The equilibrium constant `K_(c)` for the following reaction at `742^(@)`C is `6.90xx10^(-3)`. What is `K_(p)`at same temperature ?
` (1)/(2)F_(2(g)) hArr F_(g)`

Options

A. `8.64xx10^(5)`
B. `6.29xx10^(-2)`
C. `7.21xx10^(-2)`
D. `7.56xx10^(-2)`

Explanation

To find the equilibrium constant \( K_p \) from \( K_c \) for the reaction

\[ \frac{1}{2} F_2(g) \rightleftharpoons F(g) \]

at a temperature of \( 742^\circ C \) with a given \( K_c = 6.90 \times 10^{-3} \), we can follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
To convert the temperature from Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

Substituting the given temperature:

\[ T = 742 + 273 = 1015 \, K \]

### Step 2: Calculate \( \Delta n_g \)
Next, we need to calculate \( \Delta n_g \), which is the change in the number of moles of gas during the reaction.

For the reaction:

\[ \frac{1}{2} F_2(g) \rightleftharpoons F(g) \]

The number of moles of gaseous products (\( n_p \)) is 1 (from \( F(g) \)) and the number of moles of gaseous reactants (\( n_r \)) is \( \frac{1}{2} \) (from \( \frac{1}{2} F_2(g) \)). Therefore:

\[ \Delta n_g = n_p - n_r = 1 - \frac{1}{2} = \frac{1}{2} = 0.5 \]

### Step 3: Use the relationship between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the formula:

\[ K_p = K_c \cdot (R \cdot T)^{\Delta n_g} \]

Where:
- \( R \) is the ideal gas constant \( 0.0821 \, L \cdot atm/(K \cdot mol) \)
- \( T \) is the temperature in Kelvin
- \( \Delta n_g \) is the change in moles of gas

### Step 4: Substitute the values into the equation
Now, substituting the known values:

\[ K_p = (6.90 \times 10^{-3}) \cdot (0.0821 \cdot 1015)^{0.5} \]

Calculating \( R \cdot T \):

\[ R \cdot T = 0.0821 \cdot 1015 = 83.43 \]

Now, taking the square root:

\[ (83.43)^{0.5} \approx 9.12 \]

### Step 5: Calculate \( K_p \)
Now, substituting this back into the equation for \( K_p \):

\[ K_p = (6.90 \times 10^{-3}) \cdot 9.12 \approx 6.29 \times 10^{-2} \]

### Final Answer:
Thus, the value of \( K_p \) at \( 742^\circ C \) is:

\[ K_p \approx 6.29 \times 10^{-2} \]

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