Calculate the bond dissociation enthalpy for the C-H bond in methane, CH 4. C (s

Question

Calculate the bond dissociation enthalpy for the C-H bond in methane, CH4.

C (s) + 2H2 (g) <=> CH4 (g); H0 = -74.8 kJ mol-1

H2 (g) <=> 2H (g); H0 = +434.7 kJ mol-1

C (s) <=> C (g); H0 = +719.0 kJ mol-1

*Helpful Hint:* The bond disscociation enthalpy is a quarter of the enthalpy change for CH4 (g) <=> C(g) + 4H(g). Assuming that the C-H bond disscoiation enthalpy is the same in methane and ethene and that H0formation for ethene is +54.3 kJ mol-1, calculate the C = C bond dissociation enthalpy in ethene. *

Explanation / Answer

you are right to use a Hess-type cycle for this problem

we want this equation:
CH4 (g) --> C(g) & 4 H(g)

we have these:
(A)H2 (g) <=> 2H (g); H0 = +434.7 kJ mol-1

(B) C (s) <=> C (g); H0 = +719.0 kJ mol-1

(C)C (s) + 2H2 (g) <=> CH4 (g); H0 = -74.8 kJ mol-1

by subtracting (C), we get:
CH4 (g) --> C (graphite) & 2 H2 (g)
and by combing it with (B) , we get:
CH4 (g) & C (graphite) --> C (graphite) & 2 H2 (g) & C(g)
the graphites cancel out & we have:
CH4 (g) --> 2 H2 (g) & C(g)

if we we then add in 2 equation (A)'s:
2(A): 2H2 (g) -> 4H (g)

we get:
CH4 (g) & 2H2(g) --> 4 H(g) & 2 H2 (g) & C(g)

& we find that the 2H2's cancel out , giving us the equation we wished:
CH4 (g) --> C(g) & 4 H(g)
==========================

so by Hess's law if
algebraically combining "-(C)" with (B), & "2(A")'s gives us the equation that we wish,
the combining the energies of -dHC, with -dHB & 2 dHA's will give us the energy we wish

dH = -dHC, with -dHB & 2 dHA's
dH = +74.8 & +719 & (2)(+434.7)
dH = 1663.2 kJ

then as you pointed out,
CH4 (g) --> C(g) & 4 H(g)

breaks 4 moles of C-H bonds per mole CH4

so the mean bond enthalpy of the C-H bond in methane, CH4 (g),
would be 1663.2/ 4 =
your answer:
416 kJ/mol

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