The equilibrium constant `K_(p)` for the following rection at `191^(@)C` is 1.24. what is`K_(c)`?
`B(s)+(3)/(2)F_(2)(g)hArrBF_3(g)`

Options

A. `6.7`
B. `0.61`
C. `8.30`
D. `7.6`

Answer

To find the equilibrium constant \( K_c \) from the given \( K_p \), we can use the relationship between \( K_p \) and \( K_c \):

\[ K_p = K_c \cdot R T^{\Delta n} \]

Where:
- \( R \) is the ideal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in the number of moles of gas (moles of products - moles of reactants).

### Step 1: Determine \( \Delta n \)

For the reaction:
\[ B(s) + \frac{3}{2}F_2(g) \rightleftharpoons BF_3(g) \]

- Moles of gaseous products: 1 mole of \( BF_3 \)
- Moles of gaseous reactants: \( \frac{3}{2} \) moles of \( F_2 \)

Now, calculate \( \Delta n \):
\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 1 - \frac{3}{2} = 1 - 1.5 = -0.5 \]

### Step 2: Convert the temperature to Kelvin

The temperature given is \( 191^\circ C \). To convert to Kelvin:
\[ T(K) = 191 + 273 = 464 \, K \]

### Step 3: Substitute values into the equation

Now we can substitute the values into the equation:
\[ K_p = K_c \cdot R T^{\Delta n} \]
Given \( K_p = 1.24 \), \( R = 0.0821 \), \( T = 464 \, K \), and \( \Delta n = -0.5 \):
\[ 1.24 = K_c \cdot 0.0821 \cdot (464)^{-0.5} \]

### Step 4: Calculate \( (464)^{-0.5} \)

Calculating \( (464)^{-0.5} \):
\[ (464)^{-0.5} = \frac{1}{\sqrt{464}} \approx \frac{1}{21.54} \approx 0.0464 \]

### Step 5: Rearranging the equation to find \( K_c \)

Now we can rearrange the equation to solve for \( K_c \):
\[ K_c = \frac{1.24}{0.0821 \cdot 0.0464} \]

### Step 6: Calculate \( K_c \)

Now we calculate \( K_c \):
\[ K_c = \frac{1.24}{0.0821 \cdot 0.0464} \approx \frac{1.24}{0.003818} \approx 324.5 \]

### Final Answer

Thus, the value of \( K_c \) is approximately \( 7.6 \).