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Benson’s Group additivity of Cp is available for gases; what about for solid and liquid?

Benson’s Group additivity of Cp is available for gases; what about for solid and liquid?. references: “A Group Additivity Approach for the Estimation of Heat Capacities of Organic Liquids and Solids at 298 K,” Structural Chemistry 1993 , 4 , 261-269.

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Benson’s Group additivity of Cp is available for gases; what about for solid and liquid?

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  1. Benson’s Group additivity of Cp is available for gases; what about for solid and liquid?

  2. references: “A Group Additivity Approach for the Estimation of Heat Capacities of Organic Liquids and Solids at 298 K,” Structural Chemistry1993, 4, 261-269. Heat Capacity Corrections to a Standard State: A Comparison of Some New and Literature Methods for Organic Liquids and Solids,” Structural Chemistry1993, 4, 261-269. available on the web at www.umsl.edu/~jscumsl/. Go to: Access to research papers and ...

  3. Application of Group Additivity to Estimate Cp(l) and Cp(c) Protocol: 1. Group values are defined in terms of their hybridization and substitution pattern. 2. Primary, secondary, tertiary and quaternary carbons are defined in terms of the number of hydrogens attached, 3, 2, 1, 0. 3. Aromatic, aliphatic and cyclic carbons are treated separately. 4. Functional groups are treated separately.

  4. Calculate the heat capacity of [2.2] spiropentane at T = 298.15 K. Cp = quat sp3 C +4 sec sp3 Cp(l) = 18 + 4(25.9) = 121.6 J mol-1K-1 Cp(c) = 6.1 + 4(24.6) = 104.5 J mol-1K-1 Experimental value Cp(l) = 134.5 J mol-1K-1

  5. When are Cp(l) and Cp(c) values useful?

  6. Suppose you measured Hv(325 K) of spiropentane and you wanted to correct the number to T = 298.15 K. How would you do it?

  7. What do you do in the case where the group value or correction is not available? 1. Find data to identify a group value. 2. Look for another approach.

  8. At T = 298.15 K [Cp(l) - Cp(g)]expt = 10.26 + 0.26 Cp(l)calc(298)

  9. At T = 298.15 K [Cp(c) - Cp(g)]expt = 0.71 + 0.15 Cp(c)calc(298)

  10. Hfus(298) = Cp(c) [Tfus-298] + Hfus(Tfus) + Cp(l)[298-T] Hfus(298) = Hfus(Tfus) + [Cp(c) - Cp(l)][Tfus-298]

  11. [Cp(l) - Cp(g)]expt = 10.26 + 0.26 Cp(l)calc (1) [Cp(c) - Cp(g)]expt = 0.71 + 0.15 Cp(c)calc (2) Eq (2) - Eq (1) = [Cp(c) - Cp(l) ] = [0.15 Cp(c) - 0.26 Cp(l) - 9.83] (3) Hfus(298) = Hfus(Tfus) + [Cp(c) - Cp(l)][Tfus-298] Hfus(298)=Hfus(Tfus)+[0.15Cp(c) - 0.26Cp(l)-9.83][Tfus-298]

  12. How accurate are eq. (1), (2), and (3)? 95 % confidence level = 2 standard deviations For eq. (1): 30 J mol-1 K-1 For eq. (2):  1/3 the correction

  13. Suppose you measured Hv(325 K) of spiropentane and you wanted to correct the number to T = 298.15 K. How would you do it? Hv(298 K) = Hv(325 K) +[10.58 +.26Cp(l)][325-298] Cp(l) = 14 + 4(31.9) = 141.6 J mol-1K-1 Hv(298 K) = Hv(325 K) +[4100 J mol-1]

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