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)hArr(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 \( K_c \):

\[ K_p = K_c \cdot (RT)^{\Delta n} \]

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 gaseous products - moles of gaseous reactants).

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

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

- Moles of gaseous products = 1 (from \( BF_3 \))
- Moles of gaseous reactants = \(\frac{3}{2}\) (from \( \frac{3}{2}F_2 \))

Thus,
\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 1 - \frac{3}{2} = 1 - 1.5 = -0.5 \]

### Step 2: Convert temperature to Kelvin

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

### Step 3: Substitute values into the equation

Now we can substitute \( K_p \), \( R \), \( T \), and \( \Delta n \) into the equation:

Given \( K_p = 1.24 \),
\[ K_p = K_c \cdot (RT)^{\Delta n} \]

Substituting the values:
\[ 1.24 = K_c \cdot (0.0821 \cdot 464)^{-0.5} \]

### Step 4: Calculate \( RT \)

Calculating \( RT \):
\[ RT = 0.0821 \cdot 464 \approx 38.16984 \]

### Step 5: Calculate \( (RT)^{-0.5} \)

Now calculate \( (RT)^{-0.5} \):
\[ (RT)^{-0.5} = \frac{1}{\sqrt{38.16984}} \approx \frac{1}{6.17} \approx 0.162 \]

### Step 6: Solve for \( K_c \)

Now, we can solve for \( K_c \):
\[ 1.24 = K_c \cdot 0.162 \]
\[ K_c = \frac{1.24}{0.162} \approx 7.65 \]

Thus, rounding to two decimal places, we find:
\[ K_c \approx 7.6 \]

### Final Answer
The value of \( K_c \) is approximately **7.6**.