In a 7.0 L evacuated chamber, 0.50 mol `H_(2)` and 0.50 mol `I_(2)` react at `427^(@)` C.
`H_(2) (g) + I_(2) (g) hArr 2HI (g) ` + Heat , At the given temperature, `K_(c)` = 49 for the reaction.
what is the value of `K_(p)` ?

Options

A. 7
B. 49
C. 24.5
D. None of the these

Answer

To find the value of \( K_p \) for the given reaction, we can follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ H_2 (g) + I_2 (g) \rightleftharpoons 2 HI (g) + \text{Heat} \]

### Step 2: Identify the Initial Conditions
In the 7.0 L evacuated chamber, we have:
- Moles of \( H_2 \) = 0.50 mol
- Moles of \( I_2 \) = 0.50 mol

### Step 3: Determine the Change in Moles
For the reaction:
- 1 mole of \( H_2 \) reacts with 1 mole of \( I_2 \) to produce 2 moles of \( HI \).
- The total number of moles of gaseous reactants = 1 (from \( H_2 \)) + 1 (from \( I_2 \)) = 2 moles.
- The total number of moles of gaseous products = 2 moles (from \( 2 HI \)).

### Step 4: Calculate \( \Delta N \)
\[ \Delta N = \text{Moles of gaseous products} - \text{Moles of gaseous reactants} = 2 - 2 = 0 \]

### Step 5: Use the Relationship Between \( K_c \) and \( K_p \)
The relationship between \( K_c \) and \( K_p \) is given by the formula:
\[ K_p = K_c \times R^{\Delta N} \]
Where:
- \( K_c = 49 \) (given)
- \( R \) is the ideal gas constant, which is approximately \( 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( \Delta N = 0 \)

### Step 6: Substitute the Values
Since \( \Delta N = 0 \):
\[ K_p = K_c \times R^0 = K_c \times 1 = K_c \]
Thus,
\[ K_p = 49 \]

### Conclusion
The value of \( K_p \) is \( 49 \).