log `K_p/K_c`+log RT=0 is a relationship for the reaction :

Options

A. `PCl_5 hArr PCl_3+Cl_2`
B. `2SO_2 + O_2 hArr 2SO_3`
C. `H_2+I_2 hArr 2HI`
D. `N_2+3H_2 hArr 2NH_2`

Answer

To solve the question regarding the relationship \( \log \frac{K_p}{K_c} + \log RT = 0 \), we need to analyze the implications of this equation and determine which reaction exhibits this relationship.

### Step-by-Step Solution:

1. **Understanding the Given Relationship**:
The equation can be rewritten as:
\[ \log \frac{K_p}{K_c} = -\log RT \]
This implies that:
\[ \frac{K_p}{K_c} = \frac{1}{RT} \]

2. **Using the Relationship Between \( K_p \) and \( K_c \)**:
The relationship between \( K_p \) and \( K_c \) is given by the formula:
\[ K_p = K_c (RT)^{\Delta n_g} \]
where \( \Delta n_g \) is the change in the number of moles of gas (moles of gaseous products - moles of gaseous reactants).

3. **Rearranging the Equation**:
From the above equation, we can express \( \frac{K_p}{K_c} \):
\[ \frac{K_p}{K_c} = (RT)^{\Delta n_g} \]
Setting this equal to \( \frac{1}{RT} \) from our earlier step gives:
\[ (RT)^{\Delta n_g} = \frac{1}{RT} \]

4. **Equating Exponents**:
This implies:
\[ (RT)^{\Delta n_g} = (RT)^{-1} \]
Therefore, we can equate the exponents:
\[ \Delta n_g = -1 \]

5. **Finding the Correct Reaction**:
Now, we need to find a reaction where \( \Delta n_g = -1 \). This means that the number of moles of gaseous products is one less than the number of moles of gaseous reactants.

6. **Analyzing the Options**:
We will evaluate the provided options to determine which reaction has \( \Delta n_g = -1 \):
- **Option 1**: \( A \rightarrow B + C \) (Δn_g = 1)
- **Option 2**: \( A + B \rightarrow C \) (Δn_g = -1)
- **Option 3**: \( A + B \rightarrow C + D \) (Δn_g = 0)
- **Option 4**: \( A + B + C \rightarrow D \) (Δn_g = -2)

From this analysis, **Option 2** is the only reaction where \( \Delta n_g = -1 \).

### Final Answer:
The reaction that exhibits the relationship \( \log \frac{K_p}{K_c} + \log RT = 0 \) is **Option 2**.