For the reaction : `2NO(g) + Cl_(2) (g) hArr 2NOCl (g), NO` and `Cl_(2)` are initially taken in mole ratio `2 : 1`. The total pressure at equilibrium is found to be `1 atm`. If mole of `NOCl` are one fourth of that of `Cl_(2)` at equilibrium, `K_(p)` for the reaction is:

Options

A. `13//36`
B. `13//256`
C. `13//512`
D. `13//128`

Explanation

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

### Given Reaction:
\[ 2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g) \]

### Initial Conditions:
- Moles of \( NO \) and \( Cl_2 \) are in the ratio \( 2:1 \).
- Let the initial moles of \( NO \) be \( 2p \) and \( Cl_2 \) be \( p \).

### At Equilibrium:
Let \( x \) be the change in moles of \( NO \) and \( Cl_2 \) that react. The equilibrium moles will be:
- Moles of \( NO \) = \( 2p - 2x \)
- Moles of \( Cl_2 \) = \( p - x \)
- Moles of \( NOCl \) = \( 2x \)

### Total Pressure at Equilibrium:
The total pressure \( P_t \) at equilibrium is given as \( 1 \, atm \). The total number of moles at equilibrium can be expressed as:
\[ P_t = (2p - 2x) + (p - x) + 2x \]
Simplifying this:
\[ P_t = 2p - 2x + p - x + 2x = 3p - x \]
Setting this equal to the total pressure:
\[ 3p - x = 1 \quad \text{(Equation 1)} \]

### Given Condition:
It is given that the moles of \( NOCl \) at equilibrium are one fourth of that of \( Cl_2 \):
\[ 2x = \frac{1}{4}(p - x) \]
Multiplying both sides by 4:
\[ 8x = p - x \]
Rearranging gives:
\[ p = 9x \quad \text{(Equation 2)} \]

### Substitute Equation 2 into Equation 1:
Substituting \( p = 9x \) into Equation 1:
\[ 3(9x) - x = 1 \]
This simplifies to:
\[ 27x - x = 1 \implies 26x = 1 \implies x = \frac{1}{26} \]

### Finding \( p \):
Now substituting \( x \) back into Equation 2 to find \( p \):
\[ p = 9x = 9 \times \frac{1}{26} = \frac{9}{26} \]

### Finding Equilibrium Partial Pressures:
Now we can find the equilibrium partial pressures:
- \( P_{NO} = 2p - 2x = 2 \times \frac{9}{26} - 2 \times \frac{1}{26} = \frac{18}{26} - \frac{2}{26} = \frac{16}{26} = \frac{8}{13} \)
- \( P_{Cl_2} = p - x = \frac{9}{26} - \frac{1}{26} = \frac{8}{26} = \frac{4}{13} \)
- \( P_{NOCl} = 2x = 2 \times \frac{1}{26} = \frac{2}{26} = \frac{1}{13} \)

### Calculating \( K_p \):
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{NOCl})^2}{(P_{NO})^2 \cdot (P_{Cl_2})} \]
Substituting the values:
\[ K_p = \frac{\left(\frac{1}{13}\right)^2}{\left(\frac{8}{13}\right)^2 \cdot \left(\frac{4}{13}\right)} \]
Calculating:
\[ K_p = \frac{\frac{1}{169}}{\frac{64}{169} \cdot \frac{4}{13}} = \frac{1}{169} \cdot \frac{13}{256} = \frac{1}{256} \]

### Final Result:
Thus, the value of \( K_p \) is:
\[ K_p = \frac{13}{256} \]