given C12H22O11+H2O----C6H12O6+C6H12O6 heat combustion 0f 2815.8 kj/mol what is the standard enthalpy
To find the standard enthalpy change (ΔHº) for the given combustion reaction, we need to use the bond enthalpy approach or use the standard enthalpies of formation.
1. Bond Enthalpy Approach:
The bond enthalpy approach involves calculating the energy required to break the bonds in the reactants and the energy released when new bonds form in the products.
- C12H22O11: It contains carbon-carbon (C-C), carbon-hydrogen (C-H), and carbon-oxygen (C-O) bonds.
- H2O: It contains hydrogen-oxygen (O-H) and oxygen-oxygen (O=O) bonds.
- C6H12O6: It contains carbon-carbon (C-C), carbon-oxygen (C-O), and oxygen-hydrogen (O-H) bonds.
First, calculate the energy required to break the bonds in the reactants using tabulated bond enthalpy values. Then calculate the energy released when new bonds form in the products. Finally, subtract the energy required to break the bonds from the energy released in forming new bonds to determine the ΔHº.
2. Standard Enthalpy of Formation Approach:
The standard enthalpy of formation (ΔHºf) is the enthalpy change when one mole of a compound is formed from its elements in their standard states, with all reactants and products in their standard states.
Look up the standard enthalpy of formation values for all the reactants and products in the reaction. Then use the stoichiometric coefficients of the balanced equation to calculate the ΔHº for the reaction. The ΔHº for reactants must be subtracted, and the ΔHº for products must be added.
Note: The standard enthalpy of formation values can be found in thermodynamic tables or online databases.
After obtaining the ΔHº value using either approach, you can compare it to the given value of 2815.8 kJ/mol to determine if the reaction is exothermic (ΔHº < 0) or endothermic (ΔHº > 0).
To find the standard enthalpy change (∆H°) for the given combustion reaction:
C12H22O11 + H2O → C6H12O6 + C6H12O6
where the heat of combustion is 2815.8 kJ/mol, you need to calculate the enthalpy change using Hess's Law.
Hess's Law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of its individual steps.
Step 1: Combustion of glucose (C6H12O6)
C6H12O6 + 6O2 → 6CO2 + 6H2O
This reaction has a known enthalpy change of -2803 kJ/mol (combustion of 1 mole of glucose).
Step 2: Hydrolysis of sucrose (C12H22O11)
C12H22O11 + 12H2O → 12C + 12H2O
This reaction has no change in enthalpy since there are no new bonds formed or broken.
Step 3: Rearranging the reaction
12C + 12H2O → 6C6H12O6
By adding up Step 1 and Step 2, we can obtain the target reaction.
Now, we can calculate the enthalpy change using the equation:
∆H° = ∑H(products) - ∑H(reactants)
∆H° = [6(-2803 kJ/mol)] - 0
∆H° = -16818 kJ/mol
Therefore, the standard enthalpy change (∆H°) for the combustion reaction is -16818 kJ/mol.