Question

What is the enthalpy change for the hydrolysis reaction of sucrose (C12H22O11) in water, given that the molar heat of combustion for sucrose is -5632 kJ/mol and the standard enthalpy of formation for liquid water is -285.83 kJ/mol?

Answer 1

To calculate the enthalpy change for the hydrolysis reaction of sucrose, we need to use Hess's Law. Hess's Law states that the enthalpy change for a reaction is the same, whether it occurs in one step or several steps. We can use the given information to set up a series of reactions and then calculate the enthalpy change for the hydrolysis reaction.First, let's write the balanced hydrolysis reaction for sucrose:C12H22O11  s  + H2O  l   12C  s  + 11H2O  l Now, let's consider the combustion reaction for sucrose:C12H22O11  s  + 12O2  g   12CO2  g  + 11H2O  l     H_combustion = -5632 kJ/molWe also have the standard enthalpy of formation for liquid water:H2  g  + 1/2O2  g   H2O  l     H_formation = -285.83 kJ/molWe need to find the enthalpy change for the hydrolysis reaction, so we need to manipulate these reactions to get the desired reaction. First, reverse the combustion reaction:12CO2  g  + 11H2O  l   C12H22O11  s  + 12O2  g     H = +5632 kJ/molNext, multiply the enthalpy of formation reaction by 11 to match the number of water molecules in the hydrolysis reaction:11[H2  g  + 1/2O2  g   H2O  l ]    H = 11 -285.83 kJ/mol  = -3144.13 kJ/molNow, add these two manipulated reactions:12CO2  g  + 11H2O  l   C12H22O11  s  + 12O2  g     H = +5632 kJ/mol11[H2  g  + 1/2O2  g   H2O  l ]    H = -3144.13 kJ/mol--------------------------------------------------------------C12H22O11  s  + H2O  l   12C  s  + 11H2O  l     H_hydrolysis = ?Now, we can calculate the enthalpy change for the hydrolysis reaction:H_hydrolysis = H1 + H2 = 5632 kJ/mol - 3144.13 kJ/mol = 2487.87 kJ/molSo, the enthalpy change for the hydrolysis reaction of sucrose is 2487.87 kJ/mol.