For mercury, the enthalpy of vaporization is 58.51 kJ/mol and the entropy of vaporization is 92.92 J/K ? mol. What is the normal boiling point of mercury?

The Clausius-Clapeyron equation is: \(ln(P_2/P_1) = -(\Delta{H_vap}/R) (1/T_2 - 1/T_1)\) Where: P1 and P2 are the initial and final vapor pressures, ∆Hvap is the enthalpy of vaporization, R is the ideal gas constant (8.314 J/K•mol), T1 and T2 are the initial and final temperatures. Since we want to find the normal boiling point, we need to find the temperature when the vapor pressure (P2) is equal to 1 atm (101325 Pa).

We need to convert the pressure from atm to pascal (Pa) using the conversion factor: 1 atm = 101325 Pa. P1 will be equal to 100000 Pa (approximately 1 atm) and P2 will be equal to 101325 Pa (1 atm).

Now we will plug in the given values into the Clausius-Clapeyron equation as follows: \(ln(101325 / 100000) = -(\Delta{H_vap}/R) (1/T_2 - 1/T_1)\) Using the given values: Enthalpy of vaporization (∆Hvap) = 58.51 kJ/mol = 58510 J/mol Entropy of vaporization (Svap) = 92.92 J/K•mol

To calculate the boiling point, we need to solve for T2 in the equation. However, one important fact to consider is that the normal boiling point is defined at the phase transition of liquid to gas at 1 atm. Thus, we can conclude that T1 = T2. Therefore, the equation becomes: \(ln(101325 / 100000) = -(\Delta{H_vap}/R) (1/T_2 - 1/T_2)\) Now, let's simplify the equation: \(ln(1.01325) = -\frac{58510 J/mol}{8.314 J/K \cdot mol} (0)\) Since the right side of the equation is 0, the left side will be equal to 0 as well: \(ln(1.01325) = 0\) To find T2, we can use the relation between the enthalpy and entropy of vaporization: \(\Delta{H_vap} = T \cdot \Delta{S_vap}\) Solving for T2 we get: \(T_2 = \frac{\Delta{H_vap}}{\Delta{S_vap}}\) Plug in the given values: \(T_2 = \frac{58510 J/mol}{92.92 J/K \cdot mol}\) \(T_2 \approx 629.9 K\) So, the normal boiling point of mercury is approximately \(629.9 K\).