Spring 2024 CHEM 123 Recitation Activity #2 - KEY (1)

CHEM 123 Recitation Activity #2 - KEY A. The Relationship of Vapor Pressure and Temperature -- The Clausius-Clapeyron Equation Vapor pressure increases with temperature for all substances. The relationship between these two quantities is shown in the Clausius-Clapeyron equation. 𝐥𝐧 𝑷 = −∆𝑯 𝒗𝒂? 𝑹 ( ? 𝑻 ) + 𝒄 or 𝐥𝐧 𝑷 ? 𝑷 ? = −∆𝑯 𝒗𝒂? 𝑹 ( ? 𝑻 ? − ? 𝑻 ? ) Example 1: If the vapor pressure of water at 100.0 o C is 1.00 atm, what is the vapor pressure of water at 50.0 o C. Δ H vap = 40.7 kJ/mol Plan: The question asks for the vapor pressure at 50 o C. The equation that relates vapor pressure to temperature is: 𝐥𝐧 𝑷 ? 𝑷 ? = −∆𝑯 𝒗𝒂? 𝑹 ( ? 𝑻 ? − ? 𝑻 ? ) We are given one vapor pressure and its corresponding temperature and asked for the vapor pressure at a new temperature. Enter the correct values and be careful with the units. Solution: 𝐥𝐧 𝑷 ? 𝑷 ? = −∆𝑯 𝒗𝒂? 𝑹 ( ? 𝑻 ? − ? 𝑻 ? ) 𝐥𝐧 𝑷 ? ?. ?? 𝒂𝒕? = −????? ? ??? ?. ??? ? ???? ( ? ???? − ? ???? ) 𝐥𝐧 𝑷 ? = −?. ?? 𝑷 ? = 𝒆 −?.?? = ?. ?? 𝒂𝒕? 1. For any liquid, what is the vapor pressure of the substance at its normal boiling point? 1atm, 760 torr, 101.3 kPa 2. Suppose 10mL of water at 20 o C is added into a 100 mL cylinder fitted with a piston that you can slide up and down inside the cylinder, and the vapor pressure inside the cylinder is measured to be 17.5 mm Hg. Indicate what will happen to the vapor pressure of the water under the following conditions: a) 10mL of additional water is added to the system. Vapor pressure of water will stay the same . b) The volume of the piston is doubled. Vapor pressure of water will stay the same . c) The pressure inside the piston is increased by adding argon. Vapor pressure of water will stay the same . d) The temperature inside the cylinder is increased to 50 o C. Vapor pressure of water will increase .

3. If water’s vapor pressure at 100 o C is 1.00 atm, what ’ s water’s vapor pressure at 150 o C? Δ H vap = 40.7 kJ/mol ln ? 2 1 ??? = −40700 ?/??? 8.314 ?/???? ( 1 423.15 ? − 1 373.15 ? ) ? 2 = 4.71 ??? 4. If water’s vapor pressure at 100 o C is 1.00 atm, to what temperature should water be cooled to achieve a vapor pressure of 0.100 atm? ln 0.100 ??? 1 ??? = −40700 ?/??? 8.314 ?/???? ( 1 𝑇 2 − 1 373.15 ? ) 𝑇 2 = 317? 5. Calculate water’s boiling temperature on Mt. Everest (29,032 ft) where the atmospheric pressure is 253 torr. At 100 o C the vapor pressure of water is 760 torr, determine the temperature for 253 torr. ln 253 𝑡?𝑟𝑟 760 𝑡?𝑟𝑟 = −40700 ?/??? 8.314 ?/???? ( 1 𝑇 2 − 1 373.15 ? ) T 2 = 344 K 6. If a liquid has a vapor pressure of 50.0 mmHg at 25.0 o C and ΔH vap = 15.60 kJ/mol, what ’ s the liquid’s normal boiling point? Normal boiling point is when vapor pressure = standard pressure = 760 torr ln ( 760 ??𝐻? 50.0 ??𝐻? ) = −1876? −1 ( 1 𝑇 2 − 1 298.15 ) 𝑇 2 = 525 ?

The Clausius-Clapeyron equation shows the linear relationship between the natural log of vapor pressure of a substance (lnP vap ) and the inverse temperature in Kelvin (1/T). When plotted, the slope of the line is -  H vap / R. 𝐥𝐧 𝑷 = −∆𝑯 𝒗𝒂? 𝑹 ( ? 𝑻 ) + 𝒄 ?𝒓 𝐥𝐧 𝑷 ? 𝑷 ? = −∆𝑯 𝒗𝒂? 𝑹 ( ? 𝑻 ? − ? 𝑻 ? ) Example : The following graph shows the relationship between vapor pressure and temperature for a substance. Determine the normal boiling point and enthalpy of vaporization. Plan: The normal boiling point is when vapor pressure equals 1 atm/101.3kPa/760 torr. Since the pressure was in kPa, we will find the temperature at 101.3 kPa. The enthalpy of vaporization is determined from the slope of the line with the slope being -  H/R with R in standard units of 8.314 J/molK. Solution: Take the natural log of the standard pressure: ln 101.3kPa = 4.618. Find 4.618 on the y axis. It corresponds to a value of 0.00266 on the x axis. So 1/T = 0.00266 T = 376K or 103 o C. Enthalpy of vaporization: Find two points: For example, the points where the line crosses lnP = 4 and lnP = -1: y 1 = 4.0 x 1 = 0.00280 y 2 = -1.0 x 2 = 0.003748 Slope = (4.0 - -1.0)/(0.00280 K -1 – 0.003748 K -1 ) = -5274 K -5274 K = -  H vap / 8.314 J/molK  H vap = 43850 J/mol = 44 kJ/mol -1 0 1 2 3 4 5 0.0025 0.0027 0.0029 0.0031 0.0033 0.0035 0.0037 0.0039 ln P (kPa) 1/T (K -1 )

7. Suppose a plot of ln P vap vs. 1/T data for a liquid has a slope of -1876 K. What ’ s the liquid’s enthalpy of vaporization? - ΔH vap /R = -1876 K - ΔH vap /8.314J/molK = -1876 K - ΔH vap =-15597 J/mol ΔH vap = 15.60 kJ/mol 8. A plot of lnP vap versus 1/T for a liquid is shown below: a) Determine the liquid’s enthalpy of vaporization. Find two points: For example, the points where the line crosses 1/T = 0.0035 and 1/T = 0.0045 y 1 = -1.1 x 1 = 0.0035 y 2 = -4.2 x 2 = 0.0045 Slope = (-1.1 - -4.2)/(0.0035 K -1 – 0.0045 K -1 ) = -3100 K -3100 K = -  H vap / 8.314 J/molK  H vap = 25773 J/mol = 26 kJ/mol b) Determine the liquid’s normal boiling point. Normal boiling point is when vapor pressure = 1atm (ln 1atm = 0). We will use the slope -3100 K and one of our points (0.0045, -4.2) to determine this. (0 - -4.2)/(x K -1 – 0.0045 K -1 ) = -3100 K 1/x = 0.003145 K -1 x = 320K -5 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.003 0.0035 0.004 0.0045 0.005 ln P (atm) 1/T (K -1 )