Question

Given that Δ H ∘ f [ H ( g ) ] = 218.0 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CH 4 ( g ) ] = − 74.6 kJ ⋅ mol − 1 calculate the average molar bond enthalpy of the carbon‑hydrogen bond in a CH 4 molecule

Answer 1

Average molar bond enthalpy of C-H bond in
`CH_(4)` is 415.825 kJ/mol.

Step-by-step explanation:

Reaction:
`CH_(4)(g)\rightarrow C(g)+4H(g)`

Heat of reaction (energy needed to break 4 moles of C-H bond in 1 mol of
`CH_(4)`) =
`\Delta H^(0)=[1mol* \Delta H_(f)^(0)(C)_(g)]+[4mol* \Delta H_(f)^(0)(H)_(g)]-[1mol* \Delta H_(f)^(0)(CH_(4))_(g)]`

=
`[1mol* 716.7(kJ)/(mol)]+[4mol* 218.0(kJ)/(mol)]-[1mol* -74.6(kJ)/(mol)]=1663.3kJ`

1 mol of
`CH_(4)` contain 4 moles of C-H bonds.

So, average molar bond enthalpy of C-H bond in
`CH_(4)`

=
`(1663.3)/(4)kJ/mol`

= 415.825 kJ/mol