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Consider the following hypothetical reactions: $$ \begin{array}{l} \mathrm{A} \longrightarrow \mathrm{B} \quad \Delta H_{I}=+60 \mathrm{k} \mathrm{J} \\ \mathrm{B} \longrightarrow \mathrm{C} \quad \Delta H_{I}=-90 \mathrm{k} \mathrm{J} \end{array} $$ (a) Use Hess's law to calculate the enthalpy change for the reaction \(\mathrm{A} \longrightarrow \mathrm{C}\) (b) Construct an enthalpy diagram for substances \(A, B\), and \(\mathrm{C},\) and show how Hess's law applies.

Short Answer

Expert verified
(a) \( \Delta H_{A \rightarrow C} = -30 \mathrm{kJ} \). (b) Enthalpy diagram shows A to B (+60 kJ), B to C (-90 kJ), A to C (-30 kJ).

Step by step solution

01

Understand Hess's Law

Hess's Law states that the total enthalpy change for a reaction is the same, no matter how many steps the reaction takes. Thus, the enthalpy change for the overall reaction is the sum of the enthalpies of the individual steps.
02

Identify Individual Reactions

We are given two reactions: \( \mathrm{A} \rightarrow \mathrm{B} \) with \( \Delta H_I = +60 \mathrm{kJ} \) and \( \mathrm{B} \rightarrow \mathrm{C} \) with \( \Delta H_I = -90 \mathrm{kJ} \).
03

Apply Hess's Law for Overall Reaction

To find the enthalpy change for \( \mathrm{A} \rightarrow \mathrm{C} \), we sum the enthalpy changes of the individual reactions: \( \Delta H = +60 \mathrm{kJ} + (-90 \mathrm{kJ}) = -30 \mathrm{kJ} \). Thus, \( \Delta H \) for \( \mathrm{A} \rightarrow \mathrm{C} \) is \(-30 \mathrm{kJ} \).
04

Construct an Enthalpy Diagram

Create a vertical axis for enthalpy. Place \( A \) at a reference point. Position \( B \) 60 kJ higher than \( A \) (since \( \Delta H = +60 \mathrm{kJ} \)) and \( C \) 30 kJ below \( A \) (since the overall \( \Delta H = -30 \mathrm{kJ} \)). This diagram visually demonstrates how the enthalpy changes from \( A \) to \( B \) and \( B \) to \( C \), supporting Hess's Law.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
In chemistry, one of the essential principles is understanding how energy changes during reactions. The term **enthalpy change** refers to the heat energy transferred in a chemical reaction at constant pressure. It is denoted by the symbol \( \Delta H \). This energy change can either be positive or negative, indicating whether the system absorbs or releases heat.
- If \( \Delta H \) is positive, the reaction is endothermic, absorbing heat from the surroundings, as seen in the reaction \( A \rightarrow B \) with \( \Delta H_I = +60 \text{kJ} \).
- Conversely, if \( \Delta H \) is negative, the reaction is exothermic, releasing heat, like the reaction \( B \rightarrow C \) with \( \Delta H_I = -90 \text{kJ} \).

Hess's Law is particularly useful when calculating the enthalpy change for a multi-step reaction. It asserts that the total enthalpy change for a reaction is the same, irrespective of the path taken. Hence, for the reaction \( A \rightarrow C \), the overall enthalpy change is the sum of the changes from the intermediate steps, resulting in \( \Delta H = -30 \text{kJ} \).
Enthalpy Diagram
An **enthalpy diagram** is a visual tool that represents the energy changes during a reaction. It helps in understanding how the enthalpy of substances changes throughout a process.

To construct an enthalpy diagram for the reactions \( A \to B \) and \( B \to C \), we plot the enthalpy on the vertical axis:
  • Start with \( A \) at an arbitrary reference point.
  • Place \( B \) 60 kJ higher on the graph, showing the endothermic process of \( A \to B \).
  • Finally, place \( C \) 90 kJ lower than \( B \), which would be 30 kJ below \( A \), reflecting the overall exothermic process \( A \to C \).
This diagram effectively portrays how the enthalpy shifts between the substances and demonstrates the application of Hess's Law. Each segment represents an individual reaction step, correlating with the given enthalpy changes, and proves that regardless of the number of reaction steps, the initial and final enthalpy states determine the overall \( \Delta H \).
Hypothetical Reactions
**Hypothetical reactions** are a valuable concept for understanding theoretical energy changes in chemistry. These are imagined or theoretical reactions that help in analyzing and predicting the outcomes of actual reactions.

Consider our example, where we have two hypothetical reactions:
- \( A \to B \), where the enthalpy change is given as \( +60 \text{kJ} \).- \( B \to C \), with an enthalpy change of \( -90 \text{kJ} \).
The sum of these changes through Hess's Law allows us to hypothesize the enthalpy change from \( A \) to \( C \) without directly observing it. Even though these reactions may not occur in practice, they provide insight into calculating unknown \( \Delta H \) values for complex reactions. This approach is a cornerstone of thermochemistry, paving the way for predicting the feasibility and spontaneity of processes without direct experimentation.

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Most popular questions from this chapter

Consider the following reaction: $$ 2 \mathrm{CH}_{3} \mathrm{OH}(g) \longrightarrow 2 \mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \quad \Delta H=+252.8 \mathrm{~kJ} $$ (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(24.0 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}(g)\) is decomposed by this reaction at constant pressure. \((\mathbf{c})\) For a given sample of \(\mathrm{CH}_{3} \mathrm{OH},\) the enthalpy change during the reaction is \(82.1 \mathrm{~kJ} .\) How many grams of methane gas are produced? (d) How many kilojoules of heat are released when \(38.5 \mathrm{~g}\) of \(\mathrm{CH}_{4}(g)\) reacts completely with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{CH}_{3} \mathrm{OH}(g)\) at constant pressure?

Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: (a) \(q=0.763 \mathrm{~kJ}\) and \(w=-840 \mathrm{~J} .\) (b) A system releases \(66.1 \mathrm{~kJ}\) of heat to its surroundings while the surroundings do \(44.0 \mathrm{~kJ}\) of work on the system.

We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethane: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2 (g) $$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$ \begin{aligned} \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(I) & & \Delta H^{0}=-890.3 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{0} &=-571.6 \mathrm{~kJ} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(I) & \Delta H^{0}=&-3120.8 \mathrm{~kJ} \end{aligned} $$

Without doing any calculations, predict the sign of \(\Delta H\) for each of the following reactions: (a) \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) (b) \(2 \mathrm{~F}(g) \longrightarrow \mathrm{F}_{2}(g)\) (c) \(\mathrm{Mg}^{2+}(g)+2 \mathrm{Cl}^{-}(g) \longrightarrow \mathrm{MgCl}_{2}(s)\) (d) \(\operatorname{HBr}(g) \longrightarrow H(g)+B r(g)\)

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is produced by plants as follows: \(12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(I) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g)\) $$ \Delta H=5645 \mathrm{~kJ} $$ About \(4.8 \mathrm{~g}\) of sucrose is produced per day per square meter of the earth's surface. The energy for this endothermic reaction is supplied by the sunlight. About \(0.1 \%\) of the sunlight that reaches the earth is used to produce sucrose. Calculate the total energy the sun supplies for each square meter of surface area. Give your answer in kilowatts per square meter \(\left(\mathrm{kW} / \mathrm{m}^{2}\right.\) where \(\left.1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}\right)\)

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