Question

To estimate the enthalpy change for the reaction $$\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ what bond dissociation enthalpies do you need? Outline the calculation, being careful to show correct algebraic signs.

Short answer

Bond dissociation enthalpies for \(\mathrm{O=O}\), \(\mathrm{H-H}\), and \(\mathrm{O-H}\) are needed. \(\Delta H_{rxn} = -482\) kJ/mol.

Step-by-step solution

Identify the bonds in reactants and products

In the reaction \(\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), the reactants consist of \(\mathrm{O_{2}}\) with one \(\mathrm{O=O}\) double bond and \(2\ \mathrm{H_{2}}\) with two \(\mathrm{H-H}\) single bonds. The products contain \(2\ \mathrm{H_{2}O}\) molecules, each with two \(\mathrm{O-H}\) single bonds.

Write the expression for the enthalpy change

The enthalpy change for the reaction, \(\Delta H_{rxn}\), is calculated using the bond dissociation enthalpies of the bonds broken and formed. The general expression is: \[ \Delta H_{rxn} = \sum\text{(Bond Energies of Bonds Broken)} - \sum\text{(Bond Energies of Bonds Formed)}. \]

Determine the bonds broken in the reactants

The bonds broken in the reactants include 1 \(\mathrm{O=O}\) bond and 2 \(\mathrm{H-H}\) bonds. Therefore, the enthalpy change for breaking these bonds is the sum of the bond dissociation energies: \(E_{O=O} + 2E_{H-H}\).

Determine the bonds formed in the products

The bonds formed in the products include 4 \(\mathrm{O-H}\) bonds (since each of the 2 \(\mathrm{H_{2}O}\) molecules has 2 \(\mathrm{O-H}\) bonds). Therefore, the enthalpy change for forming these bonds is \(4E_{O-H}\).

Substitute bond energies and calculate \(\Delta H_{rxn}\)

Using commonly found bond enthalpy values: \(E_{O=O} = 498 \) kJ/mol, \(E_{H-H} = 436 \) kJ/mol, and \(E_{O-H} = 463 \) kJ/mol, substitute as follows: \[ \Delta H_{rxn} = (498 + 2 \times 436) - 4 \times 463 = 1370 - 1852 = -482 \text{ kJ/mol}. \] The negative sign indicates that the reaction is exothermic.

Key concepts

Bond Dissociation Enthalpy

Bond dissociation enthalpy is a crucial concept in chemistry, specifically when discussing chemical reactions. It refers to the energy required to break one mole of a specific type of bond in gaseous molecules. This parameter is essential when estimating the enthalpy change of a reaction as it involves both breaking old bonds and forming new ones.

In the context of the given reaction, the bond dissociation enthalpy values needed are from the bonds present in both the reactants and the products. For example, you will consider the energy required to break the \(O=O\) double bond in \(O_2\) and the \(H-H\) single bonds in two moles of hydrogen gas \(H_2\). On the other hand, the energy released upon forming new bonds such as the \(O-H\) single bonds in water molecules \(H_2O\) is also taken into account.

• **Bond Breaking Energy**: Identifies energy inputs required for breaking existing bonds.
• **Bond Formation Energy**: Represents energy released when new bonds are created.

This approach helps in determining whether a reaction is endothermic (requires energy) or exothermic (releases energy). A careful calculation of these energies aids in understanding the overall enthalpy change of any given chemical reaction.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products. This process entails breaking of chemical bonds in the reactants and the formation of new bonds in the products. To understand the changes entailed, it’s crucial to analyze the bonds involved in each side of the reaction equation.

For the given reaction, \(\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), it involves both breaking and forming of multiple chemical bonds:

• The reactants include one mole of oxygen gas, \(\mathrm{O}_2\), with an \(O=O\) double bond and two moles of hydrogen gas, \(\mathrm{H}_2\), each containing an \(H-H\) bond.
• The products, composed of two moles of water vapor, \(\mathrm{H}_2O\), feature two \(O-H\) bonds per water molecule.

Identifying the bonds involved helps in understanding the energy changes in terms of bond dissociation enthalpy values. It aids students in recognizing the types of bonds that need to be considered when calculating the energy demands or releases during the reaction.

Exothermic Reaction

An exothermic reaction is a chemical reaction that releases energy, primarily in the form of heat, into its surroundings. This type of reaction is characterized by a negative enthalpy change \(\Delta H_{rxn} < 0\).

In the given exercise, calculating \(\Delta H_{rxn}\) for the reaction \(\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), produced -482 kJ/mol. This negative result indicates that the reaction releases 482 kJ of energy per mole of product formed.

• This energy release is due to the greater strength (higher bond energies) of the bonds formed in the products compared to those of the bonds broken in the reactants.
• Exothermic reactions often result in a noticeable temperature increase in the surrounding area, making them easy to observe and identify.

Understanding exothermic reactions helps students predict energy changes, contributing to better handling and application of chemical processes in real-world scenarios.

Energy Calculation

Energy calculation in a chemical reaction involves determining the enthalpy change \(\Delta H_{rxn}\). This calculation assesses the balance between the energies required for breaking bonds in the reactants and the energies released upon forming new bonds in the products.

For the example reaction, the relevant calculations include:

• Breaking one \(O=O\) bond which requires 498 kJ/mol.
• Breaking two \(H-H\) bonds at 436 kJ/mol each, totaling 872 kJ.
• Forming four \(O-H\) bonds in the product, each releasing 463 kJ/mol, amassing 1852 kJ in total.

With these values, the enthalpy change \(\Delta H_{rxn}\) is calculated using the formula:
\[\Delta H_{rxn} = (498 + 872) - 1852 = -482 \, \text{kJ/mol}.\]
This equation reveals the overall energy change in the reaction and, indeed, the negative result confirms this is an exothermic reaction. Understanding how to perform such calculations is fundamental to comprehending energy transfer in chemistry.