Problem 72 - Chemistry: Principles and Reactions

Question

Given the following standard free energies at \(25^{\circ} \mathrm{C}\), $$ \begin{array}{ll} \mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow 2 \mathrm{NO}(g)+\frac{3}{2} \mathrm{O}_{2}(g) & \Delta G^{\circ}=-59.2 \mathrm{~kJ} \\ \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) & \Delta G^{\circ}=-35.6 \mathrm{~kJ} \end{array} $$ calculate \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the reaction $$ 2 \mathrm{NO}_{2}(g)+{ }_{2}^{1} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g) $$

Step-by-Step Solution

Answer

Question: Calculate the standard free energy change for the following reaction using the given data: Reaction: \(2\, \mathrm{NO}_2(g) + \frac{1}{2}\, \mathrm{O}_2(g) \longrightarrow \mathrm{N}_2\mathrm{O}_5(g)\) Given Data: 1. \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow 2 \mathrm{NO}(g) + \frac{3}{2} \mathrm{O}_{2}(g)\) ; \(\Delta G^{\circ} = -59.2 \, \mathrm{kJ}\) 2. \(\mathrm{NO}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)\) ; \(\Delta G^{\circ} = -35.6\, \mathrm{kJ}\) Answer: The standard free energy change (\(\Delta G^{\circ}\)) of the reaction \(2\, \mathrm{NO}_2(g) + \frac{1}{2}\, \mathrm{O}_2(g) \longrightarrow \mathrm{N}_2\mathrm{O}_5(g)\) at \(25^{\circ} \mathrm{C}\) is \(-12.0\) kJ.

Step 1: Write down the target reaction

We want to find \(\Delta G^{\circ}\) for the following reaction: $$ 2 \mathrm{NO}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g) $$

Step 2: Identify the equations to manipulate

We are given two reactions: $$ \begin{array}{ll} \mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow 2 \mathrm{NO}(g) + \frac{3}{2} \mathrm{O}_{2}(g) & \Delta G^{\circ} = -59.2 \mathrm{~kJ} \\ \mathrm{NO}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) & \Delta G^{\circ} = -35.6 \mathrm{~kJ} \end{array} $$

Step 3: Rearrange the first equation

Reverse the first reaction so that it aligns with the desired reaction: $$ 2 \mathrm{NO}(g) + \frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}\mathrm{O}_{5}(g) $$ Remember that reversing a reaction also reverses the sign of \(\Delta G^{\circ}\): $$ \Delta G^{\circ} = 59.2 \mathrm{~kJ} $$

Step 4: Multiply the second equation by 2

Multiply the second reaction by a factor of 2 so that we have the same number of \(\mathrm{NO}_2\) molecules on both sides of the desired reaction: $$ 2 \mathrm{NO}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ When multiplying a reaction, the value of \(\Delta G^{\circ}\) for that reaction must also be multiplied: $$ \Delta G^{\circ} = -71.2 \mathrm{~kJ} $$

Step 5: Add the manipulated equations

Add the manipulated reactions together to get the desired reaction: $$ (2 \mathrm{NO}(g) + \frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}\mathrm{O}_{5}(g)) + (2 \mathrm{NO}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)) $$ $$ \Rightarrow 2\,\mathrm{NO}_2(g) + \frac{1}{2}\,\mathrm{O}_2(g) \longrightarrow \mathrm{N}_2\mathrm{O}_5(g) $$ As we add the reactions, we also add their \(\Delta G^{\circ}\) values: $$ \Delta G^{\circ}_{\text{total}} = 59.2 \mathrm{~kJ} - 71.2 \mathrm{~kJ} = -12.0 \mathrm{~kJ} $$

Step 6: Conclusion

The standard free energy change (\(\Delta G^{\circ}\)) of the reaction $$ 2\,\mathrm{NO}_2(g) + \frac{1}{2}\,\mathrm{O}_2(g) \longrightarrow \mathrm{N}_2\mathrm{O}_5(g) $$ at \(25^{\circ} \mathrm{C}\) is \(-12.0\) kJ.

Key Concepts

Enthalpy

Enthalpy is a concept used in thermodynamics to describe the total heat content of a system. It includes the internal energy of the system as well as the product of its pressure and volume. Essentially, it represents the energy needed to create the system and maintain its pressure and volume.
Enthalpy is vital when discussing chemical reactions because it helps us understand how much heat is absorbed or released during a reaction.
Here are a few things that are important to know about enthalpy:

• It is represented by the symbol \( H \).
• Changes in enthalpy (\( \Delta H \)) are used to calculate the heat absorbed or released in a reaction.
• If \( \Delta H \) is negative, the reaction is exothermic (releases heat).
• If \( \Delta H \) is positive, the reaction is endothermic (absorbs heat).

Knowing the enthalpy change for a reaction helps predict whether the reaction will produce heat or require heat to proceed. It works in tandem with other thermodynamic properties, like entropy, to understand full reaction dynamics.

Chemical Reactions

A chemical reaction involves the transformation of one set of chemical substances into another. This process involves breaking and forming chemical bonds, leading to new products.
Chemical reactions are typically classified in a few basic types:

• Synthesis: Two or more reactants combine to form a single product.
• Decomposition: A single compound breaks down into two or more products.
• Single Replacement: An element in a compound is replaced by another element.
• Double Replacement: The ions of two compounds exchange places to form two new compounds.
• Combustion: A compound reacts with oxygen, often producing energy in the form of heat and light.

Chemical reactions are influenced by various factors:

• Concentration: Higher concentration of reactants can increase reaction rates.
• Temperature: Raising the temperature generally speeds up reactions.
• Pressure: For gases, increasing pressure can hasten the reaction.
• Catalysts: These substances increase reaction rates without being consumed.

Understanding these elements helps predict the behavior of chemical reactions under different conditions.

Thermodynamics

Thermodynamics is the branch of physical science that deals with the relations between heat and other energy forms. It is essential for understanding how energy transformations govern chemical reactions.
The laws of thermodynamics provide the framework for predicting the movement and distribution of energy:

• First Law: Energy cannot be created or destroyed, only transferred or transformed. This is also known as the principle of conservation of energy.
• Second Law: In any energy exchange, if no energy enters or leaves the system, the potential energy of the state will always be less than that of the initial state, known as entropy.
• Third Law: As temperature approaches absolute zero, the entropy of a system approaches a constant minimum.

Thermodynamics is crucial when calculating Gibbs free energy, which combines enthalpy and entropy into a single value representing the energy available to do work. In short, it allows us to predict whether a chemical reaction will occur spontaneously.
Through these concepts, thermodynamic principles help scientists and engineers design more efficient processes and products, thereby enhancing energy efficiency and sustainability.