calculate the enthalpy of formation, per mole, of sucrose, C12H2211O(s). C12H22O11(s) + 12O2(g) --> 12CO2(g) +11H2O(l)
deltaH=-5.65*10^3 kJ
Why did the sugar molecule go to the party?
Because it had "chemistry" with the other molecules!
To calculate the enthalpy of formation, we need to use the equation you provided:
C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)
The enthalpy change (ΔH) for this reaction is -5.65 * 10^3 kJ.
Now, to find the enthalpy of formation per mole of sucrose, we need to calculate the difference between the enthalpies of the products and reactants.
Since the reactant is 1 mole of C12H22O11 and the product is 12 moles of CO2, we multiply the enthalpy change by 1/12:
ΔH_f = (-5.65 * 10^3 kJ) / 12
So, the enthalpy of formation, per mole, of sucrose is approximately -471.67 kJ/mol.
Now, that's some sweet energetics!
To calculate the enthalpy of formation of sucrose (C12H22O11), you need to use the given balanced chemical equation and the enthalpies of formation of the reactants and products involved.
The enthalpy of formation (ΔHf) is the heat change when one mole of a compound is formed from its constituent elements, with all substances in their standard states. To calculate it, you can use the enthalpy of formation values for the elements in their standard states and subtract the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products.
In this case, the chemical equation is:
C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)
Given the enthalpy change (ΔH) as -5.65 × 10^3 kJ, which represents the enthalpy change per mole of the reaction. You need to divide this value by the stoichiometric coefficient of the compound of interest (sucrose), which is 1 in this case. Therefore,
ΔHf(C12H22O11) = ΔH / Stoichiometric Coefficient
= -5.65 × 10^3 kJ / 1
So, the enthalpy of formation of sucrose per mole is -5.65 × 10^3 kJ.