Answer:
The average molar bond enthalpy of the carbon-hydrogen bond (ΔHC−H) = 415.83 KJ/mol
Explanation:
The complete equation for the reaction is:
[tex]C_{(s)} + 4H_{(g)}---CH_{4(g)}[/tex] (1)
In the reaction above, we are forming 4 C-H bonds. Therefore,
Δ[tex]H_{rxn1}[/tex] = 4*(-x) kJ/mol
where 'x' is the average molar bond enthalpy of C-H bond and the negative sign signifies exothermic process (since we are forming a new compound).
The other two reactions are:
[tex]C_{(s)}[/tex]---[tex]C_{(g)}[/tex] (2)
[tex]2H_{(g)}[/tex]---[tex]4H_{(g)}[/tex] (3)
Δ[tex]H_{rxn2}[/tex] = 716.7 kJ/mol
Δ[tex]H_{rxn3}[/tex] = 4*218 = 872 kJ/mol
Using Hess' law which states that the overall enthalpy change for a reaction is independent of the paths taken. we have:
summation of equations (1), (2) and 3:
[tex]C_{(s)}[/tex]---[tex]C_{(g)}[/tex] Δ[tex]H_{rxn2}[/tex] = 716.7 kJ/mol
[tex]C_{(s)} + 4H_{(g)}---CH_{4(g)}[/tex] Δ[tex]H_{rxn1}[/tex] = 4*(-x) kJ/mol
[tex]2H_{(g)}[/tex]---[tex]4H_{(g)}[/tex] Δ[tex]H_{rxn3}[/tex] = 4*218 = 872 kJ/mol
we have:
[tex]C_{(s)}+2H_{2(g)}[/tex]---[tex]CH_{4(g)}[/tex] Δ[tex]H_{f}[/tex] = 716.7-4x+872
where:
Δ[tex]H_{f}[/tex] = -74.6 kJ/mol
Therefore: -74.6 = 716.7-4x+872 and x= 415.83 kJ/mol
Thus, the average molar bond enthalpy of the carbon-hydrogen bond in a [tex]CH_{4}[/tex] molecule is 415.83 kJ/mol
