given C12H22O11+H2O----C6H12O6+C6H12O6 heat combustion0f 5646.7kj/mol what is the standard enthalpy
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To calculate the standard enthalpy change (ΔH°) for a reaction, you need to use the standard enthalpies of formation (ΔHf°) of the reactants and products. The given equation is:
C12H22O11 + H2O → C6H12O6 + C6H12O6
First, you need to determine the balanced equation for combustion. This can be done by ensuring that the number of each type of atom is the same on both sides of the equation:
C12H22O11 + 12 O2 → 12 CO2 + 11 H2O
Then, you can use the ΔHf° values to calculate the ΔH° for this reaction. The ΔHf° values represent the enthalpy change when one mole of a compound is formed from its elements in their standard states (at 25°C and 1 atm).
The values (in kJ/mol) for the combustion reaction are:
ΔHf°(C12H22O11) = -1273.3
ΔHf°(CO2) = -393.5
ΔHf°(H2O) = -285.8
Since there are 12 moles of CO2 produced in the reaction, we multiply its ΔHf° value by 12:
12 × -393.5 kJ/mol = -4722 kJ/mol
For H2O, there are 11 moles produced, so we multiply its ΔHf° value by 11:
11 × -285.8 kJ/mol = -3144 kJ/mol
The ΔH° of the reaction is the sum of the products minus the sum of the reactants:
ΔH° = [(-4722 kJ/mol) + (-3144 kJ/mol)] - (-1273.3 kJ/mol)
= -7866 kJ/mol - (-1273.3 kJ/mol)
= -6592.7 kJ/mol
Therefore, the standard enthalpy change (ΔH°) for the combustion reaction is approximately -6592.7 kJ/mol.