Question

Carbon tetrachloride, \(\mathrm{CCl}_{4},\) has a vapor pressure of 213 torr at \(40 .^{\circ} \mathrm{C}\) and 836 torr at \(80 .^{\circ} \mathrm{C} .\) What is the normal boiling point of \(\mathrm{CCl}_{4} ?\)

Short answer

The normal boiling point of \(\mathrm{CCl}_{4}\) is approximately \(77.20^{\circ}\mathrm{C}\).

Step-by-step solution

(Step 1: Write down the Clausius-Clapeyron equation)

The Clausius-Clapeyron equation relates the vapor pressure of a substance with its temperature: \[\ln\frac{P_2}{P_1} = -\frac{L}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] where: - \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\), respectively - \(L\) is the molar heat of vaporization - \(R\) is the ideal gas constant (\(8.314\, \mathrm{J/mol \cdot K}\))

(Step 2: Convert temperatures to Kelvin)

Since the given temperatures are in Celsius, we need to convert them to Kelvin by adding 273.15: \[T_1 = 40^{\circ}\mathrm{C} + 273.15 = 313.15\,\mathrm{K}\] \[T_2 = 80^{\circ}\mathrm{C} + 273.15 = 353.15\,\mathrm{K}\]

(Step 3: Solve for the molar heat of vaporization L)

Given the vapor pressures and temperatures, we can plug the values into the Clausius-Clapeyron equation and solve for the molar heat of vaporization: \[\ln\frac{836}{213} = -\frac{L}{8.314}\left(\frac{1}{353.15} - \frac{1}{313.15}\right)\] Calculate \(L\): \[L = -8.314\,\frac{\ln\frac{836}{213}}{\frac{1}{353.15} - \frac{1}{313.15}} = 2.985 \times 10^{4}\, \mathrm{J/mol}\]

(Step 4: Solve for the normal boiling point temperature)

We can rearrange the Clausius-Clapeyron equation, substituting the vapor pressure at the normal boiling point (760 torr), to isolate the temperature: \[\ln\frac{760}{213} = -\frac{2.985 \times 10^{4}}{8.314}\left(\frac{1}{T_{nbp}} - \frac{1}{313.15}\right)\] Now we can solve for the normal boiling point temperature, \(T_{nbp}\): \[T_{nbp} = \frac{1}{\frac{1}{313.15} - \frac{8.314}{2.985 \times 10^{4}}\ln\frac{760}{213}} = 350.35\,\mathrm{K}\]

(Step 5: Convert the normal boiling point temperature back to Celsius)

Finally, convert the normal boiling point temperature, \(T_{nbp}\), from Kelvin to Celsius: \[T_{nbp} = 350.35 - 273.15 = 77.20^{\circ}\mathrm{C}\] Therefore, the normal boiling point of \(\mathrm{CCl}_{4}\) is approximately \(77.20^{\circ}\mathrm{C}\).

Key concepts

Vapor Pressure

Vapor pressure is a measure of a liquid's tendency to evaporate into a gas. It is the pressure exerted by the vapor when the liquid and vapor states are in equilibrium. At any given temperature, every liquid has a characteristic vapor pressure that increases as temperature rises, because more molecules have enough energy to overcome the liquid’s surface tension and escape into the gaseous state.

Understanding vapor pressure is crucial when studying the Clausius-Clapeyron equation because it's one of the key variables that correlate to temperature changes. In the exercise provided, we are given two vapor pressures at different temperatures which are then used to calculate the molar heat of vaporization and the normal boiling point of carbon tetrachloride.

Molar Heat of Vaporization

The molar heat of vaporization (\( L \)) is the amount of heat required to convert one mole of a liquid into vapor without a temperature change. It's a characteristic property that differs from substance to substance. This concept is at the heart of the Clausius-Clapeyron equation and plays a central role in determining how vapor pressure changes with temperature.

The step-by-step exercise exemplifies calculating the molar heat of vaporization for carbon tetrachloride by using the vapor pressures at two different temperatures. Knowing the molar heat of vaporization allows us to predict how a liquid will behave under various temperature conditions and is key in identifying the boiling point.

Normal Boiling Point

The normal boiling point of a liquid is the temperature at which its vapor pressure equals one atmosphere (101.325 kPa or 760 torr). At this point, vapor bubbles can form within the liquid, and it begins to transform into a gas (boil).

Using the Clausius-Clapeyron equation, the normal boiling point can be determined by setting the vapor pressure variable to the standard atmospheric pressure and solving for the temperature. In our exercise, we found the normal boiling point of carbon tetrachloride by using the molar heat of vaporization and the given vapor pressures at two specific temperatures. This temperature is crucial for applications involving boiling, distillation, and chemical reactions.

Temperature Conversion

Temperature conversion is necessary when using the Clausius-Clapeyron equation since it requires absolute temperature measurements in Kelvins (K). The Celsius (°C) to Kelvin conversion is one of the simplest temperature conversions, requiring only the addition or subtraction of 273.15.

In the textbook problem solution, the temperatures provided in Celsius needed to be converted to Kelvins for accurate inclusion in the Clausius-Clapeyron equation. Similarly, after finding the normal boiling point in Kelvins, the final step is to convert it back into Celsius to provide an understandable result relative to everyday temperature measurements. Proper temperature conversion avoids errors in calculations and ensures that the results are correct and meaningful.