Problem 48 - Chemistry

Question

Given the thermochemical data, \(\mathrm{A}+\mathrm{B} \longrightarrow 2 \mathrm{C} \quad \Delta H_{1}=600 \mathrm{~kJ} / \mathrm{mol}\) \(\begin{array}{ll}2 \mathrm{C}+\mathrm{D} \longrightarrow 2 \mathrm{E} & \Delta H_{1}=210 \mathrm{~kJ} / \mathrm{mol}\end{array}\) Determine the enthalpy change for each of the following: a) \(4 \mathrm{E} \longrightarrow 4 \mathrm{C}+2 \mathrm{D}\) d) \(2 C+2 E \longrightarrow 2 A+2 B+D\) b) \(\mathrm{A}+\mathrm{B}+\mathrm{D} \longrightarrow 2 \mathrm{E}\) e) \(\mathrm{E} \longrightarrow \frac{1}{2} \mathrm{~A}+\frac{1}{2} \mathrm{~B}+\frac{1}{2} \mathrm{D}\) c) \(\mathrm{C} \longrightarrow \frac{1}{2} \mathrm{~A}+\frac{1}{2} \mathrm{~B}\)

Step-by-Step Solution

Answer

a) \(-420\, \text{kJ/mol}\), b) \(810\, \text{kJ/mol}\), c) \(-300\, \text{kJ/mol}\), d) \(-990\, \text{kJ/mol}\), e) \(-405\, \text{kJ/mol}\).

Step 1: Identifying the Given Reactions

We have two given reactions with their respective enthalpy changes: \(\mathrm{A} + \mathrm{B} \rightarrow 2\mathrm{C} \quad \Delta H_1 = 600 \, \text{kJ/mol}\) and \(2\mathrm{C} + \mathrm{D} \rightarrow 2\mathrm{E} \quad \Delta H_2 = 210 \, \text{kJ/mol}\). We will use these to find the enthalpy change for the given transformations.

Step 2: Reaction a: Reverse the Given Reactions

The reaction \(4\mathrm{E} \rightarrow 4\mathrm{C} + 2\mathrm{D}\) is the reverse of the reaction \(2\mathrm{C} + \mathrm{D} \rightarrow 2\mathrm{E}\). Thus, the enthalpy change \(\Delta H\) will be the negative of \(2\times\Delta H_2\): \(\Delta H = -2 \times 210 = -420 \, \text{kJ/mol}\).

Step 3: Reaction b: Combine Given Reactions

For the reaction \(\mathrm{A} + \mathrm{B} + \mathrm{D} \rightarrow 2\mathrm{E}\), add \(\mathrm{A} + \mathrm{B} \rightarrow 2\mathrm{C}\) and \(2\mathrm{C} + \mathrm{D} \rightarrow 2\mathrm{E}\). Sum their enthalpies: \(\Delta H = 600 + 210 = 810 \, \text{kJ/mol}\).

Step 4: Reaction c: Determine Single Molecule of C

The reaction \(\mathrm{C} \rightarrow \frac{1}{2}\mathrm{A} + \frac{1}{2}\mathrm{B}\) is half of \(2\mathrm{C} \rightarrow \mathrm{A} + \mathrm{B}\). The enthalpy change is half of \(-600 \, \text{kJ/mol}\): \(\Delta H = -300 \, \text{kJ/mol}\).

Step 5: Reaction d: Adjust Reactions for Products

For \(2\mathrm{C} + 2\mathrm{E} \rightarrow 2\mathrm{A} + 2\mathrm{B} + \mathrm{D}\), first reverse \(\mathrm{A} + \mathrm{B} \rightarrow 2\mathrm{C}\) twice, and add \(2\mathrm{C} + \mathrm{D} \rightarrow 2\mathrm{E}\) as it is: \(\Delta H = -600 + -600 + 210 = -990 \, \text{kJ/mol}\).

Step 6: Reaction e: Half of Reverse Reaction

The reaction \(\mathrm{E} \rightarrow \frac{1}{2}\mathrm{A} + \frac{1}{2}\mathrm{B} + \frac{1}{2}\mathrm{D}\) is half of the reverse of \(2\mathrm{C} + \mathrm{D} \rightarrow 2\mathrm{E}\) plus half a reversal of \(\mathrm{A} + \mathrm{B} \rightarrow 2\mathrm{C}\). \(\Delta H = -210/2 + -600/2 = -405 \, \text{kJ/mol}\).

Key Concepts

Thermochemical Reactions

Thermochemical reactions excite the chemistry enthusiasts among us by focusing on both chemical transformations and the accompanying energy changes. When chemical reactions occur, bonds between atoms are broken and new ones are formed. This process either absorbs or releases energy.

- **Exothermic reactions** release energy, usually in the form of heat. This results in an increase in the temperature of the surroundings.- **Endothermic reactions** absorb energy, causing the surrounding's temperature to decrease.
A thermochemical equation explicitly shows the amount of energy change, denoted as \( \Delta H \), during a reaction. If \( \Delta H \) is positive, the reaction is endothermic. A negative \( \Delta H \) indicates an exothermic reaction. Understanding these energy changes is critical for applications in energy production and understanding natural phenomena.

Chemical Equations

Chemical equations are the language of chemistry. They use symbols and formulas to represent compounds involved in chemical reactions. Each chemical equation gives us a snapshot of the atom realignment during a reaction, showing the reactants on the left side and the products on the right.

Balanced chemical equations maintain the conservation of mass. This means the number of each type of atom is the same on both sides of the equation. When dealing with thermochemical reactions, these equations will also include the enthalpy change \( \Delta H \), indicating whether a reaction absorbs or releases heat.
Being fluent in reading and writing chemical equations is essential for reaction analysis, as demonstrated in the exercise where reactions and their respective heat changes were critically evaluated to understand the whole process.

Reaction Enthalpy Calculations

Finding the enthalpy change for reactions requires a clear understanding of how to manipulate thermochemical equations. Calculating the reaction enthalpy involves using given equations to determine energy changes.

When equations are combined, reversed, or scaled, their \( \Delta H \) values alter accordingly:

• If a reaction is reversed, the sign of \( \Delta H \) is also reversed.
• If a reaction is multiplied by a factor, \( \Delta H \) is multiplied by the same factor.

In our exercise, multiple steps revealed that reversing equations allows us to consider reactions from a new perspective. Thus, reaction enthalpy calculations facilitate understanding how different energetic dynamics interplay during chemical transformations.

Hess's Law

Hess's Law is a powerful principle in chemistry, stating that the total enthalpy change of a reaction is the same, regardless of the pathway taken, as long as the initial and final conditions are the same. This law is rooted in the conservation of energy principle.

With Hess’s Law, you can break complex reactions into simpler steps, sum their enthalpies, and reach the enthalpy change of the overall reaction. This process was utilized in the provided exercise where intermediate steps were dissected and their enthalpy changes summed up.
This flexibility is crucial for calculating enthalpies when multiple reactions are involved and when direct measurement is unfeasible. Embracing Hess’s Law can transform daunting enthalpy problems into manageable puzzles.