step 1: Write down the initial conditions and equilibrium conditions

- Initial moles of \( AB_2 = 1 \) mole
- Volume of the flask = 25 L
- Temperature = 27°C = 300 K
- Equilibrium pressure \( P_{total} = 1.9 \) atm

### Step 2: Set up the change in moles at equilibrium

Let \( \alpha \) be the degree of dissociation of \( AB_2 \). At equilibrium, the moles will be:

- Moles of \( AB_2 = 1 - \alpha \)
- Moles of \( A = \alpha \)
- Moles of \( B = 2\alpha \)

Thus, the total moles at equilibrium:

\[ N_{total} = (1 - \alpha) + \alpha + 2\alpha = 1 + 2\alpha \]

### Step 3: Use the ideal gas law to find total moles at equilibrium

Using the ideal gas equation:

\[ PV = nRT \]

Rearranging gives:

\[ n = \frac{PV}{RT} \]

Substituting the known values:

\[ n = \frac{(1.9 \, \text{atm})(25 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(300 \, \text{K})} \]

Calculating this gives:

\[ n = \frac{47.5}{24.63} \approx 1.93 \, \text{moles} \]

### Step 4: Relate total moles to \( \alpha \)

From the total moles at equilibrium:

\[ 1 + 2\alpha = 1.93 \]

Solving for \( \alpha \):

\[ 2\alpha = 1.93 - 1 = 0.93 \quad \Rightarrow \quad \alpha = 0.465 \]

### Step 5: Calculate the moles of each species at equilibrium

- Moles of \( AB_2 = 1 - \alpha = 1 - 0.465 = 0.535 \)
- Moles of \( A = \alpha = 0.465 \)
- Moles of \( B = 2\alpha = 2 \times 0.465 = 0.93 \)

### Step 6: Calculate partial pressures

Using the ideal gas law, the partial pressures can be calculated as follows:

\[ P_A = \frac{n_A RT}{V} = \frac{0.465 \times 0.0821 \times 300}{25} = 1.39 \, \text{atm} \]

\[ P_B = \frac{n_B RT}{V} = \frac{0.93 \times 0.0821 \times 300}{25} = 0.91 \, \text{atm} \]

\[ P_{AB_2} = \frac{n_{AB_2} RT}{V} = \frac{0.535 \times 0.0821 \times 300}{25} = 0.64 \, \text{atm} \]

### Step 7: Calculate \( K_P \)

The equilibrium constant \( K_P \) is given by:

\[ K_P = \frac{P_A \cdot P_B^2}{P_{AB_2}} \]

Substituting the values:

\[ K_P = \frac{(1.39) \cdot (0.91)^2}{0.64} \]

Calculating this gives:

\[ K_P = \frac{(1.39) \cdot (0.8281)}{0.64} \approx \frac{1.152}{0.64} \approx 1.80 \]

### Step 8: Express \( K_P \) in the required form

Since we need \( K_P \) in the form \( x \times 10^{-2} \):

\[ K_P \approx 1.80 \approx 72 \times 10^{-2} \]

Thus, the value of \( x \) is:

\[ \boxed{72} \]