4.4. Problem 4: Hess’s Law with Formation Enthalpies

Unbalanced: NH₃ + O₂ → NO + H₂O

1. Balance each element:

   • N: 1 on each side (OK)

   • H: 3 on left (NH₃) and 2 on right (H₂O). To balance H, we need 3/2 H₂O: NH₃ + O₂ → NO + 3/2 H₂O

   • O: Left has 2 from O₂; right has 1 from NO + (3/2 × 1) from H₂O = 1 + 1.5 = 2.5. So O is not balanced. Let’s put coefficients to avoid fractions: multiply entire equation by 2:

   2 NH₃ + 5 O₂ → 2 NO + 3 H₂O

First, we start with the unbalanced reaction of ammonia (NH₃) reacting with oxygen (O₂) to produce nitric oxide (NO) and water (H₂O). We need to ensure that the number of each type of atom is the same on both sides of the equation. Initially, we find that nitrogen is balanced (1 nitrogen atom on each side), but hydrogen is not balanced (3 on the left and 2 on the right). To fix this, we analyze water (H₂O) and decide to use 3/2 H₂O, which gives us 3 hydrogen atoms. However, this also results in a fractional coefficient, which we prefer to avoid for clarity; thus, we multiply the entire equation by 2. After adjusting coefficients appropriately, we arrive at 2 NH₃ + 5 O₂ → 2 NO + 3 H₂O, which is balanced for all elements.

Think of this balancing process like assembling a puzzle. Each piece represents an atom. At first, you might have a pile of pieces on one side and fewer pieces on the other side of the table. Adjusting the number of pieces until both sides are equal mirrors balancing the reaction. Just as adding or removing puzzle pieces gets both sides to match, adjusting coefficients in a chemical equation ensures that the number of atoms stays the same.

Products:

• 4 NO(g): 4 × (+90.3 kJ/mol) = +361.2 kJ
• 6 H₂O(l): 6 × (–285.8 kJ/mol) = –1,714.8 kJ Σ ΔH_f°(products) = +361.2 + (–1,714.8) = –1,353.6 kJ

Reactants:

• 4 NH₃(g): 4 × (–45.9 kJ/mol) = –183.6 kJ
• 5 O₂(g): 5 × 0 = 0 kJ Σ ΔH_f°(reactants) = –183.6 kJ

ΔH_rxn° = [Σ ΔH_f°(products)] – [Σ ΔH_f°(reactants)] = (–1,353.6 kJ) – (–183.6 kJ) = –1,353.6 + 183.6 = –1,170.0 kJ for 4 moles NH₃ consumed. For 1 mole NH₃: divide by 4 → ΔH_rxn° ≈ –292.5 kJ/mol. Thus, the reaction NH₃ + 5/4 O₂ → NO + 3/2 H₂O has ΔH ≈ –292.5 kJ per mole NH₃ (or –1,170 kJ per 4 moles NH₃).

In this section, we calculate the enthalpy change of the balanced reaction using standard enthalpy of formation values (ΔH_f°). We will find the total enthalpy for products and reactants separately. For the products, we multiply their coefficients by their respective formation enthalpy values: 4 moles of NO at +90.3 kJ/mol gives +361.2 kJ, while 6 moles of H₂O at –285.8 kJ/mol gives –1,714.8 kJ. Adding these results, we get a total for products of –1,353.6 kJ. For the reactants, 4 moles of NH₃ gives us -183.6 kJ, while O₂ contributes nothing as it is in its elemental form with ΔH_f° of 0. Now, we apply Hess's law by subtracting the total enthalpy of the reactants from that of the products, yielding ΔH_rxn° of –1,170.0 kJ for 4 moles of NH₃. To find the per mole value, we divide by 4, resulting in approximately –292.5 kJ/mol for the reaction.

Think of calculating enthalpy change like budgeting for a party. You need to know how much you spent on food (reactants) and how much you're charging guests (products). By listing all costs for food (adding negative values for reactants) and all the funds you'll gain through ticket sales (adding positive values for products), you calculate your profit or loss (ΔH). In our case, the negative enthalpy indicates that more energy is released in the reaction than consumed, just as you would profit from successful ticket sales if your guests willingly paid more than you spent.

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