Consider the pertial decomposition of `A` as
`2A(g)hArr2B(g)+C(g)` At equilibrium `700`mL gaseous mixture contains `100`mL of gas C at `10` atm and `300K` what is the value of `K_(p)` for the reaction ?

Options

A. `(40)/(7)`
B. `(1)/(28)`
C. `(10)/(28)`
D. `(28)/(10)`

Answer

To find the equilibrium constant \( K_p \) for the reaction \( 2A(g) \rightleftharpoons 2B(g) + C(g) \), we will follow these steps:

### Step 1: Understand the Reaction and Given Data
The reaction is:
\[ 2A(g) \rightleftharpoons 2B(g) + C(g) \]

We are given:
- Total volume of the gaseous mixture at equilibrium = 700 mL
- Volume of gas C at equilibrium = 100 mL
- Total pressure of the mixture = 10 atm
- Temperature = 300 K

### Step 2: Determine the Volume of A and B
Let the volume of A at equilibrium be \( V_A \), the volume of B be \( V_B \), and the volume of C be \( V_C \).

From the problem:
- \( V_C = 100 \) mL (given)
- Total volume = \( V_A + V_B + V_C = 700 \) mL

Using the stoichiometry of the reaction, we know that for every 2 moles of A decomposed, 2 moles of B and 1 mole of C are produced. Hence:
- \( V_B = 2 \times V_C = 2 \times 100 \) mL = 200 mL
- Therefore, \( V_A + 200 + 100 = 700 \)
- Thus, \( V_A = 700 - 300 = 400 \) mL

### Step 3: Calculate Partial Pressures
Next, we calculate the partial pressures of each gas using the ideal gas law. The total pressure is given as 10 atm.

The partial pressure of each gas can be calculated using the formula:
\[ P_i = \left( \frac{V_i}{V_{total}} \right) \times P_{total} \]

1. **Partial pressure of A**:
\[ P_A = \left( \frac{V_A}{V_{total}} \right) \times P_{total} = \left( \frac{400}{700} \right) \times 10 = \frac{4000}{700} = \frac{40}{7} \text{ atm} \]

2. **Partial pressure of B**:
\[ P_B = \left( \frac{V_B}{V_{total}} \right) \times P_{total} = \left( \frac{200}{700} \right) \times 10 = \frac{2000}{700} = \frac{20}{7} \text{ atm} \]

3. **Partial pressure of C**:
\[ P_C = \left( \frac{V_C}{V_{total}} \right) \times P_{total} = \left( \frac{100}{700} \right) \times 10 = \frac{1000}{700} = \frac{10}{7} \text{ atm} \]

### Step 4: Write the Expression for \( K_p \)
The expression for the equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_B)^2 \cdot (P_C)}{(P_A)^2} \]

Substituting the calculated partial pressures:
\[ K_p = \frac{\left(\frac{20}{7}\right)^2 \cdot \left(\frac{10}{7}\right)}{\left(\frac{40}{7}\right)^2} \]

### Step 5: Simplify the Expression
Calculating the numerator and denominator:
- Numerator:
\[ \left(\frac{20}{7}\right)^2 \cdot \left(\frac{10}{7}\right) = \frac{400}{49} \cdot \frac{10}{7} = \frac{4000}{343} \]

- Denominator:
\[ \left(\frac{40}{7}\right)^2 = \frac{1600}{49} \]

Now substituting back into the \( K_p \) expression:
\[ K_p = \frac{\frac{4000}{343}}{\frac{1600}{49}} = \frac{4000 \cdot 49}{1600 \cdot 343} \]

### Step 6: Final Calculation
Calculating \( K_p \):
\[ K_p = \frac{4000 \cdot 49}{1600 \cdot 343} = \frac{200 \cdot 49}{343} = \frac{9800}{343} \]

### Conclusion
Thus, the value of \( K_p \) for the reaction is:
\[ K_p = \frac{9800}{343} \text{ atm} \]