Question

Diethyl ether has a \(\Delta H_{\mathrm{vap}}^{\circ}\) of \(29.1 \mathrm{~kJ} / \mathrm{mol}\) and a vapor pressure of 0.703 atm at \(25.0^{\circ} \mathrm{C}\). What is its vapor pressure at \(95.0^{\circ} \mathrm{C} ?\)

Short answer

The vapor pressure at 95.0°C is 6.57 atm.

Step-by-step solution

Define the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation relates the vapor pressure and temperature of a substance and is given by \ \ \ \[\ln \left( \frac{P_2}{P_1} \right) = \frac{\Delta H_{\mathrm{vap}}^{\circ}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] where \ \( P_1 \) is the initial vapor pressure, \ \( P_2 \) is the final vapor pressure, \ \( T_1 \) is the initial temperature (in Kelvin), \ \( T_2 \) is the final temperature (in Kelvin), \ \( \Delta H_{\mathrm{vap}}^{\circ} \) is the enthalpy of vaporization, \ and \ \( R \) is the gas constant \ \(8.3145 \ \mathrm{J} /(mol \ K) \).

Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin using the formula \ \( K = ^\circ C + 273.15 \). Thus, \ \(T_1 = 25 + 273.15 = 298.15 \ K \). \ \( T_2 = 95.0 + 273.15 = 368.15 \ K \).

Set Known Values

Set up the known values from the problem: \ \( P_1 = 0.703 \ atm \), \ \( \Delta H_{\mathrm{vap}}^{\circ} = 29.1 \ \mathrm{kJ} / \ \mathrm{mol} = 29100 \ J / \ mol \), \ \( R = 8.3145 \ J / (mol \ K) \), \ \(T_1 = 298.15 \ K \), \ and \ \(T_2 = 368.15 \ K \).

Substitute Values into the Equation

Substitute the known values into the Clausius-Clapeyron equation: \ \ \ \ \[\ln \left( \frac{P_2}{P_1} \right) = \frac{29100}{8.3145} \left( \frac{1}{298.15} - \frac{1}{368.15} \right) \].

Calculate the Right-Hand Side

Calculate the value inside the bracket first: \ \ \ \ \(\frac{1}{298.15} - \frac{1}{368.15} = 0.003354016 - 0.002715266 = 0.00063875 \ K^{-1}\). \ Now multiply by \ \( \frac{29100}{8.3145} = 3500.9 \). So the complete term becomes \ \( 3500.9 \times 0.00063875 = 2.2357 \).

Solve for \(P_2\)

Now solve for \( P_2 \): \ \ \ \ \(\ln \left( \frac{P_2}{0.703} \right) = 2.2357 \) \ Exponentiate both sides to solve for \ \( P_2 \): \ \ \ \ \( \frac{P_2}{0.703} = e^{2.2357} \) \ \(P_2 = 0.703 \times e^{2.2357} \)

Compute the Final Value

Calculate \ \( e^{2.2357} \) which is approximately \ \( 9.349 \). Therefore, \ \(P_2 = 0.703 \times 9.349 = 6.57 \ \mathrm{atm} \).

Key concepts

vapor pressure

Vapor pressure is the pressure exerted by the vapor of a liquid in equilibrium with its liquid phase at a given temperature. It is an important property in thermodynamics and can tell us how easily a substance will evaporate. Higher vapor pressure means the liquid evaporates more quickly. For example, diethyl ether has a certain vapor pressure at 25°C, and by knowing this, we can predict its vapor pressure at other temperatures.

enthalpy of vaporization

Enthalpy of vaporization \(\triangle H_{\text{vap}}^{\text{°}}\) is the amount of energy required to turn 1 mole of a liquid into vapor at its boiling point under standard pressure. This energy overcomes the intermolecular forces in the liquid, allowing the molecules to enter the gas phase. In the case of diethyl ether, the \(\triangle H_{\text{vap}}^{\text{°}}\) is 29.1 kJ/mol, which helps us understand the energy needed to vaporize it at a given temperature.

temperature conversion

Temperature conversion is crucial in thermodynamic equations because many formulas, like the Clausius-Clapeyron equation, require temperatures in Kelvin. The conversion formula is simple: \(K = °C + 273.15\). So, if you have temperatures in Celsius, you must convert them to Kelvin to apply the equation correctly. For instance, converting 25°C to Kelvin results in 298.15 K.

thermodynamics

Thermodynamics is the study of energy, its conversion, and its relation to matter. In this context, the Clausius-Clapeyron equation is a thermodynamic equation that relates the change in vapor pressure with temperature. By understanding the enthalpy of vaporization and how temperature affects vapor pressure, we can predict how a substance behaves under different thermal conditions. This equation plays a vital role in fields like physical chemistry and engineering, helping to optimize processes involving phase changes.