1.7. Practice Problems for Section 1

1. Problem: Determine whether each of the following reactions is exothermic or endothermic, given these enthalpies of formation:
   • ΔH_f°[NO(g)] = +90.3 kJ/mol
   • ΔH_f°[N₂(g)] = 0
   • ΔH_f°[O₂(g)] = 0 Reaction: N₂(g) + O₂(g) → 2 NO(g)

To determine if a reaction is exothermic or endothermic, you can calculate the reaction's enthalpy change (ΔH_rxn°) using the standard enthalpies of formation (ΔH_f°) of the reactants and products.

Here’s how to do it step-by-step:

1. Write the formula for ΔH_rxn° = Σ ΔH_f°(products) – Σ ΔH_f°(reactants).
2. Calculate the total enthalpy of the products, which in this case is 2 moles of NO having ΔH_f° = +90.3 kJ/mol. Therefore, products total = 2 × 90.3 = +180.6 kJ.
3. For the reactants N₂(g) and O₂(g), look up their standard enthalpy values. Both N₂ and O₂ have ΔH_f° of 0, leading to a total of 0 kJ.
4. Now plug into the formula:
   • ΔH_rxn° = +180.6 kJ – 0 = +180.6 kJ.
5. Since ΔH is positive, this indicates the reaction is endothermic (it absorbs heat).

Think of making a smoothie. When you blend ice (solid) and fruit (which is typically at room temperature), the ice absorbs heat from the fruit and the surrounding environment to melt. In this process, the mixture feels cold because it absorbs heat. This is similar to an endothermic reaction where energy is absorbed from the surroundings.

2. Problem: Using ΔH_f° values below, calculate ΔH_rxn° for:
   • ΔH_f°[CO(g)] = –110.5 kJ/mol
   • ΔH_f°[O₂(g)] = 0
   • ΔH_f°[CO₂(g)] = –393.5 kJ/mol Reaction: CO(g) + ½ O₂(g) → CO₂(g)

To calculate the enthalpy change (ΔH_rxn°) for the reaction, we follow these steps:

1. Identify the products and their ΔH_f° values. For CO₂(g), ΔH_f° = –393.5 kJ/mol. Since we have 1 mole of CO₂ produced, total for products = –393.5 kJ.
2. Identify the reactants. For CO(g), ΔH_f° = –110.5 kJ/mol and for O₂(g), it is 0. For the reactants sum: CO = -110.5 kJ + 0 (for O₂) = -110.5 kJ.
3. Use the formula: ΔH_rxn° = Σ ΔH_f°(products) – Σ ΔH_f°(reactants).
4. Now plug in: ΔH_rxn° = -393.5 kJ – (-110.5 kJ) = -393.5 + 110.5 = -283.0 kJ.
5. Hence, ΔH_rxn° for this reaction shows it is exothermic since it is negative.

Imagine a campfire. When wood burns (similar to our reaction where CO is oxidized), it releases a significant amount of heat and light into the environment, indicating an exothermic reaction. The heat released can be very comforting on a cold night, similar to how enthalpy change in a combustion reaction releases heat energy.

3. Problem: Given ΔH_f° for CH₃OH(l) = –238.7 kJ/mol, ΔH_f° for CO₂(g) = –393.5 kJ/mol, and ΔH_f° for H₂O(l) = –285.8 kJ/mol, calculate ΔH_c° for methanol: 2 O₂(g) + CH₃OH(l) → CO₂(g) + 2 H₂O(l)

To calculate the standard enthalpy of combustion (ΔH_c°) of methanol, we need to follow similar steps:

1. Identify the products and their formation enthalpy:
   • For CO₂(g), we have ΔH_f° = –393.5 kJ, and for 2 H₂O(l), total = 2 × –285.8 kJ = –571.6 kJ.
2. For the reactants, CH₃OH(l) = –238.7 kJ and O₂(g) = 0.
3. So, the total for products = –393.5 + (–571.6) = –965.1 kJ.
4. The total for reactants = –238.7 kJ + 0 = –238.7 kJ.
5. Now calculate ΔH_c° using the formula:
   • ΔH_c° = (–965.1 kJ) - (–238.7 kJ) = –965.1 + 238.7 = –726.4 kJ per mole of methanol.
6. Therefore, the enthalpy of combustion shows that burning methanol is highly exothermic.

Think about how alcohol lamps work during camping trips. When you light the alcohol (methanol), it burns, producing heat and light, simulating a small fire. The heat produced is similar to how exothermic combustion reactions release energy, brightening your surroundings and keeping it warm.

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