Ethanol, C2H5OH, is being promoted as a clean fuel and is used as an additive in many gasoline
mixtures. Calculate the ΔH°rxn for the combustion of ethanol. ΔH°f [C2H5OH(l)] = -277.7 kJ/mol;
ΔH°f [CO2(g)] = -393.5 kJ/mol; ΔH°f [H2O(g)] = -241.8 kJ/mol
To calculate the ΔH°rxn for the combustion of ethanol, we need to determine the balanced chemical equation for the reaction.
The combustion of ethanol can be represented by the following equation:
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
According to the balanced equation, for each mole of ethanol, 2 moles of CO2 and 3 moles of H2O are produced.
Now, we can calculate the ΔH°rxn using the standard enthalpies of formation (ΔH°f) values given:
ΔH°rxn = Σ(ΔH°f, products) - Σ(ΔH°f, reactants)
Let's substitute the ΔH°f values into the equation:
ΔH°rxn = [2ΔH°f(CO2)] + [3ΔH°f(H2O)] - [ΔH°f(C2H5OH)]
ΔH°rxn = [2(-393.5 kJ/mol)] + [3(-241.8 kJ/mol)] - (-277.7 kJ/mol)
ΔH°rxn = -787.0 kJ/mol - 725.4 kJ/mol + 277.7 kJ/mol
ΔH°rxn = -787.0 kJ/mol - 725.4 kJ/mol + 277.7 kJ/mol
ΔH°rxn = -1234.7 kJ/mol
Therefore, the ΔH°rxn for the combustion of ethanol is -1234.7 kJ/mol.
To calculate the ΔH°rxn for the combustion of ethanol, we need to determine the balanced chemical equation for the combustion reaction. The combustion of ethanol can be represented by the equation:
C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(g)
In this equation, ethanol reacts with oxygen gas to produce carbon dioxide and water.
Next, we need to calculate the ΔH°rxn using the enthalpy of formation values provided.
The enthalpy of formation (ΔH°f) is the change in enthalpy that accompanies the formation of 1 mole of a compound from its constituent elements, with all substances in their standard states.
Given ΔH°f [C2H5OH(l)] = -277.7 kJ/mol, ΔH°f [CO2(g)] = -393.5 kJ/mol, and ΔH°f [H2O(g)] = -241.8 kJ/mol, we can use these values to calculate the ΔH°rxn.
To calculate ΔH°rxn, we can use the following equation:
ΔH°rxn = ΣΔH°f (products) - ΣΔH°f (reactants)
Substituting the given values into the equation, we get:
ΔH°rxn = [2(-393.5 kJ/mol) + 3(-241.8 kJ/mol)] - [-277.7 kJ/mol]
Simplifying the equation, we get:
ΔH°rxn = -787.0 kJ/mol - (-277.7 kJ/mol)
ΔH°rxn = -787.0 kJ/mol + 277.7 kJ/mol
ΔH°rxn = -509.3 kJ/mol
Therefore, the ΔH°rxn for the combustion of ethanol is -509.3 kJ/mol. This means that the reaction releases 509.3 kJ of heat per mole of ethanol combusted.
C2H5OH + 3O2 ==> 2CO2 + 3H2O
dHfrxn = (n*dHf products) - (n*dHf reactants)
Solve for dHf rxn.