To calculate the bond enthalpy of the
C−H bond in methane
(CH4), we need to follow the relationship given by Hess's Law. The method involves summing the energies associated with the formation of one mole of methane from its elements in their standard states.
Given:
- Standard enthalpy of formation of
CH4:∆Hf(CH4)=−74.8‌kJ∕mol.
- Standard enthalpy of sublimation of Carbon:
∆H‌sub ‌(C)=+719.6‌kJ∕mol.
- Bond dissociation enthalpy of Hydrogen gas:
∆H‌diss ‌(H2)=436‌kJ∕mol.
The enthalpy change for the sublimation of Carbon is the energy required to convert carbon in its solid state (graphite) to a gas of carbon atoms:
C( s)⟶C(g)‌‌∆H=+719.6‌kJ∕mol The bond dissociation enthalpy for hydrogen is the energy required to break one mole of
H2 into hydrogen atoms:
H2⟶2H(g)‌‌∆H=436‌kJ∕molSince we need 4 hydrogen atoms to form methane, the energy required to dissociate 2 moles of
H2 into 4 hydrogen atoms is:
2×436‌kJ∕mol=872‌kJ∕mol The enthalpy change for the formation of methane from gaseous carbon and hydrogen atoms is the sum of energies involved:
C( s)+2H2(g)⟶C(g)+4H(g)⟶CH4(g) Combining these steps, the total energy change should consider the sublimation of carbon, the dissociation of hydrogen gas, and the formation of methane:
∆H‌total ‌=∆H‌sub ‌(C)+2∆H‌diss ‌(H2)+4∆H‌bond ‌(C−H)
Thus, the formation of methane from its elements directly gives us:
∆H‌formation ‌(CH4)=∆H‌sub ‌(C)+2∆H‌diss ‌(H2)−4∆H‌bond ‌(C−H)
Substituting the given values:
−74.8‌kJ∕mol=719.6‌kJ∕mol+872‌kJ∕mol−4∆H‌bond ‌(C−H)
Solving for
∆H‌bond ‌(C−H) :
−74.8=1591.6−4∆H‌bond ‌(C−H) Rearranging gives:
‌4∆H‌bond ‌(C−H)=1591.6+74.8‌4∆H‌bond ‌(C−H)=1666.4Thus, the bond enthalpy of the
C−H bond in methane is:
Option B: 416.6 kJ/mol