In the reversible reaction, `2HI(g) hArr H_(2)(g)+I_(2)(g), K_(p)` is

Options

A. Greater than `K_(c)`
B. Less than `K_(c)`
C. Equal to `K_(c)`
D. Zero

Explanation

To solve the problem regarding the relationship between \( K_p \) and \( K_c \) for the reaction \( 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \), we will follow these steps:

### Step 1: Write the Reaction
The given reversible reaction is:
\[ 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \]

### Step 2: Identify the Change in Moles of Gases
To find the relationship between \( K_p \) and \( K_c \), we need to calculate \( \Delta n_g \), which is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.

- **Products:**
- \( H_2 \): 1 mole
- \( I_2 \): 1 mole
- Total moles of products = \( 1 + 1 = 2 \)

- **Reactants:**
- \( HI \): 2 moles
- Total moles of reactants = \( 2 \)

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

### Step 3: Use 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 \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.

### Step 4: Substitute \( \Delta n_g \) into the Formula
Since we found that \( \Delta n_g = 0 \), we can substitute this value into the equation:
\[ K_p = K_c (RT)^{0} \]
Since anything raised to the power of 0 is 1:
\[ K_p = K_c \cdot 1 \]
Thus, we have:
\[ K_p = K_c \]

### Conclusion
The relationship between \( K_p \) and \( K_c \) for the given reaction is:
\[ K_p = K_c \]

### Final Answer
The correct option is that \( K_p \) is equal to \( K_c \).

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