`K_(p)` for the reaction `A hArr B` is 4. If initially only A is present then what will be the partial pressure of B after equilibrium ?

Options

A. 1.2
B. 0.8
C. 0.6
D. 1

Answer

To solve the problem step by step, we can follow these calculations:

### Step 1: Write the reaction and the expression for \( K_p \)
The reaction is given as:
\[ A \rightleftharpoons B \]

The equilibrium constant \( K_p \) for this reaction is defined as:
\[ K_p = \frac{P_B}{P_A} \]

Where \( P_B \) is the partial pressure of B and \( P_A \) is the partial pressure of A at equilibrium.

### Step 2: Set up initial conditions
Initially, we have:
- 1 mole of A
- 0 moles of B

Thus, the initial partial pressures can be represented as:
- \( P_A = 1 \) atm (assuming the total pressure is 1 atm)
- \( P_B = 0 \) atm

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the number of moles of A that react to form B at equilibrium. Therefore, at equilibrium:
- The moles of A remaining = \( 1 - x \)
- The moles of B formed = \( x \)

### Step 4: Write the equilibrium partial pressures
The equilibrium partial pressures can be expressed as:
- \( P_A = 1 - x \)
- \( P_B = x \)

### Step 5: Substitute into the \( K_p \) expression
Now substituting these values into the expression for \( K_p \):
\[ K_p = \frac{P_B}{P_A} = \frac{x}{1 - x} \]

Given that \( K_p = 4 \), we can write:
\[ \frac{x}{1 - x} = 4 \]

### Step 6: Solve for \( x \)
Cross-multiplying gives:
\[ x = 4(1 - x) \]
\[ x = 4 - 4x \]
\[ x + 4x = 4 \]
\[ 5x = 4 \]
\[ x = \frac{4}{5} = 0.8 \]

### Step 7: Find the partial pressure of B
Now, substituting \( x \) back to find the partial pressure of B:
\[ P_B = x = 0.8 \, \text{atm} \]

### Conclusion
The partial pressure of B after equilibrium is:
\[ P_B = 0.8 \, \text{atm} \]