Problem 62
Question
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.
Step-by-Step Solution
Verified Answer
(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).
1Step 1: 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.
2Step 2: 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 \( \Delta H_I = -90 \mathrm{kJ} \).
3Step 3: 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 \(-30 \mathrm{kJ} \).
4Step 4: 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.
Key Concepts
Enthalpy ChangeEnthalpy DiagramHypothetical Reactions
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 \( \Delta H = -30 \text{kJ} \).
- 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:
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 \( A \to C \).
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.
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.
Other exercises in this chapter
Problem 59
Under constant-volume conditions, the heat of combustion of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is \(16.49 \mathrm{~kJ} / \
View solution Problem 60
Under constant-volume conditions, the heat of combustion of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) is \(40.18 \mathrm{~kJ} / \mathrm{g}\).
View solution Problem 64
From the enthalpies of reaction $$ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}(g) \quad \Delta H=-221.0 \mathrm{~kJ} $$ \(2 \mathrm{C}(s)+\m
View solution Problem 65
From the enthalpies of reaction $$ \begin{array}{clrl} \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) & \longrightarrow 2 \mathrm{HF}(g) & & \Delta H=-537 \mathrm{~kJ} \\
View solution