Use standard enthalpies of formation to calculate the heat produced by the combustion of one mole of ethanol by the following equation: C2H5OH (l) + 3 O2 (g) --> 2 CO2 (g) + 3 H2O (g).
deltaHrxn = (2*DHfCO2 + 3*DHfH2O)-(DHfC2H5OH)
To calculate the heat produced by the combustion of one mole of ethanol using standard enthalpies of formation, you need to follow these steps:
Step 1: Identify the compounds involved in the equation and determine their standard enthalpies of formation.
C2H5OH (l) - Ethanol
CO2 (g) - Carbon dioxide
H2O (g) - Water
The standard enthalpies of formation for these compounds are:
ΔHf° (C2H5OH) = -277.6 kJ/mol
ΔHf° (CO2) = -393.5 kJ/mol
ΔHf° (H2O) = -241.8 kJ/mol
Step 2: Determine the stoichiometric coefficients in the balanced equation.
In the balanced equation, you can see that one mole of ethanol (C2H5OH) reacts with three moles of oxygen gas (O2) to produce two moles of carbon dioxide (CO2) and three moles of water (H2O).
Step 3: Calculate the heat produced using the standard enthalpies of formation.
The heat produced during the combustion reaction is given by the difference between the sum of the standard enthalpies of formation of the products and the sum of the standard enthalpies of formation of the reactants, multiplied by the stoichiometric coefficients. Since the stoichiometric coefficients in this balanced equation are:
C2H5OH: 1
O2: 3
CO2: 2
H2O: 3
The heat produced (ΔH) can be calculated using the equation:
ΔH = [Σ(nΔHf° products)] - [Σ(nΔHf° reactants)]
Substituting the values:
ΔH = [2 × ΔHf° (CO2) + 3 × ΔHf° (H2O)] - [ΔHf° (C2H5OH) + 3 × ΔHf° (O2)]
ΔH = [2 × (-393.5 kJ/mol) + 3 × (-241.8 kJ/mol)] - [(-277.6 kJ/mol) + 3 × 0 kJ/mol]
ΔH = -787 kJ/mol
Therefore, the combustion of one mole of ethanol releases approximately -787 kJ of heat.
Note: The negative sign indicates that the reaction is exothermic, meaning it releases heat.