To calculate the enthalpy of ionization of hydrogen gas H , we need to consider the process of ionization. In this case, one mole of hydrogen gas H dissociates into two moles of hydrogen atoms H and then ionizes to form two moles of hydrogen ions H and two moles of electrons e .The overall process can be represented as:H g 2H g 2H g + 2eFirst, we need to consider the bond dissociation energy of H, which is the energy required to break one mole of H-H bonds in the gas phase. The bond dissociation energy of H is approximately 436 kJ/mol.H g 2H g ; H = +436 kJ/molNext, we need to consider the ionization energy of hydrogen atoms. The problem states that it requires 1312 kJ/mol to ionize a single hydrogen atom H in the gas phase.H g H g + e ; H = +1312 kJ/molSince there are two moles of hydrogen atoms H in one mole of hydrogen gas H , we need to multiply the ionization energy by 2.2H g 2H g + 2e ; H = 2 1312 kJ/mol = +2624 kJ/molNow, we can calculate the overall enthalpy of ionization of hydrogen gas H by adding the enthalpies of the two processes:H_total = H_dissociation + H_ionizationH_total = 436 kJ/mol + 2624 kJ/molH_total = 3060 kJ/molTherefore, the enthalpy of ionization of hydrogen gas H is 3060 kJ/mol.