To calculate the enthalpy change for the hydrolysis of sucrose, we first need to write the balanced chemical equation for the reaction:C12H22O11 s + H2O l C6H12O6 s + C6H12O6 s Sucrose + Water Glucose + FructoseNext, we need the standard enthalpy of formation values for each compound involved in the reaction:Hf C12H22O11 = -2221 kJ/molHf H2O = -285.8 kJ/molHf C6H12O6 = -1273.3 kJ/mol both glucose and fructose have the same value Now, we can use Hess's Law to calculate the enthalpy change for the reaction:H = [Hf products ] - [Hf reactants ]H = [ 2 -1273.3 - -2221 + -285.8 ]H = -2546.6 + 2506.8H = -39.8 kJ/molNow that we have the enthalpy change per mole of sucrose, we need to determine the number of moles of sucrose in 5.5 grams:Molar mass of sucrose C12H22O11 = 12 12.01 + 22 1.01 + 11 16.00 = 342.3 g/molmoles of sucrose = 5.5 g / 342.3 g/mol = 0.0161 molFinally, we can calculate the enthalpy change for the hydrolysis of 5.5 grams of sucrose:H = H moles of sucroseH = -39.8 kJ/mol 0.0161 molH = -0.641 kJThe enthalpy change for the hydrolysis of 5.5 grams of sucrose is approximately -0.641 kJ.