For which of the following reaction `K_(p)=K_(c)`?

Options

A. `N_(2(g))+3H_(2(g))hArr2NH_(3(g))`
B. `2NOCl((g))hArr2NO_((g))+Cl_(2(g))`
C. `H_(2(g))+I_(2(g))hArr2HI_((g))`
D. `CO_(2(g))+C_((s))hArr2CO_((g))`

Answer

To determine for which reaction \( K_p = K_c \), we need to understand the relationship between the two equilibrium constants. The relationship is given by the equation:

\[ K_p = K_c (RT)^{\Delta n_g} \]

where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta n_g \) is the change in the number of moles of gas, calculated as:

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

### Step-by-step Solution:

1. **Identify the Reactions**: List the reactions provided in the question and identify the gaseous reactants and products for each reaction.

2. **Calculate \( \Delta n_g \)**: For each reaction, count the number of moles of gaseous products and subtract the number of moles of gaseous reactants to find \( \Delta n_g \).

3. **Determine \( K_p \) and \( K_c \)**: According to the relationship \( K_p = K_c (RT)^{\Delta n_g} \):
- If \( \Delta n_g = 0 \), then \( K_p = K_c \).
- If \( \Delta n_g \neq 0 \), then \( K_p \neq K_c \).

4. **Select the Correct Reaction**: From the calculations, identify which reaction has \( \Delta n_g = 0 \).

### Example Calculation:

Assuming we have the following reactions:

- Reaction A: \( 2 \text{A(g)} \rightleftharpoons 2 \text{B(g)} \)
- Reaction B: \( \text{C(g)} \rightleftharpoons \text{D(g)} + \text{E(g)} \)
- Reaction C: \( \text{F(g)} + \text{G(g)} \rightleftharpoons \text{H(g)} \)
- Reaction D: \( \text{I(s)} + \text{J(g)} \rightleftharpoons \text{K(g)} \)

**Calculating \( \Delta n_g \)**:

- For Reaction A:
- Products: 2 moles of B
- Reactants: 2 moles of A
- \( \Delta n_g = 2 - 2 = 0 \)

- For Reaction B:
- Products: 2 moles (1 D + 1 E)
- Reactants: 1 mole (C)
- \( \Delta n_g = 2 - 1 = 1 \)

- For Reaction C:
- Products: 1 mole (H)
- Reactants: 2 moles (F + G)
- \( \Delta n_g = 1 - 2 = -1 \)

- For Reaction D:
- Products: 1 mole (K)
- Reactants: 1 mole (J) + 0 (I is solid)
- \( \Delta n_g = 1 - 1 = 0 \)

### Conclusion:

From the calculations, we find that:
- Reaction A and Reaction D both have \( \Delta n_g = 0 \), hence \( K_p = K_c \) for these reactions.

### Final Answer:

The reactions for which \( K_p = K_c \) are Reaction A and Reaction D.