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From the following enthalpies of reaction 2 H₂(g) + O₂(g) → 2 H₂O(l) H₂(g) + O₂(g) → H₂O2(1) calculate AH for the reaction 2 H₂O₂(1)→ 2 H₂O(l) + O₂(g). O-760 kJ O-196 kJ O-948 kJ O-384 kJ AH = -572 kJ ΔΗ = -188 kJ

From the following enthalpies of reaction 2 H₂(g) + O₂(g) → 2 H₂O(l) H₂(g) + O₂(g) → H₂O2(1) calculate AH for the reaction 2 H₂O₂(1)→ 2 H₂O(l) + O₂(g). O-760 kJ O-196 kJ O-948 kJ O-384 kJ AH = -572 kJ ΔΗ = -188 kJ

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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### Educational Content on Hess's Law and Enthalpy Diagrams **Figure 1: Understanding Hess’s Law through an Enthalpy Diagram** This enthalpy diagram illustrates Hess’s Law. The overall reaction remains equivalent to the one depicted in Figure 5.21 of the textbook. Here, we explore it using a step-by-step approach with two distinct reactions. By constructing a series of equations that sum up to the desired equation, and by knowing the values of ΔH for each intermediate reaction, we can calculate the overall change in enthalpy (ΔH). **Diagram Explanation:** - The diagram shows three steps involved in the reaction process: 1. **First Reaction:** - Reactants: CH₄(g) + 2 O₂(g) - Enthalpy change (ΔH₁): -890 kJ 2. **Second Reaction:** - Intermediates: CO(g) + 2 H₂O(l) + ½ O₂(g) - Enthalpy change (ΔH₂): -607 kJ 3. **Final Products:** - CO₂(g) + 2 H₂O(l) - Enthalpy change (ΔH₃): -283 kJ - Each reaction's enthalpy change is displayed, indicating the energy absorbed or released at each step. **Part A: Conceptual Question** Suppose the overall reaction was modified to produce 2 H₂O(g) instead of 2 H₂O(l). Would any of the enthalpy values (ΔH) in the diagram remain unchanged? - Options: - Yes - No This exercise encourages understanding of how changes in the physical states of products can affect the enthalpy changes in a chemical reaction.
Transcribed Image Text:### Educational Content on Hess's Law and Enthalpy Diagrams **Figure 1: Understanding Hess’s Law through an Enthalpy Diagram** This enthalpy diagram illustrates Hess’s Law. The overall reaction remains equivalent to the one depicted in Figure 5.21 of the textbook. Here, we explore it using a step-by-step approach with two distinct reactions. By constructing a series of equations that sum up to the desired equation, and by knowing the values of ΔH for each intermediate reaction, we can calculate the overall change in enthalpy (ΔH). **Diagram Explanation:** - The diagram shows three steps involved in the reaction process: 1. **First Reaction:** - Reactants: CH₄(g) + 2 O₂(g) - Enthalpy change (ΔH₁): -890 kJ 2. **Second Reaction:** - Intermediates: CO(g) + 2 H₂O(l) + ½ O₂(g) - Enthalpy change (ΔH₂): -607 kJ 3. **Final Products:** - CO₂(g) + 2 H₂O(l) - Enthalpy change (ΔH₃): -283 kJ - Each reaction's enthalpy change is displayed, indicating the energy absorbed or released at each step. **Part A: Conceptual Question** Suppose the overall reaction was modified to produce 2 H₂O(g) instead of 2 H₂O(l). Would any of the enthalpy values (ΔH) in the diagram remain unchanged? - Options: - Yes - No This exercise encourages understanding of how changes in the physical states of products can affect the enthalpy changes in a chemical reaction.
### Calculating Enthalpy of Reaction From the following enthalpies of reaction: 1. \(2 \, \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \, \text{H}_2\text{O}(l) \quad \Delta H = -572 \, \text{kJ}\) 2. \(\text{H}_2(g) + \text{O}_2(g) \rightarrow \text{H}_2\text{O}_2(l) \quad \Delta H = -188 \, \text{kJ}\) Calculate \(\Delta H\) for the reaction: \[2 \, \text{H}_2\text{O}_2(l) \rightarrow 2 \, \text{H}_2\text{O}(l) + \text{O}_2(g).\] ### Options: - \(-760 \, \text{kJ}\) - \(-196 \, \text{kJ}\) - \(-948 \, \text{kJ}\) - \(-384 \, \text{kJ}\) ### Instructions: Select the correct enthalpy change for the reaction above and submit your answer. **Note:** Use Hess's law to calculate the enthalpy from the given reactions.
Transcribed Image Text:### Calculating Enthalpy of Reaction From the following enthalpies of reaction: 1. \(2 \, \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \, \text{H}_2\text{O}(l) \quad \Delta H = -572 \, \text{kJ}\) 2. \(\text{H}_2(g) + \text{O}_2(g) \rightarrow \text{H}_2\text{O}_2(l) \quad \Delta H = -188 \, \text{kJ}\) Calculate \(\Delta H\) for the reaction: \[2 \, \text{H}_2\text{O}_2(l) \rightarrow 2 \, \text{H}_2\text{O}(l) + \text{O}_2(g).\] ### Options: - \(-760 \, \text{kJ}\) - \(-196 \, \text{kJ}\) - \(-948 \, \text{kJ}\) - \(-384 \, \text{kJ}\) ### Instructions: Select the correct enthalpy change for the reaction above and submit your answer. **Note:** Use Hess's law to calculate the enthalpy from the given reactions.
Expert Solution
Step 1: Defining extensive properties!

Answer:

Properties that depend on the size of the system are extensive properties and are additive in nature. Enthalpy change is an extensive property, that's why it can be added or subtracted the way chemical equations are added or subtracted.

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