Calculation of Enthalpy of CH Bond in CH4

Given data:
Enthalpy of formation of CH4(g) = 75 kJ/mol
Enthalpy of formation of C(g) = 717 kJ/mol
Enthalpy of formation of H(g) = 218 kJ/mol

To calculate the enthalpy of the CH bond in CH4, we need to use the following equation:

ΔH°f(CH4) = ΣΔH°f(products) - ΣΔH°f(reactants)

Here, the enthalpy of formation of CH4 is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.

The enthalpy of formation of CH4 can be expressed in terms of the enthalpy of the CH bond as follows:

ΔH°f(CH4) = 4 × ΔH°(CH)

We know that the enthalpy of formation of CH4 is 75 kJ/mol. Therefore, we can write:

75 kJ/mol = 4 × ΔH°(CH)

Solving for ΔH°(CH), we get:

ΔH°(CH) = 75 kJ/mol ÷ 4
ΔH°(CH) = 18.75 kJ/mol

However, we need to express the enthalpy of the CH bond in kJ/mol, not in kJ. Therefore, we need to divide the value we obtained by the number of moles of CH in one mole of CH4. Since CH4 contains four moles of H and one mole of C, it contains four moles of CH. Thus:

ΔH°(CH) = 18.75 kJ/mol ÷ 4 moles
ΔH°(CH) = 4.6875 kJ/mol

Rounding off to three significant figures, we get:

ΔH°(CH) = 4.69 kJ/mol

However, the correct answer is 416 kJ/mol, not 4.69 kJ/mol. This is because we made a mistake in our calculation above. We divided the enthalpy of formation of CH4 by the number of moles of CH in one mole of CH4, which is four. However, we should have divided it by the number of moles of CH in one mole of C, which is one. Therefore, the correct calculation is:

ΔH°(CH) = 75 kJ/mol ÷ 1 mole
ΔH°(CH) = 75 kJ/mol

Thus, the enthalpy of the CH bond in CH4 is 416 kJ/mol.