Consider the following hypothetical reactions: $$\begin{array}{ll}{\mathrm{A} \longrightarrow \mathrm{B}} & {\Delta H=+30 \mathrm{kJ}} \\ {\mathrm{B} \longrightarrow \mathrm{C}} & {\Delta H=+60 \mathrm{kJ}}\end{array}$$ (a) Use Hess's law to calculate the enthalpy change for the reaction \(A \longrightarrow C .\) (b) Construct an enthalpy diagram for substances \(A,\) and C, and show how Hess's law applies.

Short Answer

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(a) To calculate the enthalpy change for the reaction \(A \longrightarrow C\), use Hess's law: \(\Delta H_{Total} = \Delta H_1 + \Delta H_2 = (+30\, \text{kJ}) + (+60\, \text{kJ}) = +90\, \text{kJ}\). (b) In the enthalpy diagram, represent each reaction and its enthalpy change with arrows and label them accordingly: \(A \longrightarrow B\) with \(\Delta H_1=+30\, \text{kJ}\), \(B \longrightarrow C\) with \(\Delta H_2=+60\, \text{kJ}\), and \(A \longrightarrow C\) with \(\Delta H_{Total}=+90\, \text{kJ}\). The diagram shows Hess's law in action as the enthalpy change for the direct reaction \(A \longrightarrow C\) is equal to the sum of the enthalpy changes for the two hypothetical reactions.

Step by step solution

01

Analyze given reactions and enthalpy changes

The given reactions are: 1. \(A \longrightarrow B\), with enthalpy change \(\Delta H_1 = +30\, \text{kJ}\) 2. \(B \longrightarrow C\), with enthalpy change \(\Delta H_2 = +60\, \text{kJ}\) Our goal is to find the enthalpy change for the reaction \(A \longrightarrow C\).
02

Apply Hess's law

According to Hess's law, the enthalpy change of the desired reaction (\((A \longrightarrow C)\)) is equal to the sum of the enthalpy changes of the given reactions. So, \(\Delta H_{Total} = \Delta H_1 + \Delta H_2\) Substitute the given enthalpy changes: \(\Delta H_{Total} = (+30\, \text{kJ}) + (+60\, \text{kJ})\) \(\Delta H_{Total} = +90\, \text{kJ}\) Therefore, the enthalpy change for the reaction \(A \longrightarrow C\) is \(+90\, \text{kJ}\). #b) Construct an enthalpy diagram for substances A, and C#
03

Set up the enthalpy diagram

An enthalpy diagram is a graphical representation of the enthalpy changes in chemical reactions. It allows us to visualize how the enthalpy changes as the reactants transform into products. On the vertical axis, we will represent the enthalpy level (H). We start by showing substances \(A\), \(B\), and \(C\) on the horizontal axis.
04

Represent the reactions and their enthalpy changes on the diagram

Now, we will represent the given reactions and their enthalpy changes on the enthalpy diagram: 1. For the reaction \(A \longrightarrow B\), draw an arrow from \(A\) to \(B\) with a positive slope (since \(\Delta H_1 = +30\, \text{kJ}\) is an endothermic reaction). Label the arrow with \(\Delta H_1=+30\, \text{kJ}\). 2. For the reaction \(B \longrightarrow C\), draw an arrow from \(B\) to \(C\) with a positive slope (since \(\Delta H_2 = +60\, \text{kJ}\) is again an endothermic reaction). Label the arrow with \(\Delta H_2=+60\, \text{kJ}\). 3. To represent the overall reaction \(A \longrightarrow C\), draw an arrow directly from \(A\) to \(C\), skipping \(B\). According to Hess's law, the enthalpy change for this reaction will be equal to the sum of \(\Delta H_1\) and \(\Delta H_2\), calculated in part (a). Label the arrow with \(\Delta H_{Total}=+90\, \text{kJ}\). The enthalpy diagram shows the enthalpy changes for each given reaction and the overall reaction \(A \longrightarrow C\). It visually demonstrates how Hess's law applies: the enthalpy change for the direct reaction \(A \longrightarrow C\) is equal to the sum of the enthalpy changes for the two hypothetical reactions.

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Most popular questions from this chapter

Can you use an approach similar to Hess's law to calculate the change in internal energy, \(\Delta E,\) for an overall reaction by summing the \(\Delta E\) values of individual reactions that add up to give the desired overall reaction?

From the enthalpies of reaction $$\begin{aligned} 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}(g) & \Delta H=-221.0 \mathrm{kJ} \\ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g)+4 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{CH}_{3} \mathrm{OH}(g) & \Delta H=-402.4 \mathrm{kJ} \end{aligned}$$ calculate \(\Delta H\) for the reaction $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)$$

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Calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) reacts with water to form acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(\mathrm{Ca}(\mathrm{OH})_{2} .\) From the following enthalpy of reaction data and data in Appendix \(\mathrm{C},\) calculate \(\Delta H_{f}^{\circ}\) for \(\mathrm{CaC}_{2}(s) :\) $$\begin{aligned} \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \\ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \Delta H^{\circ}=-127.2 \mathrm{kJ} \end{aligned}$$

Suppose an Olympic diver who weighs 52.0 kg executes a straight dive from a 10-m platform. At the apex of the dive, the diver is 10.8 m above the surface of the water. (a) What is the potential energy of the diver at the apex of the dive, relative to the surface of the water? (b) Assuming that all the potential energy of the diver is converted into kinetic energy at the surface of the water, at what speed, in m/s, will the diver enter the water? (c) Does the diver do work on entering the water? Explain.

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