Question

The standard enthalpy of formation of solid barium oxide, \(\mathrm{BaO},\) is \(-553.5 \mathrm{kJ} / \mathrm{mol},\) and the standard enthalpy of formation of barium peroxide, \(\mathrm{BaO}_{2},\) is \(-634.3 \mathrm{kJ} / \mathrm{mol}\) (a) Calculate the standard enthalpy change for the following reaction. Is the reaction exothermic or endothermic? \(2 \mathrm{BaO}_{2}(\mathrm{s}) \rightarrow 2 \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})\) (b) Draw an energy level diagram that shows the relationship between the enthalpy change of the decomposition of \(\mathrm{BaO}_{2}\) to \(\mathrm{BaO}\) and \(\mathrm{O}_{2}\) and the enthalpies of formation of \(\mathrm{BaO}(\mathrm{s})\) and \(\mathrm{BaO}_{2}(\mathrm{s})\)

Short answer

The enthalpy change is \(161.6 \text{kJ/mol}\), and the reaction is endothermic.

Step-by-step solution

Write the Given Data

We have the standard enthalpy of formation for barium oxide, \( \Delta H_f^{\circ}(\text{BaO}) = -553.5 \text{kJ/mol} \), and for barium peroxide, \( \Delta H_f^{\circ}(\text{BaO}_2) = -634.3 \text{kJ/mol} \).

Write the Reaction Equation

The reaction is \( 2 \text{BaO}_2(s) \rightarrow 2 \text{BaO}(s) + \text{O}_2(g) \).

Use Hess's Law to Calculate Enthalpy Change

According to Hess's Law, the standard enthalpy change of the reaction \( \Delta H_{rxn}^{\circ} \) can be calculated as: \[ \Delta H_{rxn}^{\circ} = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants}) \] Substitute the given values:\[ \Delta H_{rxn}^{\circ} = \left( 2(-553.5) + 0 \right) - \left( 2(-634.3) \right) \]

Calculate the Change for Products

The total change for products is: \[ 2 \times (-553.5) = -1107 \text{kJ/mol} \] since \( \text{O}_2(g) \) is in its elemental form, its enthalpy of formation is zero.

Calculate the Change for Reactants

The total change for reactants is: \[ 2 \times (-634.3) = -1268.6 \text{kJ/mol} \]

Compute the Standard Enthalpy Change

Substitute the calculated enthalpies into the formula:\[ \Delta H_{rxn}^{\circ} = (-1107) - (-1268.6) = 161.6 \text{kJ/mol} \]

Determine the Nature of the Reaction

The positive sign of \( 161.6 \text{kJ/mol} \) indicates that the reaction is endothermic, as it absorbs energy.

Draw the Energy Level Diagram

In the energy level diagram, show \( \text{BaO}_2 \) at a lower energy state initially (-634.3), the decomposition process increases energy to reach the level of \( \text{BaO} \) and \( \text{O}_2 \) at energy -553.5. Mark the energy absorbed (161.6 \text{kJ/mol}) as the height difference between \( \text{BaO}_2 \) and the resultant \( \text{BaO} \) and \( \text{O}_2 \).

Key concepts

Hess's Law

Hess's Law is a fundamental concept in thermochemistry that states the total enthalpy change during a chemical reaction is the same, regardless of the number of steps the reaction is carried out in. This principle is based on the conservation of energy, meaning the total energy in a closed system must remain constant.

In practical terms, when calculating the enthalpy change (\( \Delta H \)) for a reaction, you can consider it as a sum of enthalpy changes of individual steps, even if those steps do not exactly mimic the reaction as it occurs naturally in a laboratory setting. By using Hess's Law, we can break down complex reactions into more manageable steps;
this is especially useful when the direct measurement of the enthalpy change of a reaction is difficult or impossible.

For example, in the original exercise, we use Hess's Law to find the standard enthalpy change (\( \Delta H_{rxn}^{\circ} \) ) of the decomposition of barium peroxide (\( \text{BaO}_2 \)). We calculate it by using the standard enthalpies of formation for both the reactants and products. This method shows how different reactions can be combined to find a solution, demonstrating the flexibility of Hess's Law.

Enthalpy of Formation

The enthalpy of formation is defined as the change in enthalpy when one mole of a compound is formed from its elements in their standard states. This is an essential concept in chemistry, as it helps us understand how much energy is either absorbed or released during the formation of a compound.

Standard enthalpies of formation (\( \Delta H_f^{\circ} \)) are frequently tabulated for a wide variety of substances at specific conditions, typically 25 degrees Celsius and 1 atmosphere pressure. These values serve as a reference point for comparing the stabilities of different compounds.

In the given exercise, the enthalpy of formation for barium oxide (\( \text{BaO} \)) is \(-553.5 \, \text{kJ/mol} \), and for barium peroxide (\( \text{BaO}_2 \)), it is \(-634.3 \, \text{kJ/mol} \). These values indicate energy is released when the compounds form from their elements, which suggests both are exothermic processes at formation. Nonetheless, when looking into the decomposition reaction, such data helps in calculating whether the overall transformation absorbs or emits energy.

Endothermic Reaction

An endothermic reaction is a chemical process in which the system absorbs energy from its surroundings, usually in the form of heat. This absorption results in a positive enthalpy change (\( \Delta H \) > 0).

Such reactions require an input of energy to proceed, often leading to a cooler environment around them as energy is absorbed. This makes it especially important to recognize endothermic reactions in both experimental and practical settings.

In the exercise, the calculated standard enthalpy change of \( 161.6 \, \text{kJ/mol} \) for the decomposition of \( \text{BaO}_2 \) into \( \text{BaO} \) and \( \text{O}_2 \) indicates an endothermic process. The positive sign tells us that energy is required for this reaction to occur, making it an absorbing process. This is visually represented in an energy level diagram by the products being at a higher energy state compared to the reactants. Recognizing whether a process is endo- or exothermic is crucial in predicting reaction behavior and energy requirements.