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

### Step-by-Step Solution:

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

2. **Define \(K_c\) and \(K_p\)**:
- \(K_c\) is the equilibrium constant expressed in terms of molar concentrations (moles per liter).
- \(K_p\) is the equilibrium constant expressed in terms of partial pressures.

3. **Determine the Change in Moles of Gas (\(\Delta n_g\))**:
- The change in the number of moles of gas is calculated as:
\[ \Delta n_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
- In this reaction, we have:
- Products: \(H_2(g) + I_2(g)\) = 1 + 1 = 2 moles
- Reactants: \(2HI(g)\) = 2 moles
- Therefore:
\[ \Delta n_g = 2 - 2 = 0 \]

4. **Relate \(K_p\) and \(K_c\)**:
- The relationship between \(K_p\) and \(K_c\) is given by the equation:
\[ K_p = K_c (R^{{\Delta n_g}}) T^{\Delta n_g} \]
- Since \(\Delta n_g = 0\), we can simplify this to:
\[ K_p = K_c (R^0) T^0 = K_c \cdot 1 \cdot 1 = K_c \]

5. **Conclusion**:
- Thus, we find that:
\[ K_p = K_c \]
- This means that the equilibrium constants \(K_p\) and \(K_c\) are equal for this reaction.

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