Chemistry - Thermochemistry Question with Solution | TestHub
Given the bond dissociation enthalpy of CH3-H bond as 103 kcal/mol and the enthalpy of formation of CH4(g) as -18 kcal/mol, find the enthalpy of formation of methyl radical. The dissociation energy of H2(g) into H (atoms) is 103 kcal/mol.
Options:
Answer:
Solution:
The dissociation of methane can be represented by the following reaction:
CH₄(g) → CH₃(g) + H(g), with Δ H = +103 kcal.
The enthalpy of formation for a hydrogen atom, Δf H(H, g), is half the bond dissociation energy of H₂, which is 103/2 = 51.5 kcal/mol.
To find the enthalpy of formation for the methyl radical, Δf H(CH₃, g), we use the equation:
Δ H = Δf H(CH₃, g) + Δf H(H, g) - Δf H(CH₄, g)
Rearranging to solve for Δf H(CH₃, g):
Δf H(CH₃, g) = Δ H - Δf H(H, g) + Δf H(CH₄, g)
Given Δf H(CH₄, g) = -18 kcal/mol, we substitute the values:
Δf H(CH₃, g) = 103 - 51.5 + (-18) = 33.5 kcal/mol
This corresponds to option (b).
