consider the following reversible gaseous reaction (at 298 K):
`(A) N_(2)O_(4)hArr` `2NO_2` (B) `2SO_(2) + O_(2)hArr2SO_(3)`
(C) `2HI hArrH_(2) + I_(2)` (D) `X + Y hArr4Z `
Highest and lowest value of `(K_(p))/(K_(c))` will be shown by the equilibrium

Options

A. (A) d,b
B. (B) a,c
C. (C) a,b
D. (D) b,c

Answer

To solve the problem of determining the highest and lowest values of \( \frac{K_p}{K_c} \) for the given reactions, we will follow these steps:

### Step 1: Understand the Relationship Between \( K_p \) and \( K_c \)

The relationship between \( K_p \) and \( K_c \) is given by the equation:

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

where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in the number of moles of gas, calculated as the difference between the moles of products and the moles of reactants.

### Step 2: Calculate \( \Delta n \) for Each Reaction

1. **Reaction A: \( N_2O_4 \rightleftharpoons 2NO_2 \)**
- Moles of products = 2 (from \( 2NO_2 \))
- Moles of reactants = 1 (from \( N_2O_4 \))
- \( \Delta n = 2 - 1 = 1 \)

2. **Reaction B: \( 2SO_2 + O_2 \rightleftharpoons 2SO_3 \)**
- Moles of products = 2 (from \( 2SO_3 \))
- Moles of reactants = 3 (from \( 2SO_2 + O_2 \))
- \( \Delta n = 2 - 3 = -1 \)

3. **Reaction C: \( 2HI \rightleftharpoons H_2 + I_2 \)**
- Moles of products = 2 (from \( H_2 + I_2 \))
- Moles of reactants = 2 (from \( 2HI \))
- \( \Delta n = 2 - 2 = 0 \)

4. **Reaction D: \( X + Y \rightleftharpoons 4Z \)**
- Moles of products = 4 (from \( 4Z \))
- Moles of reactants = 2 (from \( X + Y \))
- \( \Delta n = 4 - 2 = 2 \)

### Step 3: Calculate \( \frac{K_p}{K_c} \) for Each Reaction

Using the values of \( \Delta n \):

1. **For Reaction A:**
\[ \frac{K_p}{K_c} = R T^{1} = RT \]

2. **For Reaction B:**
\[ \frac{K_p}{K_c} = R T^{-1} = \frac{1}{RT} \]

3. **For Reaction C:**
\[ \frac{K_p}{K_c} = R T^{0} = 1 \]

4. **For Reaction D:**
\[ \frac{K_p}{K_c} = R T^{2} = RT^2 \]

### Step 4: Compare the Values

- **Reaction A:** \( RT \)
- **Reaction B:** \( \frac{1}{RT} \)
- **Reaction C:** \( 1 \)
- **Reaction D:** \( RT^2 \)

### Step 5: Determine Highest and Lowest Values

- The highest value of \( \frac{K_p}{K_c} \) is from **Reaction D** (\( RT^2 \)).
- The lowest value of \( \frac{K_p}{K_c} \) is from **Reaction B** (\( \frac{1}{RT} \)).

### Conclusion

- **Highest \( \frac{K_p}{K_c} \): Reaction D**
- **Lowest \( \frac{K_p}{K_c} \): Reaction B**