To calculate the bond dissociation energy for the single bond in methane CH4 , we need to consider the reaction where one C-H bond is broken:CH4 CH3 + HFirst, we need to find the enthalpy change H for this reaction. We can use the given data to do this:Hf CH4 = -74.8 kJ/mol enthalpy of formation of methane Hrxn = -890.4 kJ/mol enthalpy change for the complete combustion of methane The complete combustion of methane can be represented by the following reaction:CH4 + 2O2 CO2 + 2H2ONow, we can use Hess's Law to find the enthalpy change for the bond dissociation reaction. Hess's Law states that the enthalpy change for a reaction is the same, regardless of the pathway taken. We can use the following steps:1. Reverse the formation reaction of methane:-CH4 -C + 4H H = +74.8 kJ/mol 2. Add the complete combustion reaction:CH4 + 2O2 CO2 + 2H2O H = -890.4 kJ/mol 3. Add the formation reactions of CO2 and 2H2O:C + 2O2 CO2 H = -393.5 kJ/mol 2H + O2 2H2O H = -571.6 kJ/mol Now, we can sum up the reactions and their enthalpy changes:-CH4 -C + 4H H = +74.8 kJ/mol CH4 + 2O2 CO2 + 2H2O H = -890.4 kJ/mol C + 2O2 CO2 H = -393.5 kJ/mol 2H + O2 2H2O H = -571.6 kJ/mol ---------------------------------------------3H 3H2O H = ? Now, we can solve for the enthalpy change of the bond dissociation reaction:H = 74.8 - 890.4 + 393.5 + 571.6H = -74.8 + 74.8H = 0 kJ/molSince there are 4 C-H bonds in methane, we can divide the enthalpy change by 4 to find the bond dissociation energy for a single C-H bond:Bond dissociation energy = H / 4Bond dissociation energy = 0 kJ/molThe bond dissociation energy for the single bond in methane CH4 is 0 kJ/mol.