Which of the following relation(s) holds good for gaseous and reversible reactions?

Options

A. `K_(p)/K_(c )=(RT)^((Deltan)_(g))`
B. `K_(p)/K_(c)=(P)^((Deltan)_(g))`
C. `K_(c )/K_(p)=(p/(RT))^((Deltan)_(g))`
D. `K_(c )/K_(p)=(P)^(-(Deltan)_(g))`

Explanation

To solve the question regarding the relations that hold good for gaseous and reversible reactions, we will analyze the relationships between the equilibrium constants \( K_p \) and \( K_c \).

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- \( K_p \) is the equilibrium constant for gaseous reactions expressed in terms of partial pressures.
- \( K_c \) is the equilibrium constant for gaseous reactions expressed in terms of molar concentrations.

2. **Relation Between \( K_p \) and \( K_c \)**:
- For a gaseous reversible reaction, the relationship between \( K_p \) and \( K_c \) is given by the formula:
\[ K_p = K_c \cdot (RT)^{\Delta n_g} \]
- Here, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n_g \) is the change in the number of moles of gas, calculated as:
\[ \Delta n_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

3. **Rearranging the Equation**:
- Rearranging the equation \( K_p = K_c \cdot (RT)^{\Delta n_g} \) gives us:
\[ \frac{K_p}{K_c} = (RT)^{\Delta n_g} \]
- This matches with the first option provided in the question.

4. **Evaluating Other Options**:
- The second option \( \frac{K_p}{K_c} = P^{\Delta n_g} \) is incorrect as it does not include the \( RT \) term.
- The third option \( \frac{K_c}{K_p} = \frac{P}{RT}^{\Delta n_g} \) is also incorrect as it does not reflect the correct relationship.
- The fourth option \( \frac{K_c}{K_p} = P^{-\Delta n_g} \) is incorrect for the same reason.

5. **Conclusion**:
- The only correct relation that holds good for gaseous and reversible reactions is:
\[ \frac{K_p}{K_c} = (RT)^{\Delta n_g} \]
- Therefore, the correct answer is **Option 1**.