For the reaction: `2NOCI(g) hArr 2NO+CI_2(g) ,K_c "at" 427^@C` is `3xx 10^6 L mol^(-1)`. The value of `K_p` is nearly

Options

A. `7.5 xx 10^(-5)`
B. `(2.5 xx 10^(-5)`
C. `(2.5 xx 10^(-4)`
D. `17.5 xx 10^(-4)`

Answer

To find the value of \( K_p \) for the reaction:

\[ 2 \text{NOCl}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Cl}_2(g) \]

given that \( K_c = 3 \times 10^{-6} \, \text{L mol}^{-1} \) at \( 427^\circ C \), we can follow these steps:

### Step 1: Convert Temperature to Kelvin
The temperature in Celsius needs to be converted to Kelvin using the formula:

\[ T(K) = T(°C) + 273.15 \]

For \( 427^\circ C \):

\[ T(K) = 427 + 273.15 = 700.15 \, K \approx 700 \, K \]

### Step 2: Calculate \( \Delta n_g \)
Next, we need to calculate the change in the number of moles of gas (\( \Delta n_g \)). This is calculated as follows:

\[ \Delta n_g = \text{moles of products} - \text{moles of reactants} \]

From the balanced equation:

- Moles of products = 2 (from 2NO) + 1 (from Cl2) = 3
- Moles of reactants = 2 (from 2NOCl)

Thus,

\[ \Delta n_g = 3 - 2 = 1 \]

### Step 3: Use the Relationship Between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:

\[ K_p = K_c \left( R T \right)^{\Delta n_g} \]

Where:
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \) (ideal gas constant)
- \( T = 700 \, K \)
- \( \Delta n_g = 1 \)

### Step 4: Substitute Values into the Equation
Now we substitute the known values into the equation:

\[ K_p = (3 \times 10^{-6}) \left( 0.0821 \times 700 \right)^{1} \]

Calculating \( R \times T \):

\[ R \times T = 0.0821 \times 700 = 57.47 \]

Now substituting this back into the equation for \( K_p \):

\[ K_p = (3 \times 10^{-6}) \times 57.47 \]

### Step 5: Calculate \( K_p \)
Now we perform the multiplication:

\[ K_p = 3 \times 10^{-6} \times 57.47 \approx 1.7241 \times 10^{-4} \]

Thus, the value of \( K_p \) is approximately:

\[ K_p \approx 1.72 \times 10^{-4} \, \text{atm}^{-1} \]

### Summary of the Solution
The value of \( K_p \) for the reaction at \( 427^\circ C \) is approximately \( 1.72 \times 10^{-4} \, \text{atm}^{-1} \).

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