The values of `K_(p)//K_(c)` for the following reactions at 300 K are, respectively :
(At 300 K, RT = 24.62 d`m^(3)` atm `mol^(-1)`)
`N_(2)(g)+O_(2)(g)hArr 2 NO(g)`
`N_(2)O_(4)(g)hArr2 NO_(2)(g)`

Options

A. (A)`1,4.1xx10^(-2)dm^(-3)atm^(-1)mol
B. (B)`1,24.62dmatmmol^(-1)
C. (C)`24.62dm^(3)atmmol^(-1),606dm^(6)tm^(2)mol^(-2)`
D. (D)`1,24.62dm^(3)atmmol^(-1)`

Explanation

To solve the problem, we need to calculate the values of \( \frac{K_p}{K_c} \) for the two given reactions using the relationship:

\[ \frac{K_p}{K_c} = RT^{\Delta n} \]

where:
- \( R \) is the ideal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in the number of moles of gas (moles of products - moles of reactants).

### Step 1: Analyze the first reaction
The first reaction is:

\[ N_2(g) + O_2(g) \rightleftharpoons 2 NO(g) \]

**Calculate \( \Delta n \):**
- Moles of products = 2 (from \( 2 NO \))
- Moles of reactants = 1 (from \( N_2 \)) + 1 (from \( O_2 \)) = 2

So,

\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 2 = 0 \]

**Calculate \( \frac{K_p}{K_c} \):**
Using the formula,

\[ \frac{K_p}{K_c} = RT^{\Delta n} = RT^0 = 1 \]

### Step 2: Analyze the second reaction
The second reaction is:

\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]

**Calculate \( \Delta n \):**
- Moles of products = 2 (from \( 2 NO_2 \))
- Moles of reactants = 1 (from \( N_2O_4 \))

So,

\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 1 = 1 \]

**Calculate \( \frac{K_p}{K_c} \):**
Using the formula,

\[ \frac{K_p}{K_c} = RT^{\Delta n} = RT^1 = RT \]

Given \( RT = 24.62 \, \text{dm}^3 \, \text{atm} \, \text{mol}^{-1} \):

\[ \frac{K_p}{K_c} = 24.62 \]

### Final Results
- For the first reaction: \( \frac{K_p}{K_c} = 1 \)
- For the second reaction: \( \frac{K_p}{K_c} = 24.62 \)

### Summary
The values of \( \frac{K_p}{K_c} \) for the reactions are:
1. \( 1 \)
2. \( 24.62 \)