September 8 /10 2004

Fill in the table (ln(p) and 1/T) and use the table to plot. Find the slope of the graph and from the slope determine (delta)H or vaporization.

2. Choose any two temperatures and the corresponding pressure from #1. Use this to compute (delta) H of vaporization. Compare this answer to that in #1.

3. The vapor pressure of SiCl is 100 torr at 5.4 Cand the normal boiling point is 58.6 C. What is (delta) H of vaporization for the compound, in kJ/mol

4. How much energy is released when 15.3 g of steam at 115.0 C is condesned to give liquid water at 75.0 C? The heat of vaporization of water is 40.67 kJ/mol and the molar heat capacity is 75.3 J/(K mol) for liquid water and 33.6 J/(K mol) for the vapor.

5. A solution is prepared by dissolving 17.84 g of glucose (C₆H₁₂O₆) in 250 g of water. The density of the solution is 1.16 g/ml. Calculate molality, molarity and mass percent for this solution.

• 1. Data was temperature and vapor pressure.
  • table ( listed below) said to compute
    • ln(P) the natural log,
    • and 1/T where T is the temperature, in Kelvin
  • plotting this data, ln(P) on y-axis and 1/T as the x-axis
    • yields a straight line with a slope = -3634.2
  • Excel computed the slope as -3634.2 and the slope = ( -delta-H /8.314)
  • delta-H = - 8.314 * (- 3634.2)
  • delta-H= 30214.7388 J/mole or 30.2147388 KJ as enthalpy of vaporization

T(K)__ P(vapor)___ln P_____1/T

263___80.1___4.383275854___ 0.003802281

273___133___4.890349128___ 0.003663004

283___213___5.361292166___ 0.003533569

293___329.6__5.797879798___ 0.003412969

• 2. Using the first and last data points above, fit data to equation
  • lnP2 - lnP1 = 1.4146039
  • (1/T2) - (1/T1)= -0.0003893
    • slope =1.4146039 /(-0.0003893) =-3633.
  • -DH/R= -3633.
  • DH=- slope *8.314/1000 = 30.20974079 kJ/mol

• 3. Same type of problem as #2, using new data
  • T( deg C) P T(K) 1/T ln P
  • 5.4 100 278.6 0.003589375 4.605170186
  • 56.8 760 330 0.003030303 6.633318433

• slope = -3627.702209
  • 30.16071617 KJ
  • -R * slope /1000

• 4. Energy released… in three stages
  • #moles= 15.8 / 18.8 = 0.878
  • a. cool steam from 115.0 to 100 C
    • =8.78 mol x 33.6 j/K x 15 K
    • 442.512 Joule
  • b. condense steam (not KJ value)
    • =0.878 mol x 40.67 kJ/mol x 1000 J/KJ
    • 35708.26 Joules
  • c. cool the water to 75 C (25 deg change)
    • =0.878 mol x 73.3 J/K-mol x 25
    • 1608.935 Joules
  • now add ( and divide by 1000 to get KJ)
  • 37.759707 kJoule

• 5. Solution-- molarity, molality and mole fraction
  • mass of solute, 17.84g
    • FW 180
    • amount of solute =0.0991 moles

• solvent(g) 250 g (or ml)
  • FW of water is 18
  • 13.89 moles of solute

• density 1.16 g/ml
  • total mass 267.84
  • mass/density = volume 230.9 ml

• now compute concentrations
  • molarity = 0.0991 mol / 0.2309 L = 0.429 Molar
  • molality= 0.0991 mol / 0.25 kg= 0.396 molal
  • mass % 17.84 / 267.84 *100= 6.660692951 %
  • mole fr Solute C / (0.0991+13.89) = 13.89 / 13.988 = 0.007085438