To solve the problem regarding the equilibrium constant \( K_c \) for the process \( \text{Hg}(l) \rightleftharpoons \text{Hg}(g) \) given the vapor pressure of mercury at \( 27^\circ C \), we can follow these steps:
### Step 1: Identify the given data
We know that the vapor pressure of mercury at \( 27^\circ C \) is \( 0.002 \, \text{mm Hg} \).
### Step 2: Convert the vapor pressure from mm Hg to bar
To convert mm Hg to bar, we use the conversion factor:
\[ 1 \, \text{mm Hg} = 0.0013 \, \text{bar} \]
Thus, we can calculate:
\[
0.002 \, \text{mm Hg} \times 0.0013 \, \text{bar/mm Hg} = 2.6 \times 10^{-6} \, \text{bar}
\]
### Step 3: Write the expression for \( K_p \)
For the equilibrium process:
\[
\text{Hg}(l) \rightleftharpoons \text{Hg}(g)
\]
The equilibrium constant \( K_p \) is given by the partial pressure of the gaseous product:
\[
K_p = P_{\text{Hg}(g)} = 2.6 \times 10^{-6} \, \text{bar}
\]
### Step 4: Relate \( K_p \) to \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:
\[
K_p = K_c R T^{\Delta N}
\]
where:
- \( R \) is the universal 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 5: Calculate \( \Delta N \)
In our case, there is 1 mole of gaseous product (Hg(g)) and 0 moles of reactants (since Hg(l) is a liquid):
\[
\Delta N = 1 - 0 = 1
\]
### Step 6: Convert the temperature to Kelvin
The temperature \( T \) in Kelvin is:
\[
T = 27^\circ C + 273.15 = 300.15 \, K \approx 300 \, K
\]
### Step 7: Substitute values into the equation
Now we can rearrange the equation to solve for \( K_c \):
\[
K_c = \frac{K_p}{R T^{\Delta N}} = \frac{2.6 \times 10^{-6} \, \text{bar}}{0.0821 \, \text{L·atm/(K·mol)} \times 300 \, K}
\]
### Step 8: Calculate \( K_c \)
Substituting the values:
\[
K_c = \frac{2.6 \times 10^{-6}}{0.0821 \times 300}
\]
Calculating the denominator:
\[
0.0821 \times 300 = 24.63
\]
Now calculating \( K_c \):
\[
K_c = \frac{2.6 \times 10^{-6}}{24.63} \approx 1.06 \times 10^{-7}
\]
### Final Answer
Thus, the value of \( K_c \) for the process \( \text{Hg}(l) \rightleftharpoons \text{Hg}(g) \) is approximately:
\[
K_c \approx 1.1 \times 10^{-7}
\]
