What is the enthalpy of vaporization of a compound that has a vapor pressure of 151 mmHg at 247 K and 1.52 mmHg at 187 K? (R = 8.31 J/(K· mol))

Question

Quizplus · Accepted Answer

The enthalpy of vaporization can be calculated using the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at two different temperatures to its enthalpy of vaporization ($\Delta H_{vap}$). The equation in its logarithmic form is:$$\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$Given:- $P_1 = 1.52 \, \text{mmHg}$- $P_2 = 151 \, \text{mmHg}$- $T_1 = 187 \, \text{K}$- $T_2 = 247 \, \text{K}$- $R = 8.31 \, \text{J/(K·mol)}$First, convert pressures to the same units if necessary (in this case, both are already in mmHg, so no conversion is needed). Then, solve for $\Delta H_{vap}$:$$\ln\left(\frac{151}{1.52}\right) = -\frac{\Delta H_{vap}}{8.31}\left(\frac{1}{247} - \frac{1}{187}\right)$$Solving this equation for $\Delta H_{vap}$ gives a value in J/mol, which can then be converted to kJ/mol by dividing by 1000. The correct calculation leads to an enthalpy of vaporization close to 29 kJ/mol, indicating that option D is correct.

What Is the Enthalpy of Vaporization of a Compound That