4.4 - Problem 4: Hess’s Law with Formation Enthalpies
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Introduction to Hess's Law
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Today we are going to discuss Hess's Law, which helps us in calculating enthalpy changes for reactions that are hard to measure directly. Who can tell me what Hess's Law states?
I think it says that the total enthalpy change for a reaction is the same regardless of the number of steps taken?
Exactly, well done! Hess's Law is based on the principle that enthalpy is a state function. This means it depends only on the initial and final states, not the path taken to get there. Can anyone suggest why this might be useful?
It helps find the enthalpy change for complex reactions that we might not be able to measure directly.
Correct! Now let's explore this further by looking at standard enthalpies of formation.
Standard Enthalpy of Formation
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The standard enthalpy of formation, denoted as ΔH_f°, is the heat change that occurs when one mole of a compound is formed from its elements in their standard states. Why do you think we measure enthalpy in standard conditions?
This gives us a consistent basis for comparison between different reactions.
Spot on! Consistency is key. Now, let’s look at how we can use these values to calculate the enthalpy change of a reaction. Who can summarize the steps we take?
We sum the standard enthalpies of formation of the products and subtract the sum for the reactants.
Perfect! This is the core formula we will use today. Let's look at a practical example next.
Example Calculation
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We'll go through a calculation for the reaction: NH₃(g) + O₂(g) → NO(g) + H₂O(l). First, we need to balance the equation. Can someone help with that?
We start with 2 NH₃, 5 O₂ to make 2 NO and 3 H₂O.
Exactly! So the balanced equation is 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O. Now what are the ΔH_f° values we will use?
We use ΔH_f°[NH₃] = –45.9 kJ/mol, ΔH_f°[NO] = +90.3 kJ/mol, and ΔH_f°[H₂O] = –285.8 kJ/mol.
Excellent! Let's calculate the overall ΔH_rxn° together now.
Final Calculation and Recap
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Now, plugging in our values: Σ ΔH_f°(products) and Σ ΔH_f°(reactants). What would be the total for the products?
For products: 4 × (+90.3 kJ) + 6 × (–285.8 kJ) gives a total of –1,353.6 kJ.
Right! And for the reactants?
We get –183.6 kJ for the reactants.
Now calculating ΔH_rxn°, can someone give me the final answer?
ΔH_rxn° = –1,353.6 kJ – (–183.6 kJ) equals –1,170 kJ for 4 moles, which means approximately –292.5 kJ per mole.
Great job! In summary, Hess's Law and the standard enthalpy of formation provide a powerful way to calculate enthalpy changes. Any final questions?
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
Hess’s Law states that the total enthalpy change for a reaction is the same, regardless of the path taken. This section explains how to use standard enthalpies of formation to calculate enthalpy changes for chemical reactions, supported by examples and a detailed methodology.
Detailed
Detailed Summary
This section focuses on Hess’s Law, which asserts that the total enthalpy change for a chemical reaction remains constant, whether the reaction occurs in a single step or several steps. This principle is fundamental in thermodynamics because it allows chemists to calculate the enthalpy change for a reaction when direct measurement is difficult. The section details how to utilize standard enthalpies of formation (ΔH_f°) for elements in their standard states to determine the enthalpy change for a chemical reaction.
Key Concepts Covered:
1. Hess's Law: Reaffirming that enthalpy is a state function and can be calculated from known values.
2. Standard Enthalpies of Formation: Understanding how to relate the formation of products and reactants to calculate the overall enthalpy change.
3. Example Calculation: Demonstrating these principles through a concrete example involving the reaction of ammonia and oxygen, showcasing step-by-step calculations.
4. Applications: Showing how these calculations apply to complex reactions and the relevance in laboratory settings.
Through these concepts, students will understand the utility of Hess's Law in predicting and calculating thermodynamic properties of reactions.
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Balancing the Reaction
Chapter 1 of 2
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Chapter Content
Unbalanced:
NH₃ + O₂ → NO + H₂O
- Balance each element:
- N: 1 on each side (OK)
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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
Detailed Explanation
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.
Examples & Analogies
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.
Compute ΔH_rxn° Using Formation Values
Chapter 2 of 2
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Chapter Content
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).
Detailed Explanation
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.
Examples & Analogies
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.
Key Concepts
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Hess's Law: Total enthalpy change is the same regardless of the reaction path.
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Standard Enthalpy of Formation (ΔH_f°): Enthalpy change when one mole of a compound forms from its elements in standard states.
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Enthalpy Change (ΔH): Difference in enthalpy between reactants and products.
Examples & Applications
Calculating ΔH_rxn° for the reaction of NH₃(g) and O₂(g) using standard enthalpy values.
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
In Hess's Law we find,
Stories
Imagine a journey where no matter how you travel—by road, air, or sea—the distance remains the same. This is like Hess's Law, where the energy difference is always the same, regardless of the path chosen!
Memory Tools
Remember the acronym HELPS—Hess's Law, Enthalpy, Level Paths, Sum total—indicating it is the sum of the enthalpy changes.
Acronyms
Use the acronym EQUAL to remember
Enthalpy is a Quantity Undergoing Alteration in Labeled reactions
emphasizing that it's consistently defined.
Flash Cards
Glossary
- Hess's Law
The principle stating that the total enthalpy change for a chemical reaction is the same regardless of the steps taken.
- Standard Enthalpy of Formation (ΔH_f°)
The heat change associated with the formation of one mole of a compound from its elements in their standard states.
- Enthalpy Change (ΔH)
The difference in enthalpy between reactants and products in a chemical reaction.
- Reactants
The starting substances in a chemical reaction.
- Products
The substances formed as a result of a chemical reaction.
Reference links
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