Final answer:
The average molar bond enthalpy for a C-H bond in CH4 can be calculated using the standard enthalpies of formation for hydrogen, carbon, and methane, resulting in 402.2 kJ/mol.
Step-by-step explanation:
The question involves calculating the average molar bond enthalpy of the carbon-hydrogen (C-H) bond in a methane (CH4) molecule using the standard enthalpies of formation for hydrogen, carbon, and methane. The total energy required to break all four C-H bonds in methane can be calculated by subtracting the sum of the enthalpies of formation of carbon and hydrogen from the enthalpy of formation of methane, and then dividing by four to find the average energy per bond.
Given that the standard enthalpies of formation are:
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- ΔH°f[H(g)] = 218.0 kJ·mol-1
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- ΔH°f[C(g)] = 716.7 kJ·mol-1
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- ΔH°f[CH4(g)] = -74.6 kJ·mol-1
The calculation is as follows:
The total energy (in kJ) to break all the bonds in CH4 = (4 x ΔH°f[H(g)]) + (ΔH°f[C(g)]) - (ΔH°f[CH4(g)]) = (4 x 218.0) + 716.7 - (-74.6) = 1608.7 kJ
Therefore, the average molar bond enthalpy for a C-H bond in CH4 is:
D(C-H) = (1608.7 kJ/mol) / 4 = 402.2 kJ/mol