For the following gaseous reaction `H_2 +I_2 hArr 2HI`, the equilibrium constant

Options

A. `K_p gt K_c`
B. `K_p lt K_c`
C. `K_p = K_c`
D. `K_p =1// K_c`

Explanation

To solve the problem regarding the equilibrium constant for the reaction \( H_2 + I_2 \rightleftharpoons 2 HI \), we need to determine the relationship between \( K_p \) and \( K_c \).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The given reaction is:
\[ H_2 + I_2 \rightleftharpoons 2 HI \]

2. **Write the Expression for \( K_c \)**:
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

3. **Determine the Change in Moles of Gas (\( \Delta n_g \))**:
To find \( \Delta n_g \), we need to count the total number of moles of gaseous reactants and products:
- Reactants: \( H_2 + I_2 \) = 1 mole of \( H_2 \) + 1 mole of \( I_2 \) = 2 moles
- Products: \( 2 HI \) = 2 moles

Now, calculate \( \Delta n_g \):
\[ \Delta n_g = \text{moles of products} - \text{moles of reactants} = 2 - 2 = 0 \]

4. **Use the Relationship between \( K_p \) and \( K_c \)**:
The relationship between \( K_p \) and \( K_c \) is given by:
\[ K_p = K_c (RT)^{\Delta n_g} \]
Since \( \Delta n_g = 0 \):
\[ K_p = K_c (RT)^0 = K_c \cdot 1 = K_c \]

5. **Conclusion**:
Therefore, we conclude that:
\[ K_p = K_c \]

### Final Answer:
The equilibrium constant \( K_p \) is equal to \( K_c \).