Naphthalene, a white solid used to make mothballs, has a vapour pressure of 0.10 mmHg at `27^(@)C`. Hence, `K_(P)` and `K_(c)` for the equilibrium are
`C_(10)H_(8)(s) rarr C_(10)H_(8)(g)`

Options

A. 0.10,0.10
B. `0.10, 4.1 xx 10^(-3)`
C. `1.32 xx 10^(-4), 5.36 xx 10^(-6)`
D. ` 5.36 xx 10^(-6),1.32 xx 10^(-4)`

Explanation

To solve the problem, we need to determine the equilibrium constants \( K_p \) and \( K_c \) for the sublimation of naphthalene, represented by the equilibrium:

\[ C_{10}H_8(s) \rightleftharpoons C_{10}H_8(g) \]

### Step 1: Understand the Given Information
We are given the vapor pressure of naphthalene at \( 27^\circ C \), which is \( 0.10 \, \text{mmHg} \). This vapor pressure corresponds to the partial pressure of the gaseous naphthalene at equilibrium.

### Step 2: Calculate \( K_p \)
The equilibrium constant \( K_p \) is defined as:

\[ K_p = \frac{P_{products}}{P_{reactants}} \]

In our case, the product is gaseous naphthalene, and the reactant is solid naphthalene. Since we do not include pure solids in the equilibrium expression, we have:

\[ K_p = \frac{P_{C_{10}H_8(g)}}{1} \]

Given that \( P_{C_{10}H_8(g)} = 0.10 \, \text{mmHg} \):

\[ K_p = 0.10 \, \text{mmHg} \]

### Step 3: Convert \( K_p \) to Atmospheres
To convert \( K_p \) from mmHg to atmospheres, we use the conversion factor \( 1 \, \text{atm} = 760 \, \text{mmHg} \):

\[ K_p = \frac{0.10 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 1.32 \times 10^{-4} \, \text{atm} \]

### Step 4: Calculate \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:

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

Where:
- \( R \) is the ideal gas constant \( 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T \) is the temperature in Kelvin
- \( \Delta n \) is the change in the number of moles of gas (products - reactants)

In our case:
- The number of moles of gaseous products = 1 (from \( C_{10}H_8(g) \))
- The number of moles of gaseous reactants = 0 (from \( C_{10}H_8(s) \))

Thus, \( \Delta n = 1 - 0 = 1 \).

Now, convert the temperature from Celsius to Kelvin:

\[ T = 27 + 273 = 300 \, \text{K} \]

### Step 5: Rearranging the Equation for \( K_c \)
Now we can rearrange the equation to solve for \( K_c \):

\[ K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{1.32 \times 10^{-4}}{(0.0821 \times 300)^1} \]

Calculating \( RT \):

\[ RT = 0.0821 \times 300 = 24.63 \]

Now substituting back to find \( K_c \):

\[ K_c = \frac{1.32 \times 10^{-4}}{24.63} \approx 5.36 \times 10^{-6} \]

### Final Results
Thus, the values of \( K_p \) and \( K_c \) are:

\[ K_p = 1.32 \times 10^{-4} \, \text{atm} \]
\[ K_c = 5.36 \times 10^{-6} \, \text{mol/L} \]