To calculate the bond enthalpy of the $$\mathrm{C}-\mathrm{H}$$ bond in methane ($$\mathrm{CH}_4$$), 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 $$\mathrm{CH}_4$$: $$\Delta H_f(\mathrm{CH}_4) = -74.8 \, \mathrm{kJ/mol}$$.

- Standard enthalpy of sublimation of Carbon: $$\Delta H_{\text{sub}}(\mathrm{C}) = +719.6 \, \mathrm{kJ/mol}$$.

- Bond dissociation enthalpy of Hydrogen gas: $$\Delta H_{\text{diss}}(\mathrm{H}_2) = 436 \, \mathrm{kJ/mol}$$.

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:

$$\mathrm{C(s)} \rightarrow \mathrm{C(g)} \quad \Delta H = +719.6 \, \mathrm{kJ/mol}$$

The bond dissociation enthalpy for hydrogen is the energy required to break one mole of $$\mathrm{H}_2$$ into hydrogen atoms:

$$\mathrm{H}_2 \rightarrow 2 \mathrm{H(g)} \quad \Delta H = 436 \, \mathrm{kJ/mol}$$

Since we need 4 hydrogen atoms to form methane, the energy required to dissociate 2 moles of $$\mathrm{H}_2$$ into 4 hydrogen atoms is:

$$2 \mathrm{\times 436} \, \mathrm{kJ/mol} = 872 \, \mathrm{kJ/mol}$$

The enthalpy change for the formation of methane from gaseous carbon and hydrogen atoms is the sum of energies involved:

$$\mathrm{C(s)} + 2\mathrm{H}_2(g) \rightarrow \mathrm{C(g)} + 4\mathrm{H(g)} \rightarrow \mathrm{CH_4(g)}$$

Combining these steps, the total energy change should consider the sublimation of carbon, the dissociation of hydrogen gas, and the formation of methane:

$$\Delta H_{\text{total}} = \Delta H_{\text{sub}}(\mathrm{C}) + 2 \Delta H_{\text{diss}}(\mathrm{H}_2) + 4 \Delta H_{\text{bond}}(\mathrm{C-H})$$

Thus, the formation of methane from its elements directly gives us:

$$\Delta H_{\text{formation}}(\mathrm{CH}_4) = \Delta H_{\text{sub}}(\mathrm{C}) + 2 \Delta H_{\text{diss}}(\mathrm{H}_2) - 4 \Delta H_{\text{bond}}(\mathrm{C-H})$$

Substituting the given values:

$$-74.8 \, \mathrm{kJ/mol} = 719.6 \, \mathrm{kJ/mol} + 872 \, \mathrm{kJ/mol} - 4 \Delta H_{\text{bond}}(\mathrm{C-H})$$

Solving for $$\Delta H_{\text{bond}}(\mathrm{C-H})$$:

$$-74.8 = 1591.6 - 4 \Delta H_{\text{bond}}(\mathrm{C-H})$$

Rearranging gives:

$$4 \Delta H_{\text{bond}}(\mathrm{C-H}) = 1591.6 + 74.8$$

$$4 \Delta H_{\text{bond}}(\mathrm{C-H}) = 1666.4$$

$$\Delta H_{\text{bond}}(\mathrm{C-H}) = \frac{1666.4}{4} = 416.6 \, \mathrm{kJ/mol}$$

Thus, the bond enthalpy of the $$\mathrm{C}-\mathrm{H}$$ bond in methane is:

Option B: 416.6 kJ/mol

Thermodynamics

Continue this paper

Similar Thermodynamics PYQs

More practice from this question