One mole of` X_(2)`is mixed with one mole of `Y_(2)` in a flask of volume
1 litre . If at equilibrium 0.5 mole of `Y_(2)` are obtained. Then
find out `K_(P)` for reaction `X_(2(g))+Y_(2(g))hArr2XY_(g)`

Options

A. 12
B. 9
C. 4
D. 36

Answer

To solve the problem step by step, we will analyze the reaction and the information provided.

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ X_2(g) + Y_2(g) \rightleftharpoons 2XY(g) \]

### Step 2: Set up the initial conditions
Initially, we have:
- 1 mole of \( X_2 \)
- 1 mole of \( Y_2 \)
- 0 moles of \( XY \)

### Step 3: Define the changes at equilibrium
Let \( x \) be the amount of \( X_2 \) that reacts. Since the stoichiometry of the reaction indicates that 1 mole of \( X_2 \) reacts with 1 mole of \( Y_2 \) to produce 2 moles of \( XY \), we can express the changes as follows:
- Change in \( X_2 \): \(-x\)
- Change in \( Y_2 \): \(-x\)
- Change in \( XY \): \(+2x\)

### Step 4: Write the equilibrium concentrations
At equilibrium, the concentrations will be:
- Concentration of \( X_2 \): \( 1 - x \)
- Concentration of \( Y_2 \): \( 1 - x \)
- Concentration of \( XY \): \( 2x \)

### Step 5: Use the information given
It is given that at equilibrium, 0.5 moles of \( Y_2 \) are obtained. Therefore:
\[ 1 - x = 0.5 \]
Solving for \( x \):
\[ x = 1 - 0.5 = 0.5 \]

### Step 6: Calculate the equilibrium concentrations
Now substituting \( x = 0.5 \):
- Concentration of \( X_2 \): \( 1 - 0.5 = 0.5 \) moles
- Concentration of \( Y_2 \): \( 1 - 0.5 = 0.5 \) moles
- Concentration of \( XY \): \( 2 \times 0.5 = 1 \) mole

### Step 7: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[XY]^2}{[X_2][Y_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(1)^2}{(0.5)(0.5)} = \frac{1}{0.25} = 4 \]

### Step 8: Calculate \( \Delta N_g \)
To find \( K_p \), we need to calculate \( \Delta N_g \):
\[ \Delta N_g = N_p - N_r \]
Where:
- \( N_p \) = number of moles of products = 2 (from \( 2XY \))
- \( N_r \) = number of moles of reactants = 2 (from \( X_2 + Y_2 \))

Thus,
\[ \Delta N_g = 2 - 2 = 0 \]

### Step 9: Relate \( K_p \) and \( K_c \)
Using the relation:
\[ K_p = K_c (RT)^{\Delta N_g} \]
Since \( \Delta N_g = 0 \):
\[ K_p = K_c \times (RT)^0 = K_c \times 1 = K_c \]

### Step 10: Final result
Thus, we find:
\[ K_p = K_c = 4 \]