

Evaluating Environmental Fate: 


Chapter 5. Example Problem


Example problem 52 Estimate the vapor pressure at 298 K for toluene (a liquid) and naphthalene (a solid). Solution Toluene has the molecular structure CH_{3} C_{6}H_{5} and in Example 51, it’s boiling point was estimated to be 399 K. The experimental value for the boiling point is 384 K. We will estimate the vapor pressure using both the predicted and the experimental value for boiling point. Using the predicted value of 399 K: C= 18 + 0.19 T_{b }= 57.8 A = K_{F} (8.75 + R ln T_{b} ) = 1.0(8.75 + 1.987 * ln (399)) = 20.6 Ln P_{vp} = 3.83; P_{vp} = 0.021 atm =16 mm Hg Repeating the calculation for the predicted boiling point leads to a vapor pressure estimate of 19 mm Hg. Naphthalene has the formula C_{10}H_{8 } and is a solid with a melting point of 81^{o}C. The boiling point can be estimated from the methods described earlier in this section. The uncorrected group contribution estimate is: T_{b}= 198.2 + 2(45.46) + 8(28.53) = 517 K The corrected value is: T_{b}= 505 K Applying Equation 510: ln P = (4.4 + ln T_{b} )[1.803 (T_{b} /T  1)  0.803 ln(T_{b} /T)]  6.8 (T_{m} /T  1) P = 4.4 * 10^{5} atm = 0.03 mm Hg 





Chapter 5. Sample Homework Problem Estimate the properties listed in the table given below. 



For each of the properties, comment on whether molecular weight or the presence of a hydrogen bonding group has a more pronounced effect on chemical properties. 





Chapter 5. Lecture Notes 

