The vapour pressure of mercury is 0.002 mm Hg at `27^(@)C` .`K_(c)` for the process `Hg(l)hArrHg(g) `is :

Options

A. `0.002`
B. `8.12xx10^(-5)`
C. `6.48xx10^(-5)`
D. `1.068xx10^(-7)`

Answer

To solve the problem of finding the equilibrium constant \( K_c \) for the process \( \text{Hg}(l) \rightleftharpoons \text{Hg}(g) \) given the vapor pressure of mercury at 27°C, follow these steps:

### Step 1: Understand the Given Information
We are given the vapor pressure of mercury at 27°C, which is 0.002 mm Hg. This vapor pressure represents the equilibrium pressure of mercury in the gaseous state.

### Step 2: Convert Vapor Pressure to Atmospheres
To find \( K_p \), we first need to convert the vapor pressure from mm Hg to atmospheres. The conversion factor is:
\[ 1 \text{ mm Hg} = 0.0013 \text{ atm} \]
So, we calculate:
\[ \text{Vapor Pressure in atm} = 0.002 \text{ mm Hg} \times 0.0013 \text{ atm/mm Hg} = 2.6 \times 10^{-6} \text{ atm} \]

### Step 3: Write the Expression for \( K_p \)
For the equilibrium reaction \( \text{Hg}(l) \rightleftharpoons \text{Hg}(g) \), the expression for \( K_p \) is:
\[ K_p = P_{\text{Hg}(g)} \]
Since we have calculated the vapor pressure, we can say:
\[ K_p = 2.6 \times 10^{-6} \text{ atm} \]

### Step 4: Relate \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:
\[ K_p = K_c \cdot R^{\Delta N_g} \]
where \( R \) is the gas constant (0.0821 L·atm/(K·mol)), and \( \Delta N_g \) is the change in the number of moles of gas (moles of products - moles of reactants). In this case:
- Products (gaseous): 1 (from Hg(g))
- Reactants (gaseous): 0 (from Hg(l))

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

### Step 5: Calculate \( K_c \)
Now we can rearrange the equation to solve for \( K_c \):
\[ K_c = \frac{K_p}{R^{\Delta N_g}} = \frac{K_p}{R} \]
Substituting the values:
\[ K_c = \frac{2.6 \times 10^{-6} \text{ atm}}{0.0821 \text{ L·atm/(K·mol)} \times 300 \text{ K}} \]
Calculating the denominator:
\[ 0.0821 \times 300 = 24.63 \text{ L·atm/(mol·K)} \]
Now substituting this back into the equation for \( K_c \):
\[ K_c = \frac{2.6 \times 10^{-6}}{24.63} \approx 1.058 \times 10^{-7} \text{ mol/L} \]

### Final Answer
Thus, the value of \( K_c \) for the process is approximately:
\[ K_c \approx 1.068 \times 10^{-7} \text{ mol/L} \]