For reaction, `2NOCl_((g))hArr2NO_((g))+Cl_(2(g)),K_(c)` at `427^(@)C` is `3xx10^(-6)L" "mol^(-1)`. The value of `K_(p)` is nearly,

Options

A. `7.50xx10^(-5)`
B. `2.50xx10^(-5)`
C. `2.50xx10^(-4)`
D. `1.75xx10^(-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 \( K_c \) at \( 427^\circ C \) is \( 3 \times 10^{-6} \, \text{L mol}^{-1} \), we can use the relationship between \( K_p \) and \( K_c \):

\[ K_p = K_c \cdot R \cdot T^{\Delta n} \]

### Step 1: Determine \( \Delta n \)

First, we need to calculate \( \Delta n \), which is defined as the change in the number of moles of gas during the reaction:

\[ \Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

From the balanced equation:

- Products: 2 moles of NO + 1 mole of Cl2 = 3 moles
- Reactants: 2 moles of NOCl = 2 moles

Thus,

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

### Step 2: Convert Temperature to Kelvin

Next, we convert the temperature from Celsius to Kelvin:

\[ T = 427 + 273 = 700 \, \text{K} \]

### Step 3: Use the Ideal Gas Constant \( R \)

The value of the gas constant \( R \) is:

\[ R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \]

### Step 4: Substitute Values into the Equation

Now we can substitute the values into the equation for \( K_p \):

\[ K_p = K_c \cdot R \cdot T^{\Delta n} \]

Substituting the known values:

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

### Step 5: Calculate \( K_p \)

Now we perform the calculation:

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

Calculating it step by step:

1. Calculate \( 0.0821 \cdot 700 = 57.47 \)
2. Now multiply by \( 3 \times 10^{-6} \):

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

### Final Result

Thus, rounding this value gives us:

\[ K_p \approx 1.75 \times 10^{-4} \]