pure nitrosyl chloride (NOCl) gas was heated to `240^(@) C` in a 1.0 L container. At equilibrium the total pressure was 1.0 atm and the NOCl pressure was 0.64 atm . What would be the value of `K_(P)` ?

Options

A. `1.02 atm`
B. `16.875xx10^(-3)`atm
C. `16xx10^(-2)`atm
D. none of these

Answer

To find the equilibrium constant \( K_p \) for the dissociation of nitrosyl chloride (NOCl), we can follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation of nitrosyl chloride can be represented as:
\[ 2 \text{NOCl} (g) \rightleftharpoons 2 \text{NO} (g) + \text{Cl}_2 (g) \]

### Step 2: Define initial conditions
Let the initial pressure of NOCl be \( P \) atm. At the start, the pressures of NO and Cl2 are both 0 atm.

### Step 3: Define changes at equilibrium
Let \( x \) be the amount of NOCl that dissociates at equilibrium. Therefore, at equilibrium:
- The pressure of NOCl will be \( P - 2x \)
- The pressure of NO will be \( 2x \)
- The pressure of Cl2 will be \( x \)

### Step 4: Set up the equations based on given data
From the problem, we know:
1. The total pressure at equilibrium is 1.0 atm:
\[ P - 2x + 2x + x = 1 \]
Simplifying gives:
\[ P + x = 1 \quad \text{(Equation 1)} \]

2. The pressure of NOCl at equilibrium is given as 0.64 atm:
\[ P - 2x = 0.64 \quad \text{(Equation 2)} \]

### Step 5: Solve the equations
From Equation 1:
\[ P = 1 - x \]

Substituting \( P \) into Equation 2:
\[ (1 - x) - 2x = 0.64 \]
\[ 1 - 3x = 0.64 \]
\[ 3x = 1 - 0.64 \]
\[ 3x = 0.36 \]
\[ x = 0.12 \text{ atm} \]

Now substituting \( x \) back into Equation 1 to find \( P \):
\[ P + 0.12 = 1 \]
\[ P = 1 - 0.12 \]
\[ P = 0.88 \text{ atm} \]

### Step 6: Calculate the equilibrium pressures
Now we can find the equilibrium pressures:
- Pressure of NOCl:
\[ P_{NOCl} = P - 2x = 0.88 - 2(0.12) = 0.88 - 0.24 = 0.64 \text{ atm} \]
- Pressure of NO:
\[ P_{NO} = 2x = 2(0.12) = 0.24 \text{ atm} \]
- Pressure of Cl2:
\[ P_{Cl2} = x = 0.12 \text{ atm} \]

### Step 7: Write the expression for \( K_p \)
The expression for \( K_p \) for the reaction is:
\[ K_p = \frac{(P_{NO})^2 \cdot (P_{Cl2})}{(P_{NOCl})^2} \]

Substituting the equilibrium pressures:
\[ K_p = \frac{(0.24)^2 \cdot (0.12)}{(0.64)^2} \]

### Step 8: Calculate \( K_p \)
Calculating the above expression:
\[ K_p = \frac{0.0576 \cdot 0.12}{0.4096} \]
\[ K_p = \frac{0.006912}{0.4096} \]
\[ K_p = 0.016875 \text{ atm}^{-1} \]

### Final Answer
Thus, the value of \( K_p \) is:
\[ K_p = 16.875 \times 10^{-3} \text{ atm}^{-1} \]

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