For the reaction `2HIhArrH_(2)(g)+I_(2)(g)`

Options

A. `K_(p)=K_(c)`
B. `K_(p)gtK_(c)`
C. `K_(p)ltL_(c)`
D. `L_(c)=sqrtK_(P)`

Answer

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

### Step-by-step Solution:

1. **Write the Equilibrium Expression for \(K_c\)**:
The equilibrium constant \(K_c\) for the reaction is given by the formula:
\[ K_c = \frac{[H_2][I_2]}{[HI]^2} \]
Here, \([H_2]\) and \([I_2]\) are the molar concentrations of hydrogen and iodine gases, respectively, and \([HI]\) is the molar concentration of hydrogen iodide.

2. **Determine the Change in Moles of Gas (\(\Delta n_g\))**:
\(\Delta n_g\) is calculated as:
\[ \Delta n_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
In this reaction, there are 2 moles of gaseous reactants (2 moles of \(HI\)) and 2 moles of gaseous products (1 mole of \(H_2\) and 1 mole of \(I_2\)):
\[ \Delta n_g = (1 + 1) - 2 = 0 \]

3. **Relate \(K_p\) to \(K_c\)**:
The relationship between \(K_p\) and \(K_c\) is given by the equation:
\[ K_p = K_c \cdot (RT)^{\Delta n_g} \]
Substituting \(\Delta n_g = 0\):
\[ K_p = K_c \cdot (RT)^0 \]
Since any term raised to the power of zero is equal to 1:
\[ K_p = K_c \cdot 1 = K_c \]

4. **Conclusion**:
From the calculations, we conclude that:
\[ K_p = K_c \]
Therefore, the correct option is that \(K_p\) is equal to \(K_c\).

### Final Answer:
The answer to the question is: \(K_p = K_c\).

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