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What is the standard enthalpy of formation of methane (CH4) given the following bond dissociation energies (in kJ/mol): C-H = 413, C-C = 348, C=C = 614, C≡C = 837, H-H = 436?

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To calculate the standard enthalpy of formation of methane  CH4 , we need to consider the bond dissociation energies of the reactants and products involved in the formation reaction. The formation reaction for methane can be written as:C  graphite  + 2 H2  g   CH4  g Now, we need to calculate the energy required to break the bonds in the reactants and the energy released when the bonds in the product are formed.For the reactants, we have:1 C-C bond in graphite: We don't need to consider this bond since it's not broken in the formation of methane.2 H-H bonds in H2: 2  436 kJ/mol = 872 kJ/molFor the product, we have:4 C-H bonds in CH4: 4  413 kJ/mol = 1652 kJ/molNow, we can calculate the standard enthalpy of formation  Hf  using the following equation:Hf =   Bond energies of reactants  -   Bond energies of products Hf =  872 kJ/mol  -  1652 kJ/mol  = -780 kJ/molSo, the standard enthalpy of formation of methane  CH4  is -780 kJ/mol.
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