Chemistry - Thermochemistry Question with Solution | TestHub
Standard enthalpy of formation of gaseous ethane, ethene and benzene from gaseous atoms are -2839, -2275 and -5506 kJ/mol, respectively. The bond enthalpy of C-H bond is 412 kJ/mol. The magnitude of resonance energy of benzene, compared with one Kekulé structure is
Options:
Answer:
Solution:
For ethane, with six C–H bonds and one C–C bond, the bond energy of the C–C bond, denoted as BE(C–C), can be calculated as:
BE(C–C) = 2839 - 6(412) = 367 kJ/mol
For ethene, with four C–H bonds and one C=C bond, the bond energy of the C=C bond, denoted as BE(C=C), is:
BE(C=C) = 2275 - 4(412) = 627 kJ/mol
A single Kekulé structure of benzene contains six C–H bonds, three C–C bonds, and three C=C bonds. The total energy for one Kekulé structure is:
6(412) + 3(367) + 3(627) = 2472 + 1101 + 1881 = 5454 kJ/mol
Given the actual energy of benzene as 5506 kJ/mol, the resonance energy is calculated as:
5506 - 5454 = 52 kJ
This corresponds to option (b).
