Consider the following data :$\Delta_f \text{H}^\ominus (\text{methane, g}) = -X \text{ kJ mol}^{-1}$Enthalpy of sublimation of graphite $= Y \text{ kJ mol}^{-1}$Dissociation enthalpy of $\text{H}_2 = Z \text{ kJ mol}^{-1}$The bond enthalpy of $\text{C}-\text{H}$ bond is given by :

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Question

Consider the following data :
$\Delta_f \text{H}^\ominus (\text{methane, g}) = -X \text{ kJ mol}^{-1}$
Enthalpy of sublimation of graphite $= Y \text{ kJ mol}^{-1}$
Dissociation enthalpy of $\text{H}_2 = Z \text{ kJ mol}^{-1}$
The bond enthalpy of $\text{C}-\text{H}$ bond is given by :

$\frac{X + Y + 4Z}{2}$

$\frac{X + Y + 2Z}{4}$

$\frac{-X + Y + Z}{4}$

$X + Y + Z$

The correct answer is

$\frac{X + Y + 2Z}{4}$

Bond Enthalpy Calculation for Methane

This solution explains how to calculate the C-H bond enthalpy in methane ($\text{CH}_4$) using the provided thermochemical data and Hess's Law. The key is to relate the enthalpy of formation to the energies of bond breaking and formation.

Thermochemical Data Provided

Standard enthalpy of formation of methane ($\Delta_f H^\ominus(\text{methane, g})$): $-X \text{ kJ mol}^{-1}$
Enthalpy of sublimation of graphite ($\text{C}(s) \rightarrow \text{C}(g)$): $Y \text{ kJ mol}^{-1}$
Dissociation enthalpy of $\text{H}_2$ ($\text{H}_2(g) \rightarrow 2\text{H}(g)$): $Z \text{ kJ mol}^{-1}$

Applying Hess's Law to Methane Formation

The standard enthalpy of formation of methane ($\text{CH}_4$) involves the reaction:

$ \text{C}(s) + 2\text{H}_2(g) \rightarrow \text{CH}_4(g) \quad \Delta H_{\text{formation}} = -X \text{ kJ mol}^{-1} $

We can break this process down into steps involving atomization:

Sublimation of carbon: $\text{C}(s) \rightarrow \text{C}(g)$; $\Delta H_1 = Y \text{ kJ mol}^{-1}$
Dissociation of hydrogen molecules: $2\text{H}_2(g) \rightarrow 4\text{H}(g)$; $\Delta H_2 = 2 \times Z \text{ kJ mol}^{-1}$ (since the dissociation enthalpy of $\text{H}_2$ is $Z$ for 1 mole of $\text{H}_2$, and we need 4 moles of H atoms).
Formation of methane from gaseous atoms: $\text{C}(g) + 4\text{H}(g) \rightarrow \text{CH}_4(g)$; $\Delta H_3$. This step involves forming 4 C-H bonds. If the bond enthalpy of a C-H bond is $B$, then $\Delta H_3 = -4B$.

According to Hess's Law, the total enthalpy of formation is the sum of the enthalpy changes for these steps:

$ \Delta H_{\text{formation}} = \Delta H_1 + \Delta H_2 + \Delta H_3 $

Substituting the given values and the expression for $\Delta H_3$:

$ -X = Y + (2Z) + (-4B) $

$ -X = Y + 2Z - 4B $

Solving for C-H Bond Enthalpy

Rearrange the equation to solve for $B$, the bond enthalpy of the C-H bond:

$ 4B = X + Y + 2Z $

$ B = \frac{X + Y + 2Z}{4} $

The bond enthalpy of the C-H bond is $\frac{X + Y + 2Z}{4} \text{ kJ mol}^{-1}$.

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