4.1. Problem 1: Enthalpy of Reaction via Formation Values

Given Standard Heats of Formation (all at 298.15 K):

● ΔH_f°[C₂H₆(g)] = –84.0 kJ/mol ● ΔH_f°[NO₂(g)] = +33.2 kJ/mol ● ΔH_f°[H₂O(l)] = –285.8 kJ/mol ● ΔH_f°[CO₂(g)] = –393.5 kJ/mol ● ΔH_f°[NH₃(g)] = –45.9 kJ/mol ● ΔH_f°[C₂H₅OH(l)] = –277.0 kJ/mol

This chunk lists the standard heats of formation for various compounds. These values indicate how much energy is absorbed or released during the formation of one mole of each substance from its elements in their standard states, typically at a pressure of 1 bar and a temperature of 298.15 K.

Think of these standard heats of formation as a 'price tag' for each compound. Just like you might pay energy to create something (like building a new toy from scratch), these values tell you how much energy is involved in forming each substance.

Calculate ΔH_rxn° for the combustion of ethanol:

C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)

This chunk introduces a specific combustion reaction of ethanol. To find the enthalpy change (ΔH_rxn°) for this reaction, we will use the heats of formation listed earlier. The overall change in enthalpy for a reaction can be determined by summing the heats of formation of the products and subtracting the sum for the reactants. This reflects the energy balance of the reaction.

Imagine you have a recipe where you know the 'caloric content' (energy equivalent of the heats of formation) of each ingredient (reactants and products). By summing their caloric values, you can determine how much 'energy' is consumed or released by cooking (the reaction itself)!

Solution Steps:

1. Write ΔH_rxn° = Σ ΔH_f°(products) – Σ ΔH_f°(reactants).
2. Products: ○ 2 CO₂(g): 2 × (–393.5) = –787.0 kJ ○ 3 H₂O(l): 3 × (–285.8) = –857.4 kJ Sum products = –787.0 + (–857.4) = –1,644.4 kJ
3. Reactants: ○ C₂H₅OH(l): –277.0 kJ ○ 3 O₂(g): 3 × 0 = 0 Sum reactants = –277.0 kJ
4. ΔH_rxn° = (–1,644.4) – (–277.0) = –1,644.4 + 277.0 = –1,367.4 kJ per mole ethanol burned.

Here's the step-by-step calculation process. First, we set up the equation for ΔH_rxn°. Then we compute the total for products (2 CO₂ and 3 H₂O) and the total for reactants (C₂H₅OH and O₂). We combine these to find the overall enthalpy change for the combustion of ethanol. The final result, ΔH_rxn° = –1,367.4 kJ, indicates that this amount of energy is released for each mole of ethanol burned.

This process is akin to budgeting your monthly expenses. You add up all the costs (heats of formation for products) and subtract any income you earn (heats of formation for reactants) to see how much you spent (the energy released or absorbed).

Hence, the standard enthalpy of combustion of ethanol is –1,367.4 kJ/mol.

In summation, this final result tells us that burning one mole of ethanol releases a significant amount of energy (–1,367.4 kJ). This information is crucial for understanding energy content in fuels and their impact on thermochemistry.

Consider this energy release like the warmth you feel when you burn fuel for cooking or heating. The amount of energy calculated helps us understand which fuels give off more heat when burned, similar to how some woods burn hotter than others.

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