You think you know;but do you really know…

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Presentation transcript:

You think you know;but do you really know… 1. Are you familiar with the various types of intermolecular bonds? Y or N?

You think you know; but do you really know… 2. Can you differentiate which would be the dominant force of attraction in various situations comparing molecules? Y or N?

Dipole-Dipole You think you know; but do you really know… 3. What would the dominant force of attraction be in each of the following situations? A. Two polar molecules with small and similar molar masses that do not contain hydrogen… Dipole-Dipole

You think you know; but do you really know… 3. What would the dominant force of attraction be in each of the following situations? B. Two polar molecules with small and similar molar masses; one containing hydrogen… Dipole-Dipole and perhaps H-bonding!

H-bonding! You think you know; but do you really know … 3. What would the dominant force of attraction be in each of the following situations? C. Two polar molecules with small and similar molar masses; one containing hydrogen bonded to nitrogen the other to chlorine. (note Cl and N have identical E.N.’s) H-bonding!

LDF You think you know; but do you really know … 3. What would the dominant force of attraction be in each of the following situations? D. Two nonpolar molecules that may or may not contain hydrogen but definitely do not contain N, O or F. LDF

Look at the ball and stick models of dimethyl ether and ethyl alcohol Look at the ball and stick models of dimethyl ether and ethyl alcohol. Then determine which has the lowest volatility. dimethyl ether CH3OCH3 ethyl alcohol C2H5OH Due to H-bonding!

Q:Why and How Does Evaporation Take Place? A:In every sample of a liquid the kinetic energy of the particles is described by a distribution. Some particles have low KE some have high KE and most have around the average KE. (this is called a Maxwell-Boltzman Distribution) Most particles don’t have enough energy to break away, but these particles do! # of particles K.E. Threshold Energy: energy needed for particles to breakaway from the surface of a liquid evaporation = (at some point in time) rate (This means an equilibrium has been established) condensation time = represents particles that have broken away from the surface of the liquid

K.E. evaporation = rate condensation time Important Point #1:The higher the attractive force strength the lower the vapor pressure! # of particles K.E. evaporation = rate condensation time When the condensation rate catches up to the evaporation rate, the # of vapor particles above the liquid is constant; pressure from those particles (i.e. vapor pressure) can then be measured. Another property of a liquid related back to the attractive force strength!

T2>T1 K.E. evaporation = rate condensation time vp temp Particles with enough energy to break away increases in number with higher temp. # of particles K.E. Threshold energy needed for particles to breakaway evaporation = rate condensation time Typical Vapor Pressure Curve Heat up the liquid # of particles above the liquid is constant; vapor pressure can be measured. vp (higher at higher temp.s) temp

Important Point #2: Equilibrium Vapor Pressure Curve for Water Notice that the vapor pressure of water is 760 mm of Hg at 100 ° C. Recall that normal atmospheric pressure is 760 mm of Hg. Therefore…Boiling occurs when the vapor pressure of a liquid is equal to the atmospheric pressure… … and the boiling point is the temperature at which this occurs.

Since (for a straight line) y = mx + b ln vp = (-DHv/R)(1/T) + b At two different temperatures: ln vp1 = (-DHv/R)(1/T1) + b These y intercepts are equal ln vp2 = (-DHv/R)(1/T2) + b Since the equations have equal y intercepts, through algebraic manipulation the equations can be combined to give: ln (vp1/vp2) = (DHv/R)(1/T2-1/T1) The Clausius-Clapeyron Eq. The slope of the line turns out the be equal to –DHv/R ln vp vp temp 1 / T This exponential relationship can be straightened out using natural logs Since (for a straight line) y = mx + b ln vp = (-DHv/R)(1/T) + b

ln (vp1/vp2) = (DHv/R)(1/T2-1/T1) The Clausius-Clapeyron Eq. Important Point #3: This equation shows the relationship between vapor pressure and heat of vaporization. Both properties that are related the attractive force strength between the particles!

Calculate the enthalpy of vaporization of methanol using the following data: Temp.(°C) V.P. (in mm Hg) You can see VP increase with temp. -6.0 20.0 5.0 40.0 12.1 60.0 21.2 100.0 Use these two temps and vapor pressures and solve for the heat of vaporization using the C-C equation

Find DHv by manipulation of C-C equation: ln(vp1/vp2) R DHv = (1/T2 – 1/T1) DHv = ln(40mm/60mm)8.31J/mol K (1/285.1 K – 1/278 K) DHv = -3.37J/mol K (-8.96 x 10-5 K) DHv = 37600 J/mol DHv = 37.6 kJ/mol