Calculate the average bond energy in kJ per mol of bonds for the C-H bond from t

Question

Calculate the average bond energy in kJ per mol of bonds for the C-H bond from the following data: C (graphite) + 2H2 (g) --> CH4 (g) ?H^0rxn = -74.81 kJ ?H^0f for H(g) = 218 KJ and for C(g) = 716.7 kJ

Explanation / Answer

Calculate the mean bond enthalpy of the C-H bond in methane, CH4 (g), given the following information: ½ H2 (g) -> H (g) +218 kJ mol-1 C (graphite) -> C (g) +717 kJ mol-1 [Delta]H[theta]f [CH4 (g)] -75 kJ mol-1 (standard enthalpy of formation of methane) You may need to use a Hess-type cycle for this problem. Note that there are four C-H bonds in the methane molecule. ----------- I've tried using the Hess cycle: C + 2H2 -> CH4 both combust in 2O2 to give CO2 + 2H2O, t Hess-type cycle for this problem we want this equation: CH4 (g) --> C(g) & 4 H(g) we have these: (A) ½ H2 (g) -> H (g) dH = +218 kJ mol-1 (B) C (graphite) -> C (g) dH = +717 kJ mol-1 (C) C (graphite) & 2 H2 (g) -> CH4 (g) dH = -75 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 4 equation (A)'s: 4(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), & "4(A")'s gives us the equation that we wish, the combining the energies of -dHC, with -dHB & 4 dHA's will give us the energy we wish dH = -dHC, with -dHB & 4 dHA's dH = +75 & +717 & (4)(+218) dH = 1664 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 1664 / 4 = your answer: 416 kJ/mol & that is extremely close to textbook for the generic C-H

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