Table sugar consists mostly of sucrose, C12H22O11. The standard enthalpy of combustion for sucrose is the standard state delta H for the reaction:
C12H22O11 + 12 O 2 ---> 12 CO2 + 11 H2O
Calculate this standard state delta H. Give answer in units of kJ to five significant figures. Use the following values of standard enthalpy of formation (in kJ/mol): CO2 = -393.51, H2O = -285.83, C12H22O11 = -2221.2.
To calculate the standard enthalpy of combustion (delta H) for the given reaction, you need to use the standard enthalpy of formation values.
The standard enthalpy of formation is the change in enthalpy when one mole of a substance is formed from its elements in their standard states.
In the given reaction, you have:
C12H22O11 + 12 O2 --> 12 CO2 + 11 H2O
First, you need to calculate the standard enthalpy of formation for the reactant (C12H22O11) and the products (CO2 and H2O). Then, multiply them by the stoichiometric coefficients in the balanced equation.
The given standard enthalpy of formation values are:
CO2 = -393.51 kJ/mol
H2O = -285.83 kJ/mol
C12H22O11 = -2221.2 kJ/mol
Now, let's calculate the standard enthalpy of combustion:
Standard enthalpy of formation for the reactant (C12H22O11):
= (-2221.2 kJ/mol) × 1 mole
= -2221.2 kJ
Standard enthalpy of formation for the products (CO2 and H2O):
= (12 CO2: 12 × -393.51 kJ/mol) + (11 H2O: 11 × -285.83 kJ/mol)
= -4722.12 kJ + (-3144.13 kJ)
= -7866.25 kJ
Now, calculate the difference between the standard enthalpy of formation of the reactant and the products:
Standard enthalpy of combustion (delta H) = Products - Reactants
= -7866.25 kJ - (-2221.2 kJ)
= -5645.05 kJ
Therefore, the standard enthalpy of combustion (delta H) for the given reaction is -5645.05 kJ to five significant figures.