For the reaction `2NO_(2(g))hArrN_(2)O_(4(g)),K_(p)//K_(c)` is equal to

Options

A. `(1)/(RT)`
B. `sqrt(RT)`
C. `RT`
D. `(RT)^(2)`

Explanation

To solve the problem of finding the ratio \( \frac{K_p}{K_c} \) for the reaction \( 2NO_2(g) \rightleftharpoons N_2O_4(g) \), we can follow these steps:

### Step 1: Write the expression for \( K_p \) and \( K_c \)
The equilibrium constants \( K_p \) and \( K_c \) are defined as follows:
- \( K_p \) is the equilibrium constant in terms of partial pressures.
- \( K_c \) is the equilibrium constant in terms of molar concentrations.

### Step 2: Identify the change in moles of gas (Δn)
To find \( \Delta n \), we need to calculate the difference between the number of moles of gaseous products and the number of moles of gaseous reactants.

- **Products**: There is 1 mole of \( N_2O_4 \).
- **Reactants**: There are 2 moles of \( NO_2 \).

So, we calculate \( \Delta n \):
\[ \Delta n = \text{(moles of products)} - \text{(moles of reactants)} = 1 - 2 = -1 \]

### Step 3: Use the relationship between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the formula:
\[ K_p = K_c (RT)^{\Delta n} \]
Where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol)).
- \( T \) is the temperature in Kelvin.

### Step 4: Substitute \( \Delta n \) into the equation
Since we found \( \Delta n = -1 \), we substitute this into the equation:
\[ K_p = K_c (RT)^{-1} \]
This can be rearranged to find \( \frac{K_p}{K_c} \):
\[ \frac{K_p}{K_c} = \frac{1}{RT} \]

### Step 5: Final answer
Thus, the ratio \( \frac{K_p}{K_c} \) is:
\[ \frac{K_p}{K_c} = \frac{1}{RT} \]