Problem 47
Question
Given the thermochemical data,$$\begin{array}{ll}\mathrm{A}+6 \mathrm{~B} \longrightarrow 4 \mathrm{C} & \Delta H_{1}=-1200 \mathrm{~kJ} / \mathrm{mol} \\\\\mathrm{C}+\mathrm{B} \longrightarrow \mathrm{D} & \Delta H_{1}=-150 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ Determine the enthalpy change for each of the following: a) \(\mathrm{D} \longrightarrow \mathrm{C}+\mathrm{B}\) d) \(2 \mathrm{D} \longrightarrow 2 \mathrm{C}+2 \mathrm{~B}\) b) \(2 \mathrm{C} \longrightarrow \frac{1}{2} \mathrm{~A}+3 \mathrm{~B}\) e) \(6 \mathrm{D}+\mathrm{A} \longrightarrow 10 \mathrm{C}\) c) \(3 \mathrm{D}+\frac{1}{2} \mathrm{~A} \stackrel{\longrightarrow}{\longrightarrow} \mathrm{C}\)
Step-by-Step Solution
Verified Answer
a) +150 kJ/mol, d) +300 kJ/mol, b) +600 kJ/mol, e) -300 kJ/mol, c) +150 kJ/mol.
1Step 1: Calculate Reverse Reaction Entropy
For the reaction given by \( \mathrm{D} \rightarrow \mathrm{C} + \mathrm{B} \), the reverse of \( \mathrm{C} + \mathrm{B} \rightarrow \mathrm{D} \), the enthalpy change \( \Delta H \) is the opposite of the forward reaction. Therefore, \( \Delta H = +150 \, \mathrm{kJ/mol} \).
2Step 2: Scale Up Reaction Entropy
For the reaction \( 2 \mathrm{D} \rightarrow 2 \mathrm{C} + 2 \mathrm{B} \), we double the reaction \( \mathrm{D} \rightarrow \mathrm{C} + \mathrm{B} \). So the enthalpy change is \( 150 \, \mathrm{kJ/mol} \times 2 = 300 \, \mathrm{kJ/mol} \).
3Step 3: Rearrange and Balance Equation
For the reaction \( 2\mathrm{C} \rightarrow \frac{1}{2} \mathrm{~A} + 3 \mathrm{~B} \), notice that \( 4 \mathrm{C} \rightarrow \mathrm{A} + 6 \mathrm{~B} \) has an enthalpy change \( \Delta H = +1200 \, \mathrm{kJ/mol} \) (opposite sign). So, dividing the entire equation by 2, the enthalpy change is \( +600 \, \mathrm{kJ/mol} \).
4Step 4: Combine and Balance Enthalpies
For the reaction \( 6\mathrm{D} + \mathrm{A} \rightarrow 10\mathrm{C} \), we use reactions: \( 6(\mathrm{D} \rightarrow \mathrm{C} + \mathrm{B}) \) which gives \( 6 \times 150 = +900 \, \mathrm{kJ/mol} \) and the reverse \( \mathrm{A} + 6\mathrm{B} \rightarrow 4\mathrm{C} \) with \( \Delta H = +1200 \, \mathrm{kJ/mol} \). Adding these gives \( \Delta H = 900 - 1200 = -300 \, \mathrm{kJ/mol} \).
5Step 5: Combine Steps for Final Enthalpy
For \( 3\mathrm{D} + \frac{1}{2} \mathrm{A} \rightarrow \mathrm{C} \), use \( 3(\mathrm{D} \rightarrow \mathrm{C} + \mathrm{B}) \) which gives \( 3 \times 150 = +450 \, \mathrm{kJ/mol} \) and reverse \( \frac{1}{2} \mathrm{A} + 3\mathrm{B} \rightarrow 2\mathrm{C} \) which gives \( +600 \, \mathrm{kJ/mol} \). Combine both: \( +450 - 300 = +150 \, \mathrm{kJ/mol} \).
Key Concepts
ThermochemistryChemical ReactionsHess's LawReaction Enthalpy Calculations
Thermochemistry
Thermochemistry is a branch of chemistry that deals with studying the heat involved during chemical reactions. It is particularly concerned with energy changes that accompany reactions and phase changes. Understanding these energy changes is important for predicting whether a reaction will occur spontaneously. In thermochemistry, energy changes are often expressed as enthalpy changes (\(\Delta H\)). The enthalpy change reflects the heat absorbed or released during a reaction at constant pressure. If \(\Delta H\) is negative, the reaction is exothermic, meaning it releases heat into its surroundings. In contrast, a positive \(\Delta H\) indicates an endothermic reaction, which absorbs heat from the surroundings. These concepts are crucial when studying reaction mechanisms and predicting the behavior of chemical processes. Thermochemistry provides insight into how chemical reactions occur and evolve by focusing on energy. It emphasizes the natural tendency of systems to progress toward lower energy states, which is a guiding principle in understanding chemical behavior.
Chemical Reactions
Chemical reactions involve the transformation of one or more substances into different substances through the breaking and forming of chemical bonds. This can involve elements, compounds, or both, and is a fundamental process that occurs in chemistry.
In any chemical reaction, the reactants undergo changes to become products, and the reaction can either release or absorb energy.
Key points to understand about chemical reactions include:
- Conservation of mass: The total mass of reactants equals the total mass of products.
- Reaction rate: This dictates how fast the reactants are converted into products and can be influenced by factors like temperature, pressure, and concentration.
- Equilibrium: Many reactions reach a state of balance where the rate of the forward reaction equals the rate of the reverse reaction.
Hess's Law
Hess's Law is a principle in chemistry stating that the total enthalpy change for a chemical reaction is the same, no matter how many steps the reaction is carried out in. This law is based on the conservation of energy and is vital for calculating enthalpy changes indirectly.Using Hess's Law, chemists can:
- Determine the enthalpy change for reactions that are difficult to study directly.
- Break down complex reactions into simpler steps which can be calculated and summed.
- Apply algebraic manipulations (reversing, scaling, etc.) to known reactions to find unknown enthalpies.
Reaction Enthalpy Calculations
Reaction enthalpy calculations involve determining the change in enthalpy (\(\Delta H\)) for a given chemical reaction. These calculations are essential in thermochemistry as they predict how much heat is absorbed or released. Calculations can be made using standard enthalpy changes or applying Hess's Law for more complex reactions.Common steps involved in reaction enthalpy calculations include:
- Identifying the reaction stages and basic enthalpy changes.
- Reversing reactions when needed, which changes the sign of \(\Delta H\).
- Applying stoichiometry to scale reactions accordingly, adjusting \(\Delta H\) proportionally.
- Combining individual enthalpy changes for multi-step reactions.
Other exercises in this chapter
Problem 45
State Hess's law. Explain, with one example, the usefulness of Hess's law in thermochemistry.
View solution Problem 46
Describe how chemists use Hess's law to determine the \(\Delta H_{\mathrm{f}}^{\circ}\) of a compound by measuring its heat (enthalpy) of combustion.
View solution Problem 48
Given the thermochemical data, \(\mathrm{A}+\mathrm{B} \longrightarrow 2 \mathrm{C} \quad \Delta H_{1}=600 \mathrm{~kJ} / \mathrm{mol}\) \(\begin{array}{ll}2 \m
View solution Problem 49
From these data, $$\begin{array}{l}\mathrm{S} \text { (rhombic) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) \\\\\qquad \begin{aligned}\Delta H_{\math
View solution