What is the normal boiling point of mercury?
Given : `DeltaH _(f)^(@)(Hg,l)=0,S^(@)(Hg,l)=77.4 J//K-"mol"`
`DeltaH _(f)^(@)(Hg,g)= 60.8 kJ//"mol", S^(@)(Hg, g)=174.4 J//K-"mol"`

Options

A. 624.8 K
B. 626.8 K
C. 636.8 K
D. None of these

Explanation

To find the normal boiling point of mercury, we can use the relationship between Gibbs free energy (ΔG), enthalpy (ΔH), and entropy (ΔS). At the boiling point, the system is in equilibrium, which means ΔG = 0. Therefore, we can set up the equation:

\[ \Delta G = \Delta H - T \Delta S = 0 \]

From this, we can rearrange the equation to find the temperature (T):

\[ T = \frac{\Delta H}{\Delta S} \]

### Step 1: Calculate ΔS (Change in Entropy)
We need to find the change in entropy (ΔS) when mercury transitions from liquid to gas. This can be calculated as follows:

\[ \Delta S = S^\circ(Hg, g) - S^\circ(Hg, l) \]

Substituting the given values:

\[ \Delta S = 174.4 \, \text{J/K·mol} - 77.4 \, \text{J/K·mol} = 97 \, \text{J/K·mol} \]

### Step 2: Calculate ΔH (Change in Enthalpy)
Next, we calculate the change in enthalpy (ΔH) for the transition from liquid to gas. This is given by:

\[ \Delta H = \Delta H_f^\circ(Hg, g) - \Delta H_f^\circ(Hg, l) \]

Since the enthalpy of formation of liquid mercury is zero, we have:

\[ \Delta H = 60.8 \, \text{kJ/mol} - 0 = 60.8 \, \text{kJ/mol} \]

### Step 3: Convert ΔH to J/mol
To match the units of ΔS, we need to convert ΔH from kJ/mol to J/mol:

\[ \Delta H = 60.8 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 60800 \, \text{J/mol} \]

### Step 4: Calculate the Boiling Point (T)
Now we can substitute ΔH and ΔS into the equation for T:

\[ T = \frac{\Delta H}{\Delta S} = \frac{60800 \, \text{J/mol}}{97 \, \text{J/K·mol}} \]

Calculating this gives:

\[ T = 626.8 \, \text{K} \]

### Conclusion
The normal boiling point of mercury is approximately **626.8 K**.

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