Question

Calculate the enthalpy change for the hydrolysis of sucrose given that 10.0 g of sucrose was completely hydrolyzed in excess water. The reaction takes place at 25°C and 1 atm pressure. The following reaction is involved: C12H22O11(s) + H2O(l) → 12C(s) + 11H2O(l) The enthalpy of formation of C12H22O11(s) is -2226.8 kJ/mol and the enthalpy of formation of H2O(l) is -285.8 kJ/mol.

Answer 1

To calculate the enthalpy change for the hydrolysis of sucrose, we will use the following equation:H =  Hf products  -  Hf reactants First, we need to determine the number of moles of sucrose in 10.0 g. The molar mass of sucrose  C12H22O11  is: 12 x 12.01  +  22 x 1.01  +  11 x 16.00  = 144.12 + 22.22 + 176.00 = 342.34 g/molNumber of moles of sucrose = mass / molar mass = 10.0 g / 342.34 g/mol = 0.0292 molNow, we can calculate the enthalpy change for the reaction:H = [12Hf C  + 11Hf H2O ] - [Hf sucrose  + Hf H2O ]Since the enthalpy of formation of carbon  C  in its standard state is 0 kJ/mol, the equation becomes:H = [11Hf H2O ] - [Hf sucrose  + Hf H2O ]H = [11 -285.8 kJ/mol ] - [ -2226.8 kJ/mol  +  -285.8 kJ/mol ]H = [-3143.8 kJ/mol] - [-2512.6 kJ/mol]H = -631.2 kJ/molNow, we need to find the enthalpy change for 0.0292 mol of sucrose:H = -631.2 kJ/mol x 0.0292 mol = -18.42 kJThe enthalpy change for the hydrolysis of 10.0 g of sucrose is -18.42 kJ.