To calculate the standard enthalpy change for the sublimation of 5 moles of solid iodine, we need to first find the enthalpy change for the sublimation of 1 mole of iodine. Sublimation is the process of a solid turning directly into a gas, bypassing the liquid phase. In this case, we can consider the sublimation of iodine as a two-step process: first, the solid iodine melts fusion , and then the liquid iodine vaporizes vaporization .The enthalpy change for the sublimation of 1 mole of iodine can be calculated by adding the enthalpy of fusion and the enthalpy of vaporization:H_sublimation = H_fusion + H_vaporizationH_sublimation = 15.52 kJ/mol + 41.57 kJ/molH_sublimation = 57.09 kJ/molNow that we have the enthalpy change for the sublimation of 1 mole of iodine, we can calculate the enthalpy change for the sublimation of 5 moles of iodine:H_sublimation 5 moles = 5 * H_sublimation 1 mole H_sublimation 5 moles = 5 * 57.09 kJ/molH_sublimation 5 moles = 285.45 kJTherefore, the standard enthalpy change for the sublimation of 5 moles of solid iodine is 285.45 kJ.