Problem 100 - General Chemistry

Question

The vapor pressure of water is \(17.5 \mathrm{mmHg}\) at \(20.0^{\circ} \mathrm{C}\) and \(355.1 \mathrm{mmHg}\) at \(80.0^{\circ} \mathrm{C}\). Calculate the boiling point of water at \(760.0 \mathrm{mmHg}\)

Step-by-Step Solution

Answer

The boiling point of water at 760.0 mmHg is approximately 100.0°C.

Step 1: Understanding the Clausius-Clapeyron Equation

To determine the boiling point of water at a new pressure, we can utilize the Clausius-Clapeyron equation, which is \[\ln \left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]where \(P_1\) and \(P_2\) are the initial and final pressures, \(T_1\) and \(T_2\) are the initial and final temperatures (in Kelvin), and \(\Delta H_{vap}\) is the enthalpy of vaporization.

Step 2: Converting Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin using the formula: \[ \text{Kelvin} = \text{Celsius} + 273.15 \]Thus, \(20.0^{\circ} \text{C} = 293.15\, \text{K}\) and \(80.0^{\circ} \text{C} = 353.15\, \text{K}\).

Step 3: Calculating ΔH_vap

To find \(\Delta H_{vap}\), we use the known pressures and temperatures:\[ \ln \left(\frac{355.1}{17.5}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{353.15} - \frac{1}{293.15}\right) \]Assume \(R = 8.314\, \text{J/mol}\, \text{K}\). Solve this equation for \(\Delta H_{vap}\).

Step 4: Setting Up the Final Equation for New Conditions

With \(\Delta H_{vap}\) known, rearrange the Clausius-Clapeyron equation to solve for the temperature \(T_2\) at \(P_2 = 760.0\, \text{mmHg}\):\[ \ln \left(\frac{760.0}{17.5}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{293.15}\right) \]Solve for \(\frac{1}{T_2}\), and invert to find \(T_2\).

Step 5: Converting Temperature Back to Celsius

Using the Kelvin value obtained for \(T_2\), convert back to Celsius using: \[ \text{Celsius} = \text{Kelvin} - 273.15 \]

Key Concepts

Vapor Pressure

Vapor pressure is a key concept when dealing with the phase transition from liquid to vapor. It represents the pressure exerted by the vapor in equilibrium with its liquid at a given temperature. This delicate balance means that the amount of liquid turning into vapor equals the amount of vapor condensing back into liquid. The vapor pressure of a substance increases as the temperature rises.
Imagine molecules in a liquid jostling around; heating them gives these molecules more energy, and more of them escape into the vapor phase. At 20°C, water has a vapor pressure of 17.5 mmHg, while at 80°C, it's a much higher 355.1 mmHg. This jump illustrates how sensitive vapor pressure is to temperature changes.
Vapor pressure is crucial for determining when a substance will boil, which occurs when the vapor pressure equals atmospheric pressure.

Enthalpy of Vaporization

Enthalpy of vaporization (\(\Delta H_{vap}\)) is the amount of energy required to convert a liquid into vapor without changing its temperature. This energy is necessary to overcome the molecular forces binding the liquid. It's typically measured in Joules per mole (J/mol).
The Clausius-Clapeyron equation uses the \(\Delta H_{vap}\) to connect temperature, pressure, and phase change. With the vapor pressures and temperatures given in the exercise, you can calculate \(\Delta H_{vap}\) by rearranging the equation:\[\ln \left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]where \(R\) is the universal gas constant. Determining \(\Delta H_{vap}\) helps predict boiling points under different pressures.

Temperature Conversion

Temperature conversion is often necessary for scientific calculations, particularly in equations like the Clausius-Clapeyron. The standard unit for temperature in these equations is Kelvin because it is an absolute scale that begins at absolute zero.
To convert Celsius to Kelvin, add 273.15 to the Celsius temperature:\[\text{Kelvin} = \text{Celsius} + 273.15\]Conversely, to convert Kelvin back to Celsius for easier interpretation:\[\text{Celsius} = \text{Kelvin} - 273.15\]In our example, 20°C is converted to 293.15 K, and 80°C is converted to 353.15 K. Making these conversions ensures accuracy when plugging values into the Clausius-Clapeyron equation.

Boiling Point Calculation

Boiling point calculation under varying pressures can be efficiently conducted using the Clausius-Clapeyron equation. It allows us to understand the relationship between pressure and temperature at phase changes. By calculating the boiling point, we determine the temperature at which the vapor pressure of the liquid equals the ambient pressure.
The exercise requires calculating the boiling point of water at 760 mmHg using known vapor pressures and temperatures. Reorganize the Clausius-Clapeyron equation to solve for the unknown temperature when the pressure is 760 mmHg:\[\ln \left(\frac{760.0}{17.5}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{293.15}\right)\]Solving for \(T_2\), followed by converting back to Celsius, gives the desired boiling point. This calculation is essential in understanding real-world applications like cooking and industrial processes.