For the reaction: `2NOCl(g) hArr 2NO(g) +Cl_(2)(g), K_(c)` at `427^(@)C` is `3xx10^(-6) L mol^(-1)`. The value of `K_(p)` is

Options

A. `7.5`
B. `2.5xx10^(-5)`
C. `2.0xx10^(-4)`
D. `1.72xx10^(-4)`

Explanation

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 \( K_c \) at \( 427^\circ C \) is \( 3 \times 10^{-6} \, \text{L mol}^{-1} \), we can use the relationship between \( K_p \) and \( K_c \):

\[ K_p = K_c \times R T^{\Delta N_g} \]

### Step 1: Identify the values needed for the calculation
- \( K_c = 3 \times 10^{-6} \, \text{L mol}^{-1} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \) (universal gas constant)
- Temperature \( T = 427^\circ C = 427 + 273 = 700 \, \text{K} \)

### Step 2: Calculate \( \Delta N_g \)
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
In this reaction:
- Moles of gaseous products = 2 (from \( 2 \text{NO} \)) + 1 (from \( \text{Cl}_2 \)) = 3
- Moles of gaseous reactants = 2 (from \( 2 \text{NOCl} \))

Thus,
\[ \Delta N_g = 3 - 2 = 1 \]

### Step 3: Substitute the values into the equation for \( K_p \)
Now we can substitute the values into the equation:

\[ K_p = K_c \times R \times T^{\Delta N_g} \]
\[ K_p = (3 \times 10^{-6}) \times (0.0821) \times (700)^{1} \]

### Step 4: Calculate \( K_p \)
Calculating \( (700)^{1} \):
\[ K_p = (3 \times 10^{-6}) \times (0.0821) \times (700) \]
\[ K_p = (3 \times 10^{-6}) \times (57.47) \quad \text{(since \( 0.0821 \times 700 \approx 57.47 \))} \]
\[ K_p \approx 1.7241 \times 10^{-4} \, \text{L}^2 \text{atm} \text{mol}^{-2} \]

### Step 5: Round to appropriate significant figures
Thus, rounding to two significant figures, we get:
\[ K_p \approx 1.72 \times 10^{-4} \, \text{L}^2 \text{atm} \text{mol}^{-2} \]

### Final Answer
The value of \( K_p \) is \( 1.72 \times 10^{-4} \).