Question

Mercury is the only metal that is a liquid at room temperature. When mercury vapor is inhaled, it is readily absorbed by the lungs, causing significant health risks. The enthalpy of vaporization of mercury is 59.1 \(\mathrm{kJ} / \mathrm{mol}\) . The normal boiling point of mercury is \(357^{\circ} \mathrm{C}\) . What is the vapor pressure of mercury at \(25^{\circ} \mathrm{C} ?\)

Short answer

The vapor pressure of mercury at 25°C is approximately \(3.86 \times 10^{-6}\,\mathrm{kPa}\).

Step-by-step solution

Clausius-Clapeyron equation

The Clausius-Clapeyron equation is given by: \[ \ln{\frac{P_2}{P_1}} = -\frac{\Delta H_\text{vap}}{R}(\frac{1}{T_2} - \frac{1}{T_1}) \] where \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\) respectively, \(\Delta H_\text{vap}\) is the enthalpy of vaporization, and R is the ideal gas constant (8.314 J/mol·K). Step 2: Convert temperatures to Kelvin

Convert temperatures to Kelvin

To use the Clausius-Clapeyron equation, we need to convert the given temperatures from degrees Celsius to Kelvin. \(T_1 = 357^{\circ}\mathrm{C} + 273.15 = 630.15\,\mathrm{K}\) \(T_2 = 25^{\circ}\mathrm{C} + 273.15 = 298.15\,\mathrm{K}\) Step 3: Normal boiling point vapor pressure

Normal boiling point vapor pressure

At the normal boiling point, the vapor pressure is equal to 1 atmosphere (101.3 kPa). Therefore, \(P_1 = 101.3\, \mathrm{kPa}\). Step 4: Enthalpy of vaporization

Enthalpy of vaporization

The given enthalpy of vaporization for mercury is \(\Delta H_\text{vap} = 59.1\,\mathrm{kJ/mol}\). Convert this value to J/mol: \(\Delta H_\text{vap} = 59.1\, \mathrm{kJ/mol} \times 1000\,\mathrm{J/kJ} = 59100\,\mathrm{J/mol}\) Step 5: Calculate the vapor pressure at 25°C

Calculate the vapor pressure at 25°C

Now, we can substitute the known values into the Clausius-Clapeyron equation and solve for the vapor pressure \(P_2\) at 25°C (298.15 K): \[ \ln{\frac{P_2}{101.3\, \mathrm{kPa}}} = -\frac{59100\, \mathrm{J/mol}}{8.314\, \mathrm{J/mol}\cdot\mathrm{K}}(\frac{1}{298.15\, \mathrm{K}} - \frac{1}{630.15\, \mathrm{K}}) \] First, solve the equation inside the parentheses and then multiply it by the enthalpy of vaporization divided by the gas constant: \[ \ln{\frac{P_2}{101.3\, \mathrm{kPa}}} = -12.42 \] Now, to find \(P_2\), take the exponential of both sides: \[ \frac{P_2}{101.3\, \mathrm{kPa}} = e^{-12.42} \] Finally, multiply both sides by the normal boiling point vapor pressure to obtain the vapor pressure at 25°C: \[ P_2 = 101.3\, \mathrm{kPa} \times e^{-12.42} \approx 3.86 \times 10^{-6}\,\mathrm{kPa} \] The vapor pressure of mercury at 25°C is approximately \(3.86 \times 10^{-6}\,\mathrm{kPa}\).

Key concepts

Vapor Pressure

Vapor pressure is an important concept in understanding how substances evaporate and interact with their surroundings. It refers to the pressure exerted by the vapor of a liquid or solid substance when it is in equilibrium with its liquid or solid form in a closed system. This equilibrium is achieved when the rate of evaporation equals the rate of condensation.
At the normal boiling point, vapor pressure equals atmospheric pressure, making the substance boil. For mercury, the normal boiling point is 357°C. However, when calculating vapor pressure at a temperature like 25°C, the measurement is much lower because less of the liquid evaporates at this lower temperature.

• At higher temperatures, vapor pressure increases because more molecules have enough energy to break free from the liquid and become gas.
• Vapor pressure is specific to temperature and substance.
• It helps determine how volatile a substance is, which is crucial for chemical processes and safety considerations.

Enthalpy of Vaporization

The enthalpy of vaporization is the amount of energy required to transform a given quantity of a substance from a liquid into a gas at constant pressure. This energy input breaks the intermolecular forces that hold the liquid together. For mercury, the enthalpy of vaporization is 59.1 kJ/mol. This is quite high compared to many other liquids, which reflects the strong cohesive forces within mercury.
Understanding enthalpy of vaporization allows us to grasp how much energy is needed to convert a liquid to gas, which is essential for processes such as distillation and chemical reactions.

• It's usually expressed in units of kJ/mol.
• The higher the enthalpy of vaporization, the more energy is needed for the phase transition from liquid to gas.
• This property varies among substances because different materials have various strengths of intermolecular forces.

Temperature Conversion

Temperature plays a pivotal role in calculations involving vapor pressure and phase changes, such as those described by the Clausius-Clapeyron equation. Converting temperature from Celsius to Kelvin is essential because many equations in chemistry and physics, like the Clausius-Clapeyron equation, require absolute temperature values in Kelvin for accurate computations.

To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, 25°C becomes 298.15 K, and 357°C becomes 630.15 K.

• Kelvin is the SI unit for temperature and is used in many scientific equations.
• It provides an absolute scale where zero Kelvin is the absolute lowest temperature, meaning no thermal energy remains.
• Converting temperatures accurately ensures precision in any related calculations, such as those needed for equilibrium or reaction rate computations.

Ideal Gas Constant

The ideal gas constant (symbolized as R) is a key component in many fundamental chemical equations, including the Clausius-Clapeyron equation. The value commonly used is 8.314 J/mol·K, which helps relate energy, temperature, and moles of a substance in idealized conditions.

Understanding R's role is crucial for solving problems related to gas laws and states of matter.

• R is the proportionality factor that connects terms like pressure, volume, temperature, and moles in the ideal gas law \(PV = nRT\).
• It demonstrates how energy inputs correlate with temperature and exponential changes in gas behavior.
• Using the correct unit for R ensures that calculations are consistent with other constants and measures used in chemical equations.