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KU CHEM 170 - Vapour Pressure, Clausius Clapeyron Equation, Boiling Point

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Chem170 1st Edition Lecture 9 Outline of Last Lecture I. Upcoming Lab: Measuring the Molecular Weight of AirII. Last Lecture: VocabularyIII. Vapor Pressure of LiquidIV. Clausius Clapeyron EquationV. Boiling PointOutline of Current Lecture VI. Last Lecture: Vapour Pressure, Clausius Clapeyron Equation, Boiling PointVII. VocabularyVIII. ExamplesCurrent LectureCh. 12 – Liquids and SolidsLast Lecture: Clausius Clapeyron Equationln(P)=(∆ HvapR)1T+CVapor pressure is the pressure that the water vapor exerts when the condensation reaction and the vaporization reaction are at equilibrium.Vapor pressure is temperature dependent, because as the temperature increases energy is added into the system which allows more molecules to break free of the intermolecular forces more frequently.Boiling Point is the temperature at which the vapor pressure is equal to the pressure of the surrounding atmosphere. The boiling point is directly related to the strength of the intermolecular forces in a liquid.The normal Boiling Point occurs when the vapor pressure of the liquid equals the atmospheric pressure.There are 2 definitions because the atmospheric pressure varies from location to location.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.ex. At 1 atm (or 760 mmHg) water usually boils at 100°C. In Colorado the atmospheric pressure was 624.79mmHg. What was the boiling point?ln(624.79 mmHg)=−44KJmol0.082L× atmk ×mol(1T1)+C1+−[ln(760 mmHg)=−44KJmol0.082L× atmk × mol(1T2)+C2]¿ln(624.79 mmHg)−ln (760 mmHg)=−44KJmol0.082L ×atmk ×mol(1T1)−−44KJmol0.082L ×atmk × mol(1T2)¿ln(624.79 mmHg)−ln (760 mmHg)=−44KJmol0.082L ×atmk ×mol(1T1−1T2)¿ln(624.79760mmHg)=−44KJmol0.082L ×atmk × mol(1372−1T2)T2=368 K =95 ℃ex. Would C7H16 or C8H18 have the higher boiling point?C8H18 because it has stronger London dispersion forces. It has stronger London dispersion forces, because it is a larger moleculeFun fact: C7H16 boils at 98°C and C8H18 boils at 126°Cex. Would NH2NH2 and CH3CH3 have the higher boiling point?NH2NH2 would have a higher boiling point, because it has stronger intermolecularforces. NH2NH2 has hydrogen bonding and London dispersion forces acting between molecules while CH3CH3 only has London dispersion forces acting between molecules.Vocabulary:VaporizationH2O(l )→ H2O(g)ΔHvapCondensationH2O(g)→ H2O(l)FusionH2O(s)→ H2O(l)ΔHfusionFreezingH2O(l )→ H2O(s)Sublimation H2O(s)→ H2O(g )ΔHsup∆ E=q+w=q+P ∆ V → qp=∆ E+P ∆ V =∆ H∆ Hvap=Hvapor−Hliquid-A phase change is a physical change in the substance at a constant temperature.-Evaporation can cool surfaces, because as the high energy molecules leave the liquid, the molecules that remain behind are low energy and thus a lower temperature.-Note that ΔH is a state function so you can add different ΔH together.ex.H2O(s)→ H2O(l)∆ Hfusion+−[H2O(l)→ H2O(g)]∆ HvapH2O(s)→ H2O(g )∆ H¿therefore, ∆ H¿=∆ Hfusion+∆ Hvapex. How much heat is required to melt 100g of H2O(s) at 0°C? Given that ΔHfusion=6.01KJmol.Fusion H2O(s)→ H2O(l)ΔHfusionq=(100 g)(1 mol18 g)(6.01 KJ1 mol)=33.4 KJex. How much heat is required to raise 100g of H2O(s) from 25°C to 50°C?q=ms ∆ T =(100 g)(4.18Jg ×℃)(50 ℃ −25 ℃)=10.4 KJex. How much heat is required to change 100g of ice from 0°C to 70°C?Fusion H2O(s)→ H2O(l)ΔHfusionq=(100 g)(1 mol18 g)(6.01 KJ1 mol)=33.4 KJChange in temperatureq=ms ∆ T =(100 g)(4.18Jg ×℃)(70 ℃ −0 ℃)=29 KJTotal: 33.4 KJ +29 KJ=62.7


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