Calculate the heat of the following reaction:C6H12O6 → 2C2H5OH + 2CO2Given the following heats of formation:C6H12O6 ΔH°f= -1260 kJ/molC2H5OH ΔH°f= -277.7 kJ/molCO2 ΔH°f= -393.5 kJ/molGroup of answer choices-336.1 kJ/mol-1266 kJ/mol-22.6 kJ/mol-82.4 kJ/mol
Question
Calculate the heat of the following reaction:C6H12O6 → 2C2H5OH + 2CO2Given the following heats of formation:C6H12O6 ΔH°f= -1260 kJ/molC2H5OH ΔH°f= -277.7 kJ/molCO2 ΔH°f= -393.5 kJ/molGroup of answer choices-336.1 kJ/mol-1266 kJ/mol-22.6 kJ/mol-82.4 kJ/mol
Solution
To calculate the heat of the reaction, we need to use the formula:
ΔH°rxn = Σ ΔH°f (products) - Σ ΔH°f (reactants)
The heat of formation (ΔH°f) of the products are:
- For 2 moles of C2H5OH: 2 * -277.7 kJ/mol = -555.4 kJ/mol
- For 2 moles of CO2: 2 * -393.5 kJ/mol = -787 kJ/mol
So, the total ΔH°f for the products is -555.4 kJ/mol - 787 kJ/mol = -1342.4 kJ/mol
The heat of formation (ΔH°f) of the reactant (C6H12O6) is -1260 kJ/mol.
Now, we can substitute these values into the formula:
ΔH°rxn = (-1342.4 kJ/mol) - (-1260 kJ/mol) = -82.4 kJ/mol
So, the heat of the reaction is -82.4 kJ/mol.
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To find the enthalpy of a reaction using heats of formation…
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