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) \(3 \mathrm{D}+\frac{1}{2} \mathrm{~A} \stackrel{\longrightarrow}{\longrightarrow} \mathrm{C}\)

Short answer

a) +150 kJ/mol, d) +300 kJ/mol, b) +600 kJ/mol, e) -300 kJ/mol, c) +150 kJ/mol.

Step-by-step solution

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, \( \Delta H = +150 \, \mathrm{kJ/mol} \).

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 \( 150 \, \mathrm{kJ/mol} \times 2 = 300 \, \mathrm{kJ/mol} \).

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 \( +600 \, \mathrm{kJ/mol} \).

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 \( \Delta H = 900 - 1200 = -300 \, \mathrm{kJ/mol} \).

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

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.

Understanding these basics helps in predicting the behavior and outcome of chemical reactions, making these concepts crucial for reaction enthalpy calculations.

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.

This is particularly useful for reactions where direct measurement of \(\Delta H\) is challenging or impossible. By using known reactions and their enthalpy changes as building blocks, Hess's Law allows for the construction and analysis of new, complex reactions, simplifying enthalpy calculations.

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.

These calculations aid in understanding the energy profile of reactions, helping predict whether a reaction will be spontaneous, the conditions required, and the feasible energy applications of a reaction.