Problem 75 - General Chemistry: Principles and Modern Applications

Question

Use Hess's law and the following data $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ \Delta H^{\circ}=-802 \mathrm{kJ} \end{aligned}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&+247 \mathrm{kJ} \end{aligned}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&+247 \mathrm{kJ} \end{aligned}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&+206 \mathrm{kJ} \end{aligned}$$ to determine \(\Delta H^{\circ}\) for the following reaction, an important source of hydrogen gas $$\mathrm{CH}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g})$$

Step-by-Step Solution

Answer

The value of \( \Delta H^{\circ} \) for the reaction \( \mathrm{CH}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \) is \( +607 \) kJ.

Step 1: Identify the Target Reaction

The reaction for which we need to determine the change in heat reaction (\( \Delta H^{\circ} \)) is: \( \mathrm{CH}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \)

Step 2: Manipulate the given reactions

To obtain the desired reaction, the given reactions need to be manipulated in such a way that, when added, they result in the target reaction. Starting with the reaction \( \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \), with \( \Delta H^{\circ} = -802 \) kJ, we should reverse it and halve it to get \( \frac{1}{2} \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{CH}_{4}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \), with \( \Delta H^{\circ} = 401 \) kJ. Then, considering reaction \( \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \), with \( \Delta H^{\circ} = +206 \) kJ, we should use it as given.

Step 3: Sum the manipulated reactions

When we add these two manipulated reactions, we get the desired reaction. If we add up the enthalpy change of these two reactions, we will get the \( \Delta H^{\circ} \) for the desired reaction. After calculation, \( \Delta H^{\circ} = 401 \) kJ (for the first reaction) + \( +206 \) kJ (for the second reaction) = \( +607 \) kJ.

Key Concepts

Enthalpy Change

Understanding enthalpy change is central to grasping chemical reactions and energy transformations in chemistry. Enthalpy change, denoted as \( \Delta H \), is the heat absorbed or evolved in a reaction at constant pressure. It indicates whether a process is endothermic (\( \Delta H > 0 \), absorbing heat) or exothermic (\( \Delta H < 0 \), releasing heat).

In the exercise provided, we focused on calculating the \( \Delta H \) for a specific chemical reaction involving methane \((\mathrm{CH}_4)\). This involved using Hess's Law to determine the overall enthalpy change from individual steps. By manipulating individual reactions with known \( \Delta H \) values, we can calculate the overall enthalpy change for a reaction pathway by simply adding these values. This is because enthalpy is a state function, meaning its change depends only on the initial and final states, not on the path taken.

Thermochemical Equations

A thermochemical equation is a balanced chemical equation that includes the enthalpy change as part of the equation. This communicates not only the substances involved in the reaction but also the associated energy changes, as seen in the equations from the original exercise.

When dealing with thermochemical equations, it's essential to note the phase of each substance (e.g., \( \mathrm{g} \) for gas, \( \mathrm{l} \) for liquid). It impacts the enthalpy because different phases require or release different amounts of energy during the transformation. For example, a reaction might appear similar but can have different enthalpy values if water changes phases from liquid to gas.

• If a reaction is reversed, the sign of \( \Delta H \) is also reversed.
• If a reaction is multiplied by a coefficient, \( \Delta H \) must be multiplied by the same coefficient.

Correct manipulation and adding of thermochemical equations allow accurate modeling of complex chemical reactions and their energetics.

Chemical Reactions

Chemical reactions involve the breaking and forming of bonds between atoms to create new substances. These rearrangements are accompanied by energy changes expressed in enthalpy changes.

In the context of Hess's Law, chemical reactions can be broken down into individual steps whose enthalpy changes can be summed to find the overall reaction's enthalpy change. This is highly practical as it allows us to compute \( \Delta H \) for reactions that are difficult to measure directly.

Consider the combustion of methane \((\mathrm{CH}_4)\) in our exercise, which breaks the bonds of \( \mathrm{CH}_4 \) and \( \mathrm{O}_2 \) molecules while forming \( \mathrm{CO}_2 \) and \( \mathrm{H}_2 \mathrm{O} \). These bond changes are reflected in the enthalpy changes, showcasing how energy evolves or is consumed during the process.

• Every chemical reaction involves breaking and forming bonds.
• Energy is required to break bonds, thus absorbed from surroundings.
• Energy is released when forming bonds, thus liberated to surroundings.