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Ch.5 - Thermochemistry
Chapter 5, Problem 62b

Consider the following hypothetical reactions: A → B ΔHI = +60 kJ B → C ΔHII = -90 kJ (b) Construct an enthalpy diagram for substances A, B, and C, and show how Hess's law applies.

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Identify the enthalpy changes for each reaction: \( \Delta H_{I} = +60 \text{kJ} \) for \( A \rightarrow B \) and \( \Delta H_{II} = -90 \text{kJ} \) for \( B \rightarrow C \).
Draw an enthalpy diagram with the y-axis representing enthalpy (H) and the x-axis representing the reaction progress.
Place substance A at the starting point on the y-axis. Since \( \Delta H_{I} = +60 \text{kJ} \), draw an upward arrow to represent the conversion of A to B, indicating an increase in enthalpy.
From B, draw a downward arrow to represent the conversion of B to C, since \( \Delta H_{II} = -90 \text{kJ} \), indicating a decrease in enthalpy.
Apply Hess's Law by adding the enthalpy changes of the two steps to find the overall enthalpy change for the reaction \( A \rightarrow C \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Enthalpy (ΔH)

Enthalpy is a thermodynamic quantity that represents the total heat content of a system. It is often expressed in kilojoules (kJ) and can indicate whether a reaction is exothermic (releases heat, ΔH < 0) or endothermic (absorbs heat, ΔH > 0). In the given reactions, the positive and negative values of ΔH indicate the heat changes associated with the transformations between substances A, B, and C.
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Hess's Law

Hess's Law states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps taken to achieve the reaction. This principle allows us to calculate the enthalpy change for a reaction by summing the enthalpy changes of individual steps, making it useful for constructing enthalpy diagrams and understanding the overall energy changes in a series of reactions.
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Enthalpy Diagram

An enthalpy diagram visually represents the enthalpy changes of reactants and products in a chemical reaction. It typically shows the relative energy levels of the substances involved, allowing for a clear understanding of how energy is absorbed or released during the reaction process. In this case, the diagram would illustrate the transitions from A to B and then from B to C, highlighting the respective enthalpy changes.
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Related Practice
Textbook Question

Under constant-volume conditions, the heat of combustion of benzoic acid (C6H5O6) is 15.57 kJ/g. A 3.500-g sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 20.94 to 24.72 °C. (b) If the size of the sucrose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

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Textbook Question

Under constant-volume conditions, the heat of combustion of naphthalene (C10H8) is 40.18 kJ/g. A 2.50-g sample of naphthalene is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.50 to 28.83 °C. (c) Suppose that in changing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?

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Open Question
Can you use an approach similar to Hess’s law to calculate the change in internal energy, _x001F_E, for an overall reaction by summing the _x001F_E values of individual reactions that add up to give the desired overall reaction?
Textbook Question

Calculate the enthalpy change for the reaction P4O6(s) + 2 O2(g) → P4O10(s) given the following enthalpies of reaction: P4(s) + 3 O2(g) → P4O6(s) ΔH = -1640.1 kJ P4(s) + 5 O2(g) → P4O10(s) ΔH = -2940.1 kJ

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Textbook Question

From the enthalpies of reaction 2 C(s) + O2(g) → 2 CO(g) ΔH = -221.0 kJ 2 C(s) + O2(g) + 4 H2(g) → 2 CH3OH(g) ΔH = -402.4 kJ Calculate ΔH for the reaction CO(g) + 2 H2(g) → CH3OH(g)

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Textbook Question

From the enthalpies of reaction H2(g) + F2(g) → 2 HF(g) ΔH = -537 kJ C(s) + 2 F2(g) → CF4(g) ΔH = -680 kJ 2 C(s) + 2 H2(g) → C2H4(g) ΔH = +52.3 kJ Calculate H for the reaction of ethylene with F2: C2H4(g) + 6 F2(g) → 2 CF4(g) + 4 HF(g)

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