Question

The bombardier beetle uses an explosive discharge as a defensive measure. The chemical reaction involved is the oxidation of hydroquinone by hydrogen peroxide to produce quinone and water: $$ \mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ Calculate \(\Delta H\) for this reaction from the following data: \(\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(a q)+\mathrm{H}_{2}(g)\) $$ \begin{aligned} \Delta H &=+177.4 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q) & \Delta H=-191.2 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H=-241.8 \mathrm{~kJ} \\ \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta H=-43.8 \mathrm{~kJ} \end{aligned} $$

Short answer

The enthalpy change for the bombardier beetle's defensive reaction is \(764.6\,\text{kJ}\).

Step-by-step solution

Identify the target reaction

The reaction we need to calculate the enthalpy change for is: \(\mathrm{C}_6 \mathrm{H}_4(\mathrm{OH})_2(aq) + \mathrm{H}_2 \mathrm{O}_2(aq) \longrightarrow \mathrm{C}_6 \mathrm{H}_4 \mathrm{O}_2(aq) + 2 \mathrm{H}_2 \mathrm{O}(l)\)

Apply Hess's Law

Hess's Law states that the enthalpy change of a reaction depends only on the initial and final states of the reactants and products, and not on the pathway taken. Therefore, we can use the given reactions to create a path to the target reaction and sum their enthalpy changes to find the enthalpy change for the target reaction.

Rearrange the given reactions

We need to arrange and/or reverse the given reactions in such a way that, when combined, they result in the target reaction. To do this, we can follow these steps: 1. Leave Reaction 1 as it is: \(\mathrm{C}_6 \mathrm{H}_4(\mathrm{OH})_2(aq) \longrightarrow \mathrm{C}_6 \mathrm{H}_4 \mathrm{O}_2(aq) + \mathrm{H}_2(g)\) \(\Delta H = +177.4\,\text{kJ}\) 2. Reverse Reaction 2: \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g)\) \(\Delta H = +191.2\,\text{kJ}\) 3. Multiply Reaction 3 by 2 to match the number of water molecules in the target reaction, and then reverse it: \(2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g)\) \(\Delta H = +483.6\,\text{kJ}\) 4. Multiply Reaction 4 by 2 to match the number of water molecules in the target reaction: \(2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H = -87.6\,\text{kJ}\)

Add the rearranged reactions and their enthalpy changes

Add the four modified reactions together: \(\mathrm{C}_6 \mathrm{H}_4(\mathrm{OH})_2(aq) + \mathrm{H}_2 \mathrm{O}_2(aq) \longrightarrow \mathrm{C}_6 \mathrm{H}_4 \mathrm{O}_2(aq) + 2 \mathrm{H}_2 \mathrm{O}(l)\) Add their enthalpy changes: \(\Delta H = (+177.4) + (+191.2) + (+483.6) + (-87.6) = 764.6\, \text{kJ}\)

Write the answer

The enthalpy change for the bombardier beetle's defensive reaction is \(764.6\,\text{kJ}\).

Key concepts

Hess's Law

Understanding Hess's Law is crucial when tackling problems involving enthalpy change calculations in chemical reactions. Enthalpy, a measure of the total energy of a thermodynamic system, is a state function, meaning it depends only on the initial and final states of a system and not the path taken to get there. Hess's Law takes advantage of this property by allowing us to determine the enthalpy change of a complex reaction by breaking it down into a series of simpler steps, whose enthalpies are known.

When applying Hess's Law, we can either reverse or multiply the enthalpy changes of known reactions to construct a pathway that represents the overall reaction of interest. This approach is helpful because it removes the need to directly measure the enthalpy change of the overall reaction, which might be difficult or impossible in some cases. Instead, we can rely on tabulated standard enthalpies of formation or transition states from databases.

In the case of the bombardier beetle's defense mechanism, the enthalpy change calculation is made possible by Hess's Law. We take the known reactions with their enthalpy changes, and by reversing and scaling them properly, we are able to sum the changes and thus find the overall reaction's enthalpy change. It is a powerful tool for students and chemists alike, turning a potentially arduous empirical task into a more manageable, theoretical calculation.

Chemical Thermodynamics

Chemical thermodynamics is the branch of chemistry that deals with the study of the interrelation of heat and work with chemical reactions or with physical changes of state within the confines of the laws of thermodynamics. It involves looking at the flow and transformation of energy within chemical processes, with enthalpy being a central concept. Enthalpy changes inform us about the heat absorbed or evolved during reactions at constant pressure.

Considering our example, the explosive defensive reaction of the bombardier beetle can be understood from a thermodynamic perspective. The reactants are transformed into products, and there is a flow of energy in the form of heat. By calculating the enthalpy change of the reaction using the given data, we gain insights into the energy dynamics of the beetle's unique defense mechanism and the thermodynamic feasibility of the reaction.

To improve students' grasp of these calculations, it is important to recognize that thermodynamics is more than a set of abstract equations; it's a description of physical reality. Students should be encouraged to connect the mathematics of thermodynamics with the tangible concepts of energy and temperature they experience in their daily lives.

Oxidation-Reduction Reactions

The oxidation-reduction (redox) reactions are a fundamental class of chemical reactions involving the transfer of electrons from one substance to another. Understanding redox reactions is essential because they play a major role in energy production, chemical synthesis, and biological processes.

In redox terms, the bombardier beetle's defense mechanism entails the oxidation of hydroquinone into quinone, with hydrogen peroxide being the oxidizing agent. During this reaction, hydroquinone loses electrons (is oxidized), whereas hydrogen peroxide gains electrons (is reduced). The movement of electrons corresponds to a flow of energy, which is captured in the enthalpy change of the reaction. Making these electron transfer processes clear can help students link the concept of redox to the observable heat and energy changes during chemical reactions.

By tying the enthalpy change calculations to redox concepts, educators can foster a deeper understanding of the underlying principles behind these reactions, thereby enabling students to predict and explain the outcomes of a wide range of chemical processes.