Chemistry - Thermochemistry Question with Solution | TestHub
The enthalpy of formation of liquid methyl alcohol in kJ/mol, using the following data (in kJ/mol):
Heat of vaporization of liquid methyl alcohol = 38.
Heat of formation of gaseous atoms from the elements in their standard states: H, 218; C, 715; O, 249.
Average bond energies: C-H, 415; C-O, 356; O-H, 463.
Options:
Answer:
Solution:
To determine the enthalpy of formation for CH₃OH, we first atomize the elements and then form the gaseous CH₃OH from these atoms.
The atomization of elements for CH₃OH is represented by the reaction:
C(s) + 2H₂(g) + 1/2O₂(g) → C(g) + 4H(g) + O(g)
The energy required for this process is calculated as:
715 + 4(218) + 249 = 1836 kJ
Next, we calculate the energy released when forming gaseous CH₃OH from its constituent atoms. This involves the formation of 3 C-H bonds, 1 C-O bond, and 1 O-H bond:
3(C-H) + 1(C-O) + 1(O-H) = 3(415) + 356 + 463 = 2064 kJ
The enthalpy of formation for gaseous CH₃OH, Δ Hf(g), is the difference between the energy absorbed during atomization and the energy released during bond formation:
Δ Hf(g) = 1836 - 2064 = -228 kJ/mol
Finally, to find the enthalpy of formation for liquid CH₃OH, we consider the condensation process, which releases an additional 38 kJ/mol:
Δ Hf(liquid) = -228 - 38 = -266 kJ/mol
This corresponds to option (d).
