Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) has been proposed as an alternative fuel. Calculate the standard enthalpy of combustion per gram of liquid ethanol.

The combustion of ethanol is a chemical reaction where ethanol (\( C_2H_5OH \) reacts with oxygen (\( O_2 \) to produce carbon dioxide (\( CO_2 \) and water (\( H_2O \) as products. This process releases energy in the form of heat, which is why ethanol can be used as a fuel. When we consider the combustion of ethanol for energy, knowing the amount of energy released per gram is vital. This helps us determine its effectiveness and efficiency as a fuel compared to other energy sources.

Understanding the balanced chemical equation of the combustion process is the starting point. The equation \[ C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O \] is important because it provides information on the stoichiometry of the reaction, indicating how much reactant is needed to produce a certain amount of products, key for calculating the enthalpy change associated with the combustion.

Hess's Law states that the total enthalpy change during a chemical reaction is the same regardless of the number of steps in which the reaction is carried out. This principle is essential in thermodynamics, especially in calculating the enthalpy changes for reactions like the combustion of ethanol. In practice, this means we can calculate the enthalpy change for a complex reaction by summing the enthalpy changes of individual simpler steps that lead to the same final state.

Using Hess's Law, we can pinpoint the enthalpy of combustion of ethanol by taking the standard enthalpy of formation values for the reactants and products. The equation used in our example reflects applying Hess's Law to find the standard enthalpy of combustion by incorporating these values and considering the stoichiometry of the balanced reaction.

Standard enthalpy values, denoted as \( \Delta H^\circ \) for the formation of compounds, are crucial in calculating the energy changes in chemical reactions. These values refer to the amount of heat absorbed or released when one mole of a substance is formed from its elements under standard conditions, which are 1 atmosphere of pressure and a specified temperature, usually 25°C (298 K).

The standard enthalpy of formation is zero for elemental substances in their standard state, such as elemental oxygen (\( O_2 \)) in the case of our reaction. In contrast, compounds have specific standard enthalpy values that are determined experimentally and tabulated for reference. These values allow us to calculate enthalpy changes in reactions using Hess's Law by considering the chemical quantities and the direction of heat flow (endothermic or exothermic) during the process.

To comprehend the energy content per gram of a fuel such as ethanol, we must be able to calculate its molar mass. The molar mass is calculated by adding the atomic masses of each element that makes up the compound, weighted by the number of atoms of each element in the molecular formula. The atomic masses are measured in grams per mole (\( g/mol \)).

For ethanol (\( C_2H_5OH \)), we calculate its molar mass by adding the masses of 2 moles of Carbon (\( 12.01 \ g/mol \)), 6 moles of Hydrogen (\( 1.008 \ g/mol \)), and 1 mole of Oxygen (\( 16.00 \ g/mol \)), resulting in a molar mass of 46.07 g/mol. This is instrumental in providing us with the final piece of the puzzle, allowing us to deduce the enthalpy of combustion per gram of ethanol, hence revealing its energy potential as a fuel.