Ethanol, C₂H₅OH, is promoted as a clean fuel and is used as an additive in many gasoline mixtures. Calculate the ΔH°_rxn for the combustion of ethanol.

Question

Quizplus · Accepted Answer

To calculate the ΔH°rxn of the combustion of ethanol, we need to know the balanced chemical equation for the reaction:

C2H5OH + 3O2 → 2CO2 + 3H2O

From this equation, we can see that the coefficients for ethanol and oxygen are 1 and 3, respectively. We also need to know the standard enthalpies of formation for each compound involved:

ΔH°f(C2H5OH) = -277.6 kJ/mol
ΔH°f(CO2) = -393.5 kJ/mol
ΔH°f(H2O) = -285.8 kJ/mol

Using Hess's Law, we can calculate ΔH°rxn by subtracting the sum of the standard enthalpies of formation for the reactants from the sum of the standard enthalpies of formation for the products:

ΔH°rxn = [2(ΔH°f(CO2)) + 3(ΔH°f(H2O))] - [ΔH°f(C2H5OH) + 3(ΔH°f(O2))]
ΔH°rxn = [2(-393.5 kJ/mol) + 3(-285.8 kJ/mol)] - [-277.6 kJ/mol + 3(0 kJ/mol)]
ΔH°rxn = -1318.2 kJ/mol + 277.6 kJ/mol
ΔH°rxn = -1040.6 kJ/mol

However, we need to divide this value by 2 to get the ΔH° for the combustion of one mole of ethanol:

ΔH°rxn = -1040.6 kJ/mol ÷ 2
ΔH°rxn = -520.3 kJ/mol

Finally, we need to convert from kJ/mol to kJ/g, using the molar mass of ethanol:

Molar mass of C2H5OH = 46.07 g/mol

ΔH°rxn = -520.3 kJ/mol ÷ 46.07 g/mol
ΔH°rxn = -11.29 kJ/g

However, the question is asking for ΔH°S1U1B1rxnS1U1B0, which is the enthalpy change for the reaction per mole of ethanol burned, with all reactants and products in their standard states at 25°C and 1 atm. This means we need to account for the enthalpy change associated with changing the temperature and pressure of the reactants and products to their standard states. This enthalpy change is called the standard state enthalpy, denoted by ΔH°S1U1B1. The ΔH°S1U1B1 values for the reactants and products can be found in standard thermodynamic tables.

Using the value for ΔH°S1U1B1 for each compound, we can calculate ΔH°S1U1B1rxn. The standard state enthalpy for each compound is:

ΔH°S1U1B1(C2H5OH) = -277.69 kJ/mol
ΔH°S1U1B1(CO2) = -393.51 kJ/mol
ΔH°S1U1B1(H2O) = -285.83 kJ/mol
ΔH°S1U1B1(O2) = 0 kJ/mol

Therefore:

ΔH°S1U1B1rxn = [2(ΔH°S1U1B1(CO2)) + 3(ΔH°S1U1B1(H2O))] - [ΔH°S1U1B1(C2H5OH) + 3(ΔH°S1U1B1(O2))]
ΔH°S1U1B1rxn = [2(-393.51 kJ/mol) + 3(-285.83 kJ/mol)] - [-277.69 kJ/mol + 3(0 kJ/mol)]
ΔH°S1U1B1rxn = -1318.17 kJ/mol + 277.69 kJ/mol
ΔH°S1U1B1rxn = -1040.48 kJ/mol

Again, we need to divide by 2 to get the ΔH°S1U1B1 for the combustion of one mole of ethanol:

ΔH°S1U1B1rxn = -1040.48 kJ/mol ÷ 2
ΔH°S1U1B1rxn = -520.24 kJ/mol

Therefore, the answer is A, -1235.4 kJ. Note that this value is negative because the reaction is exothermic (it releases heat). The negative sign indicates that the reaction is releasing more heat than it is absorbing, which is why it is favorable for the reaction to occur.

Ethanol, C2H5OH, Is Promoted as a Clean Fuel and Is

Ethanol, C₂H₅OH, Is Promoted as a Clean Fuel and Is