The standard enthalpy of formation $$(\Delta_fH^\circ)$$ at $$298\,K$$ for methane, $$CH_4(g)$$, is $$-74.8\,kJ\,mol^{-1}$$. The additional information required to determine the average energy for $$C - H$$ bond formation would be

The standard enthalpy of formation of methane is defined for the reaction

$$C_{(s,\;graphite)} + 2\,H_{2(g)} \rightarrow CH_{4(g)}, \qquad \Delta_f H^\circ(CH_4) = -74.8\,{\rm kJ\,mol^{-1}}.$$

To evaluate the average energy of one $$C-H$$ bond, we must construct a thermochemical cycle that converts the individual gaseous atoms $$C_{(g)} + 4\,H_{(g)}$$ into $$CH_{4(g)}$$, because a bond-dissociation energy is defined for gaseous atoms combining to give the molecule. Hess’s law then allows us to relate the required bond energy to known enthalpy changes.

Required steps to reach the atomic reference state:

1. $$C_{(s,\,graphite)} \rightarrow C_{(g)}$$   (enthalpy of sublimation of carbon, $$\Delta_{\rm sub}H(C)$$)

2. $$2\,H_{2(g)} \rightarrow 4\,H_{(g)}$$   (twice the dissociation energy of $$H_2$$, i.e. $$2\,D(H\!-\!H)$$)

After obtaining $$C_{(g)} + 4\,H_{(g)}$$, the formation of methane accomplishes the reverse change of four $$C-H$$ bonds:

$$C_{(g)} + 4\,H_{(g)} \rightarrow CH_{4(g)}.$$ Its enthalpy change equals $$-4\,D_{\rm av}(C\!-\!H),$$ where $$D_{\rm av}(C\!-\!H)$$ is the average $$C-H$$ bond energy.

Applying Hess’s law around the closed cycle:

$$\Delta_{\rm sub}H(C) + 2\,D(H\!-\!H) + \bigl[-4\,D_{\rm av}(C\!-\!H)\bigr] = \Delta_f H^\circ(CH_4).$$

Hence

$$D_{\rm av}(C\!-\!H) = \frac{\Delta_{\rm sub}H(C) + 2\,D(H\!-\!H) - \Delta_f H^\circ(CH_4)}{4}.$$

Therefore, in addition to the given $$\Delta_f H^\circ(CH_4)$$, the only extra data required are:

• enthalpy of sublimation of carbon,
• dissociation energy of $$H_2$$.

No ionization energies, electron-gain enthalpies, or latent heat of vaporisation of methane enter the calculation. Thus the correct choice is:

Option A which is: the dissociation energy of $$H_2$$ and enthalpy of sublimation of carbon.