The equilibrium constant `K_(c)` for the following reaction at `842^(@)C` is `7.90xx10^(-3)` .What is `K_(p)` at same temperature ?
`(1)/(2)F_(2)(g)hArrF(g)`

Options

A. `8.64xx10^(-5)`
B. `8.26xx10^(-4)`
C. `7.90xx10^(-2)`
D. `7.56xx10^(-2)`

Answer

To find the equilibrium constant \( K_p \) from \( K_c \) for the given reaction at \( 842^\circ C \), we can follow these steps:

### Step 1: Write the equilibrium expression
The reaction given is:
\[ \frac{1}{2} F_2(g) \rightleftharpoons F(g) \]

### Step 2: Identify \( K_c \)
The equilibrium constant \( K_c \) is provided as:
\[ K_c = 7.90 \times 10^{-3} \]

### Step 3: Determine \( \Delta n \)
Calculate \( \Delta n \), which is the difference between the number of moles of gaseous products and reactants.
- Products: 1 mole of \( F(g) \)
- Reactants: \( \frac{1}{2} \) mole of \( F_2(g) \)

Thus,
\[ \Delta n = \text{(moles of products)} - \text{(moles of reactants)} = 1 - \frac{1}{2} = \frac{1}{2} = 0.5 \]

### Step 4: Convert temperature to Kelvin
Convert the temperature from Celsius to Kelvin:
\[ T(K) = 842 + 273 = 1115 \, K \]

### Step 5: Use the formula to calculate \( K_p \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:
\[ K_p = K_c \cdot R \cdot T^{\Delta n} \]
Where:
- \( R \) (the ideal gas constant) = \( 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1} \)

Substituting the values:
\[ K_p = (7.90 \times 10^{-3}) \cdot (0.0821) \cdot (1115)^{0.5} \]

### Step 6: Calculate \( K_p \)
First, calculate \( (1115)^{0.5} \):
\[ (1115)^{0.5} \approx 33.4 \]

Now substitute back into the equation:
\[ K_p = (7.90 \times 10^{-3}) \cdot (0.0821) \cdot (33.4) \]

Calculating:
\[ K_p \approx (7.90 \times 10^{-3}) \cdot (2.744) \approx 0.0217 \approx 7.56 \times 10^{-2} \]

### Final Answer
Thus, the value of \( K_p \) at \( 842^\circ C \) is:
\[ K_p \approx 7.56 \times 10^{-2} \]