For the reversible system : `X_((g))hArrY_((g))+Z_((g))`, a quantity of `X` was heated at constant pressure `P` at a certain temperature. The equilibrium partial pressure of `X` was found to be `P//7`. What is the value of `K_(p)` at given temperature?

Options

A. `6P//7`
B. `9P//7`
C. `36P//7`
D. `6P`

Answer

To find the equilibrium constant \( K_p \) for the given reversible reaction \( X_{(g)} \rightleftharpoons Y_{(g)} + Z_{(g)} \), we will follow these steps:

### Step 1: Write the reaction and initial conditions
The reaction is:
\[ X_{(g)} \rightleftharpoons Y_{(g)} + Z_{(g)} \]
Initially, we have:
- Moles of \( X \) = \( a \)
- Moles of \( Y \) = 0
- Moles of \( Z \) = 0

### Step 2: Define the change in moles at equilibrium
Let \( x \) be the amount of \( X \) that dissociates at equilibrium. Therefore, at equilibrium:
- Moles of \( X \) = \( a - x \)
- Moles of \( Y \) = \( x \)
- Moles of \( Z \) = \( x \)

### Step 3: Calculate the total moles at equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = (a - x) + x + x = a + x \]

### Step 4: Use the given partial pressure of \( X \)
We are given that the equilibrium partial pressure of \( X \) is \( \frac{P}{7} \). The partial pressure of a gas can be expressed as:
\[ P_X = \text{Mole fraction of } X \times P \]
Thus:
\[ P_X = \frac{(a - x)}{(a + x)} \times P = \frac{P}{7} \]

### Step 5: Set up the equation and solve for \( x \)
From the equation above, we can set up:
\[ \frac{(a - x)}{(a + x)} \times P = \frac{P}{7} \]
Dividing both sides by \( P \) (assuming \( P \neq 0 \)):
\[ \frac{(a - x)}{(a + x)} = \frac{1}{7} \]

Cross-multiplying gives:
\[ 7(a - x) = a + x \]
Expanding this:
\[ 7a - 7x = a + x \]
Rearranging the terms:
\[ 7a - a = 7x + x \]
\[ 6a = 8x \]
Thus:
\[ x = \frac{3a}{4} \]

### Step 6: Calculate the partial pressures of \( Y \) and \( Z \)
At equilibrium:
- Moles of \( Y \) = \( x = \frac{3a}{4} \)
- Moles of \( Z \) = \( x = \frac{3a}{4} \)

The partial pressures of \( Y \) and \( Z \) can be calculated as follows:
\[ P_Y = \frac{x}{(a + x)} \times P = \frac{\frac{3a}{4}}{(a + \frac{3a}{4})} \times P = \frac{\frac{3a}{4}}{\frac{7a}{4}} \times P = \frac{3}{7}P \]

Similarly, for \( P_Z \):
\[ P_Z = \frac{x}{(a + x)} \times P = \frac{3}{7}P \]

### Step 7: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_Y \times P_Z}{P_X} \]
Substituting the values we found:
\[ K_p = \frac{\left(\frac{3}{7}P\right) \times \left(\frac{3}{7}P\right)}{\frac{P}{7}} \]
\[ K_p = \frac{\frac{9}{49}P^2}{\frac{P}{7}} \]
\[ K_p = \frac{9P^2}{49} \times \frac{7}{P} \]
\[ K_p = \frac{63P}{49} = \frac{9P}{7} \]

### Final Answer
Thus, the value of \( K_p \) at the given temperature is:
\[ K_p = \frac{9P}{7} \]