StreakPeaked· Practice

ExamsJEE AdvancedChemistry

The enthalpy change for the aqueous mutarotation of glucose, alpha-D-glucose(aq) -> beta-D-glucose(aq), was measured using a microcalorimeter and found to be -1.16 kJ/mol. The enthalpies of dissolving each solid form in water were also measured: dissolving solid alpha-D-glucose gives DeltaH = +10.72 kJ/mol, and dissolving solid beta-D-glucose gives DeltaH = +4.68 kJ/mol. Using Hess's law, determine DeltaH (in kJ/mol) for the solid-state conversion: alpha-D-glucose(s) -> beta-D-glucose(s).

  1. +4.88 kJ/mol
  2. -4.88 kJ/mol
  3. -2.44 kJ/mol
  4. +2.44 kJ/mol

Correct answer: +4.88 kJ/mol

Solution

Combining the three equations: DeltaH = +10.72 + (-1.16) + (-4.68) = +4.88 kJ/mol. The solid alpha form has higher enthalpy than solid beta, so the conversion is endothermic.

Related JEE Advanced Chemistry questions

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →