Question

The boiling point of methanol is $65.0^{\circ} \mathrm{C}$ and the standard enthalpy of formation of methanol vapor is $-201.2 \mathrm{~kJ} / \mathrm{mol} .$ Calculate the vapor pressure of methanol (in $\mathrm{mmHg}$ ) at $25^{\circ} \mathrm{C}$. (Hint: See Appendix 2 for other thermodynamic data of methanol.)

Name: The boiling point of methanol is 65.0^∘C and the standard enthalpy of formation of methanol vapor is -201.2 kJ / mol . Calculate the vapor pressure of methanol (in mmHg ) at 25^∘C. (Hint: See Appendix 2 for other thermodynamic data of methanol.)
Uploaded: 2022-01-28T16:59:09-08:00
Duration: 2 min 15 s
Channel: Amita Prajapat
Description: The boiling point of methanol is 65.0^∘C and the standard enthalpy of formation of methanol vapor is -201.2 kJ / mol . Calculate the vapor pressure of methanol (in mmHg ) at 25^∘C. (Hint: See Appendix 2 for other thermodynamic data of methanol.)

          The boiling point of methanol is $65.0^{\circ} \mathrm{C}$ and the standard enthalpy of formation of methanol vapor is $-201.2 \mathrm{~kJ} / \mathrm{mol} .$ Calculate the vapor pressure of methanol (in $\mathrm{mmHg}$ ) at $25^{\circ} \mathrm{C}$. (Hint: See Appendix 2 for other thermodynamic data of methanol.)

Added by Shane H.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Amita Prajapat

01/28/2022

Step 1

Step 1: Use the Clausius-Clapeyron equation: \[\ln\left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\] Show more…

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The boiling point of methanol is $65.0^{\circ} \mathrm{C}$ and the standard enthalpy of formation of methanol vapor is $-201.2 \mathrm{~kJ} / \mathrm{mol} .$ Calculate the vapor pressure of methanol (in $\mathrm{mmHg}$ ) at $25^{\circ} \mathrm{C}$. (Hint: See Appendix 2 for other thermodynamic data of methanol.)