Question

Given the following data: $$ \begin{array}{ll}{\mathrm{NO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}(g)} & {\Delta H=233 \mathrm{kJ}} \\ {2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)} & {\Delta H=-427 \mathrm{kJ}}\end{array} $$ $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta H=-199 \mathrm{kJ} $$ Calculate the bond energy for the \(\mathrm{O}_{2}\) bond, that is, calculate \(\Delta H\) for: $$ \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}(g) \qquad \Delta H=? $$

Short answer

The enthalpy change for the dissociation of \(O_{2}(g)\) to \(2O(g)\) is \(-2.5\, kJ\).

Step-by-step solution

Identify the target equation

The exercise asks for the enthalpy change of: \( O_{2}(g) \longrightarrow 2 O(g) \qquad \Delta H=? \)

Manipulate the given equations

First, we notice that in the given equations, we do not have O2 as a reactant and O as a product. However, we have O2 as a product and O as a reactant in some of them. To manipulate the given equations, we can reverse them to have the correct atoms on both reactant and product sides. By doing so, we will also need to change the sign of the enthalpy change for each reversed equation. We reverse the first two equations: 1. \( NO_{2}(g) \longrightarrow NO(g) + O(g) \quad \Delta H = 233 \, kJ \) (given equation) \( NO(g) + O(g) \longrightarrow NO_{2}(g) \quad \Delta H = -233 \, kJ \) (reversed equation) 2. \( 2 O_{3}(g) \longrightarrow 3 O_{2}(g) \quad \Delta H = -427 \, kJ \) (given equation) \( 3 O_{2}(g) \longrightarrow 2 O_{3}(g) \quad \Delta H = 427 \, kJ \) (reversed equation) The third equation remains the same: 3. \( NO(g) + O_{3}(g) \longrightarrow NO_{2}(g) + O_{2}(g) \quad \Delta H = -199 \, kJ \)

Add the manipulated equations

Now, let's add the reversed equations from Step 2 together: 1. \( NO(g) + O(g) \longrightarrow NO_{2}(g) \quad \Delta H = -233 \, kJ \) 2. \( 3 O_{2}(g) \longrightarrow 2 O_{3}(g) \quad \Delta H = 427 \, kJ \) 3. \( NO(g) + O_{3}(g) \longrightarrow NO_{2}(g) + O_{2}(g) \quad \Delta H = -199 \, kJ \) For better clarity, we rewrite these equations as: A. \( NO(g) + O(g) + 3 O_{2}(g) \longrightarrow NO_{2}(g) + 2 O_{3}(g) + NO(g) + O_{3}(g) \) By adding these equations, the following terms cancel out: NO(g), O(g), NO2(g), and O3(g) Resulting in: \( 3 O_{2}(g) \longrightarrow 2 O_{3}(g) + O_{2}(g) \) Which can be simplified as: \( 2 O_{2}(g) \longrightarrow 2 O_{3}(g) \)

Calculate the enthalpy change for the target equation

We added the equations to get the simplified equation, so now we must also add their enthalpy changes: \(\Delta H_{total} = -233\, kJ + 427\, kJ - 199\, kJ = -5\, kJ \) Now we have the enthalpy change for \( 2 O_{2}(g) \longrightarrow 2 O_{3}(g) \), which is -5 kJ. To find the enthalpy change for the target equation, \( O_{2}(g) \longrightarrow 2 O(g) \), we will divide the enthalpy change we just calculated by 2: \(\Delta H = -\frac{5 \, kJ}{2} = -2.5 \, kJ\) So, the enthalpy change for the dissociation of \( O_{2}(g) \) to \( 2 O(g) \) is \(-2.5 \, kJ\).

Key concepts

Bond Energy

Bond energy is a central concept in chemistry. It relates to the energy needed or released when a bond between two atoms is formed or broken.
This concept is particularly useful in understanding chemical reactions and the energy changes that accompany them. Let's dive deeper into this:

• **Definition**: Bond energy refers to the amount of energy required to break one mole of the bond in a gaseous substance.
• **Measuring Bond Energy**: It is typically measured in kilojoules per mole (kJ/mol). When bonds are broken, energy is absorbed (endothermic process). When bonds form, energy is released (exothermic process).
• **Role in Chemical Reactions**: Bond energy helps us understand reaction mechanisms. By comparing bond energies of reactants and products, we can predict whether a reaction will release or absorb energy.

When calculating enthalpy changes in reactions, considering the bond energies involved gives a more detailed perspective on the energy dynamics.

Thermodynamics

Thermodynamics is a fundamental scientific principle that describes how energy is transferred and transformed. It is crucial for understanding chemical reactions and their associated energy changes.
Here are the main aspects that relate to our discussion:

• **First Law of Thermodynamics**: Energy cannot be created or destroyed. In chemical reactions, this means the total energy of reactants equals the total energy of products.
• **Enthalpy (\( \Delta H \))**: This is a measure of the total heat content in a chemical system. Enthalpy changes accompany chemical reactions, indicating how much heat is absorbed or released.
• **Calculating with Enthalpy**: For any reaction, knowing the enthalpy change can tell us whether it's exothermic (releases heat, \( \Delta H < 0 \)) or endothermic (absorbs heat, \( \Delta H > 0 \)).

Understanding these aspects of thermodynamics helps in predicting and explaining the direction and nature of chemical reactions.

Chemical Reactions

Chemical reactions are processes in which substances interact to form new products. They are the core of chemistry and are governed by various principles and dynamics.
To better understand these reactions, consider the following:

• **Reactants and Products**: In a chemical reaction, reactants are transformed into products. This involves breaking bonds in reactants and forming new bonds in products.
• **Enthalpy Changes**: Each reaction has an associated enthalpy change. By calculating the difference in enthalpy between reactants and products, we can predict the reaction's energy dynamics.
• **Role of Bond Energies**: Breaking and forming bonds require energy changes, which are calculated using bond energies. This concept allows chemists to predict whether a reaction is likely to occur spontaneously.
• **Example**: Consider the breakdown of \( O_2 \). This is endothermic because energy must be supplied to break the strong \( O=O \) double bond, highlighting the importance of bond energy in reactions.

Understanding chemical reactions requires a clear grasp of these components and their interplay in a reaction's progress.