Question

The enthalpy change for the oxidation of styrene, \(\mathbf{C}_{8} \mathrm{H}_{8}\) is measured by calorimetry. $$\begin{aligned}\mathrm{C}_{8} \mathrm{H}_{8}(\ell)+10 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 8 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\ell) & \\\& \Delta H_{\mathrm{rsn}}^{\circ}=-4395.0 \mathrm{kJ}\end{aligned}$$ Use this value, along with the standard heats of formation of \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\ell),\) to calculate the enthalpy of formation of styrene, in \(\mathrm{kJ} / \mathrm{mol}\).

Short answer

The enthalpy of formation for styrene is 103.8 kJ/mol.

Step-by-step solution

Write the Equation for Enthalpy Change

The given reaction is \( \mathrm{C}_{8} \mathrm{H}_{8}(\ell) + 10 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 8 \mathrm{CO}_{2}(\mathrm{g}) + 4 \mathrm{H}_{2} \mathrm{O}(\ell) \) with \( \Delta H_{\mathrm{rsn}}^{\circ} = -4395.0 \mathrm{kJ} \). The formula to find the enthalpy of formation \( \Delta H_{f}^{\circ} \) of a reactant is: \( \Delta H_{\mathrm{rsn}}^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \sum \Delta H_{f}^{\circ}(\text{reactants}) \).

List Given Values and Constants

The standard heats of formation for carbon dioxide \( \mathrm{CO}_{2}(\mathrm{g}) \) and water \( \mathrm{H}_{2} \mathrm{O}(\ell) \) are \(-393.5\, \mathrm{kJ/mol} \) and \(-285.8\, \mathrm{kJ/mol} \) respectively. We need to determine \( \Delta H_{f}^{\circ}(\mathrm{C}_{8}\mathrm{H}_{8}) \).

Calculate Enthalpy of Products

Calculate \( \Delta H_{\text{f}}^{\circ} \) for the products using their standard enthalpies of formation: \( 8(-393.5\, \mathrm{kJ/mol}) + 4(-285.8\, \mathrm{kJ/mol}) \). This equals \(-3148.0\, \mathrm{kJ} + (-1143.2\, \mathrm{kJ}) = -4291.2\, \mathrm{kJ} \).

Rearrange the Formula and Solve for Styrene

Using the rearranged formula \( \Delta H_{\mathrm{rsn}}^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \Delta H_{f}^{\circ}(\mathrm{C}_{8} \mathrm{H}_{8}) - 10 \Delta H_{f}^{\circ}(\mathrm{O}_{2}(\mathrm{g})) \), and knowing \( \Delta H_{f}^{\circ}(\mathrm{O}_{2}(\mathrm{g})) = 0 \), solve for \( \Delta H_{f}^{\circ}(\mathrm{C}_{8} \mathrm{H}_{8})\). This gives \( -4395.0 = -4291.2 - \Delta H_{f}^{\circ}(\mathrm{C}_{8} \mathrm{H}_{8}) \).

Simplify and Calculate Result

Simplify the equation: \( \Delta H_{f}^{\circ}(\mathrm{C}_{8} \mathrm{H}_{8}) = -4291.2 + 4395.0 \). The solution is \( \Delta H_{f}^{\circ}(\mathrm{C}_{8} \mathrm{H}_{8}) = 103.8\, \mathrm{kJ/mol} \).

Key concepts

Calorimetry

Calorimetry is an experimental technique used to measure the amount of heat released or absorbed during a chemical reaction. It plays a crucial role in determining the enthalpy change, which is the change in heat content, of a reaction. In this exercise, calorimetry is used to measure the enthalpy change for the oxidation of styrene. By understanding this concept, you can evaluate the energy changes that occur during a chemical process and gain insights into the reactive properties of substances.
Whether in a laboratory setting with sophisticated instruments or with simpler equipment for educational purposes, calorimetry encompasses the principles of heat transfer within a system. It requires careful isolation of the reaction environment to ensure accurate measurements, minimizing the influence of external temperatures.
Knowing how to apply calorimetry properly helps chemists calculate the heat involved in reactions, contributing to fields such as thermochemistry and energy science. This makes it an indispensable tool in both theoretical studies and practical applications.

Oxidation Reaction

An oxidation reaction is a chemical process wherein a substance loses electrons. In the context of this exercise, the oxidation reaction involves styrene, a hydrocarbon, reacting with oxygen to produce carbon dioxide and water.
Oxidation reactions are a type of redox (reduction-oxidation) reaction, where one substance undergoes oxidation while another is reduced, essentially resulting in a transfer of electrons. Here, styrene donates electrons to oxygen, which is reduced as it gains these electrons, forming by-products like carbon dioxide and water.
The oxidation of hydrocarbons such as styrene often releases energy, as is evident by the negative enthalpy change in the exercise, signifying an exothermic process. Understanding how oxidation reactions work is fundamental in studying combustion, metabolism, and even rusting, offering insights into how energy operates in chemical transformations.

Standard Heats of Formation

The standard heats of formation, denoted as \( Delta H_{f}^{\circ}\) are defined as the change in enthalpy when one mole of a substance is formed from its elements in their standard states. This thermodynamic quantity is pivotal in understanding chemical reactions.
In this exercise, the reaction makes use of standard heats of formation values for carbon dioxide and water, which are given as \(-393.5 \, \mathrm{kJ/mol}\) and \(-285.8 \, \mathrm{kJ/mol}\) respectively. These values apply under specific conditions: 1 atm pressure and a defined temperature, often 298 K (25 °C).

• By knowing the standard heats of formation, chemists can predict the enthalpy change of a reaction using Hess's Law, by determining the total enthalpy of products and subtracting the enthalpy of reactants.
• This information is fundamental for constructing enthalpy diagrams and conducting energy balance calculations in chemical engineering.

Standard heats of formation provide a crucial baseline for comparing the stability and reactivity of compounds, facilitating efficient reaction design and development.

Enthalpy Change

Enthalpy change, symbolized as \( Delta H_{rsn}^{\circ}\), represents the difference in the enthalpy of products and reactants during a chemical reaction. It's a key parameter in understanding energy flow in chemical processes.
In the exercise, the enthalpy change for the oxidation of styrene is given as \(-4395.0 \, \mathrm{kJ}\), indicating the amount of heat released.
This negative value signals an exothermic reaction, where energy is liberated to its surroundings. Calculating the enthalpy change is crucial for determining whether a process is energetically favorable and feasible.
It informs decisions in industries that rely on chemical processes, like pharmaceuticals and petrochemicals, aide in optimizing energy efficiency, and reduce costs. Understanding this concept helps predict reaction behavior, allowing chemists to control and manipulate reaction pathways more effectively.

• Enthalpy change calculations take into account not only reaction equilibria but also how temperature and pressure conditions influence the energy landscape of a reaction.

Therefore, grasping the concept of enthalpy change equips students and professionals with the ability to anticipate the energetic consequences of chemical transformations.