To find the equilibrium constants \( K_p \) and \( K_c \) for the equilibrium reaction of naphthalene, we can follow these steps:
### Step 1: Understand the Reaction
The equilibrium reaction is given as:
\[
C_{10}H_8(s) \rightleftharpoons C_{10}H_8(g)
\]
Here, naphthalene (C₁₀H₈) is in solid form on the left and in gaseous form on the right.
### Step 2: Identify the Vapor Pressure
The vapor pressure of naphthalene at \( 27^\circ C \) is given as \( 0.10 \, \text{mmHg} \). This pressure represents the partial pressure of the gaseous naphthalene at equilibrium.
### Step 3: Calculate \( K_p \)
The equilibrium constant \( K_p \) is defined as:
\[
K_p = \frac{P_{C_{10}H_8(g)}}{P_{C_{10}H_8(s)}}
\]
Since the solid does not contribute to the pressure, we have:
\[
K_p = P_{C_{10}H_8(g)} = 0.10 \, \text{mmHg}
\]
### Step 4: Convert \( K_p \) to Atmospheres
To convert \( K_p \) from mmHg to atmospheres, we use the conversion factor \( 760 \, \text{mmHg} = 1 \, \text{atm} \):
\[
K_p = \frac{0.10 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = \frac{0.10}{760} \approx 1.32 \times 10^{-4} \, \text{atm}
\]
### Step 5: Calculate \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by:
\[
K_p = K_c \cdot R T^{\Delta n_g}
\]
Where:
- \( R \) is the gas constant \( 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T \) is the temperature in Kelvin
- \( \Delta n_g \) is the change in moles of gas, which is the moles of gaseous products minus the moles of gaseous reactants.
For our reaction:
- Moles of gaseous products = 1 (from \( C_{10}H_8(g) \))
- Moles of gaseous reactants = 0 (from \( C_{10}H_8(s) \))
Thus, \( \Delta n_g = 1 - 0 = 1 \).
### Step 6: Substitute Values to Find \( K_c \)
Now we can rearrange the equation to solve for \( K_c \):
\[
K_c = \frac{K_p}{R T}
\]
Convert the temperature:
\[
T = 27^\circ C + 273 = 300 \, \text{K}
\]
Now substitute the values:
\[
K_c = \frac{1.32 \times 10^{-4} \, \text{atm}}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(300 \, \text{K})}
\]
Calculating the denominator:
\[
K_c = \frac{1.32 \times 10^{-4}}{24.63} \approx 5.36 \times 10^{-6} \, \text{mol/L}
\]
### Final Results
Thus, we have:
- \( K_p \approx 1.32 \times 10^{-4} \, \text{atm} \)
- \( K_c \approx 5.36 \times 10^{-6} \, \text{mol/L} \)
