In the equilibrium reaction, `2HI(g)hArr H_(2)(g)+I_(2)(g)`, which of the following expressions is true?

Options

A. `K_(P)=K_(C)`
B. `K_(C)=2K_(P)`
C. `K_(P)=(K_(C))/(2)`
D. `K_(C)=K_(P)(RT)^(2)`'

Answer

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

### Step-by-Step Solution:

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

2. **Determine the Number of Moles**:
- **Reactants**: There are 2 moles of \( HI \) on the reactant side.
- **Products**: There is 1 mole of \( H_2 \) and 1 mole of \( I_2 \) on the product side, totaling 2 moles.

3. **Calculate \( \Delta N_G \)**:
- \( \Delta N_G \) is defined as the difference between the number of moles of gaseous products and the number of moles of gaseous reactants:
\[ \Delta N_G = \text{(moles of products)} - \text{(moles of reactants)} = 2 - 2 = 0 \]

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 \( \Delta N_G = 0 \), we can substitute this into the equation:
\[ K_p = K_c (RT)^0 \]
- Anything raised to the power of 0 is equal to 1, therefore:
\[ K_p = K_c \cdot 1 \]
- This simplifies to:
\[ K_p = K_c \]

5. **Conclusion**: The correct expression that holds true for the given equilibrium reaction is:
\[ K_p = K_c \]

### Final Answer:
The correct expression is:
\[ K_p = K_c \]