Problem 55 - Chemistry and Chemical Reactivity

Question

The standard enthalpy of formation of solid barium oxide, \(\mathrm{BaO},\) is \(-553.5 \mathrm{kJ} / \mathrm{mol},\) and the 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? $$\mathrm{BaO}_{2}(\mathrm{s}) \longrightarrow \mathrm{BaO}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ (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 \(\mathrm{BaO}_{2}(\mathrm{s})\).

Step-by-Step Solution

Answer

(a) 80.8 kJ/mol, endothermic. (b) Energy diagram shows increase from \(\mathrm{BaO}_2\) to \(\mathrm{BaO} + \frac{1}{2} \mathrm{O}_2\).

Step 1: Understand the given data

We are provided with the standard enthalpies of formation for barium oxide \( \mathrm{BaO} \) and barium peroxide \( \mathrm{BaO}_2 \). These values are \(-553.5 \mathrm{kJ/mol}\) and \(-634.3 \mathrm{kJ/mol}\), respectively.

Step 2: Write the chemical equation

The reaction given is the decomposition of barium peroxide into barium oxide and oxygen: \[\mathrm{BaO}_2(\mathrm{s}) \longrightarrow \mathrm{BaO}(\mathrm{s}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g})\]

Step 3: Apply Hess's Law

According to Hess's Law, the change in enthalpy \( \Delta H \) for a reaction is given by the difference between the enthalpies of formation of the products and the reactants. \[ \Delta H = [ \Delta H_f(\mathrm{BaO}) + \frac{1}{2}\Delta H_f(\mathrm{O}_2)] - \Delta H_f(\mathrm{BaO}_2) \] Given that standard enthalpy of formation for \( \mathrm{O}_2(g) \) is 0, the equation simplifies to: \[ \Delta H = [ -553.5 + 0] - (-634.3)\]

Step 4: Calculate the enthalpy change

Substitute the given enthalpy values into the formula: \[ \Delta H = (-553.5 + 0) + 634.3 \] \[ \Delta H = 80.8 \mathrm{kJ/mol} \]

Step 5: Determine the reaction type

Since the enthalpy change is positive (\( 80.8 \mathrm{kJ/mol} \)), the reaction absorbs heat, making it endothermic.

Step 6: Represent on the energy level diagram

Draw an energy level diagram with the enthalpy level of \( \mathrm{BaO}_2 \) lower than that of \( \mathrm{BaO} \) by 80.8 kJ/mol. The diagram should show an increase in enthalpy from \( \mathrm{BaO}_2 \) to \( \mathrm{BaO} + \frac{1}{2} \mathrm{O}_2 \), indicating an endothermic process.

Key Concepts

Hess's Law

Hess's Law is an important principle in thermochemistry. This law tells us that if a chemical reaction can be expressed as a series of steps, then the enthalpy change of the overall reaction is just the sum of the enthalpy changes of the individual steps. The beauty of Hess's Law is that it allows us to calculate the enthalpy change for reactions where direct measurement is difficult.
To apply Hess's Law, we use known enthalpies of formation, which are the changes in enthalpy when one mole of a compound is formed from its elements in their standard states. The enthalpy change for a reaction can be determined by the formula:
\[ \Delta H_{reaction} = \sum \Delta H_f (\text{products}) - \sum \Delta H_f (\text{reactants}) \]
For example, to find the enthalpy change for the decomposition of barium peroxide, we use the enthalpy of formation for the products and reactants. This allows calculation of the enthalpy change even when we can't directly measure it during the reaction.

Enthalpy of Formation

Enthalpy of formation is a key concept that helps us understand energy changes in chemical reactions. It is the heat absorbed or released when one mole of a compound is formed from its constituent elements under standard conditions (usually 1 atm pressure and 25°C). These enthalpies are crucial because they provide a reference point for comparing different substances.
When we talk about standard enthalpy of formation, we mean the enthalpy change when forming a substance from the elements in their most stable states. A negative enthalpy of formation indicates that forming the compound from its elements releases heat, making it an exothermic process. Conversely, a positive value would suggest an endothermic formation.
For example, the enthalpy of formation for barium oxide (\(\mathrm{BaO}\)) is given as \(-553.5 \mathrm{kJ/mol}\). This value indicates that forming \(\mathrm{BaO}\) from barium and oxygen is an exothermic process. Using these values in Hess's Law helps us compute the enthalpy change for reactions, even if they're not directly observable.

Endothermic Reaction

An endothermic reaction is one that absorbs heat from its surroundings. This means the products have higher energy than the reactants. In terms of enthalpy change, an endothermic reaction has a positive \(\Delta H\). Simply put, energy is required to drive the reaction forward.
In the exercise given, the decomposition of barium peroxide to barium oxide and oxygen is an endothermic process. We calculated an enthalpy change of \(80.8 \mathrm{kJ/mol}\), which means this amount of energy is absorbed during the reaction. This absorption of heat can be visualized in an energy level diagram as an upward shift, indicating that the products of the reaction are at a higher energy level than the reactants.
Understanding whether a reaction is endothermic or exothermic helps in predicting how it will behave in different conditions, like temperature changes. This can guide practical applications in laboratories and industries where energy costs and safety are important concerns.