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Chapter 7: Problem 75

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 $$\mathrm{CH}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g})$$

Short Answer

Expert verified
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 by step solution

01

Identify the Target Reaction

The reaction for which we need to determine the change in heat reaction (\( \Delta H^{\circ} \)) is: \( \mathrm{CH}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \)
02

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.
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A 125 \(g\) stainless steel ball bearing \((\mathrm{spht}=\) \(0.50 \mathrm{Jg}^{-1}\) \(\left.^{\circ} \mathrm{C}^{-1}\right)\) at \(525^{\circ} \mathrm{C}\) is dropped into \(75.0 \mathrm{mL}\) of water at \(28.5^{\circ} \mathrm{C}\) in an open Styrofoam cup. As a result, the water is brought to a boil when the temperature reaches \(100.0^{\circ} \mathrm{C} .\) What mass of water vaporizes while the boiling continues? \(\left(\Delta H_{\mathrm{vap}}^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\right)\).

An alternative approach to bomb calorimetry is to establish the heat capacity of the calorimeter, exclusive of the water it contains. The heat absorbed by the water and by the rest of the calorimeter must be calculated separately and then added together. A bomb calorimeter assembly containing \(983.5 \mathrm{g}\) water is calibrated by the combustion of \(1.354 \mathrm{g}\) anthracene. The temperature of the calorimeter rises from 24.87 to \(35.63^{\circ} \mathrm{C} .\) When \(1.053 \mathrm{g}\) citric acid is burned in the same assembly, but with 968.6 g water, the temperature increases from 25.01 to \(27.19^{\circ} \mathrm{C}\). The heat of combustion of anthracene, \(\mathrm{C}_{14} \mathrm{H}_{10}(\mathrm{s}),\) is \(-7067 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{C}_{14} \mathrm{H}_{10} \cdot\) What is the heat of combustion of citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7},\) expressed in \(\mathrm{kJ} / \mathrm{mol} ?\)

Thermite mixtures are used for certain types of welding, and the thermite reaction is highly exothermic. $$\begin{array}{r} \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Fe}(\mathrm{s}) \\ \Delta H^{\circ}=-852 \mathrm{kJ} \end{array}$$ \(1.00 \mathrm{mol}\) of granular \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(2.00 \mathrm{mol}\) of granular Al are mixed at room temperature \(\left(25^{\circ} \mathrm{C}\right),\) and a reaction is initiated. The liberated heat is retained within the products, whose combined specific heat over a broad temperature range is about \(0.8 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} .\) (The melting point of iron is \(1530^{\circ} \mathrm{C} .\) ) Show that the quantity of heat liberated is more than sufficient to raise the temperature of the products to the melting point of iron.

The internal energy of a fixed quantity of an ideal gas depends only on its temperature. A sample of an ideal gas is allowed to expand at a constant temperature (isothermal expansion). (a) Does the gas do work? (b) Does the gas exchange heat with its surroundings? (c) What happens to the temperature of the gas? (d) What is \(\Delta U\) for the gas?

In the Are You Wondering \(7-1\) box, the temperature variation of enthalpy is discussed, and the equation \(q_{P}=\) heat capacity \(\times\) temperature change \(=C_{P} \times \Delta T\) was introduced to show how enthalpy changes with temperature for a constant-pressure process. Strictly speaking, the heat capacity of a substance at constant pressure is the slope of the line representing the variation of enthalpy (H) with temperature, that is $$C_{P}=\frac{d H}{d T} \quad(\text { at constant pressure })$$ where \(C_{P}\) is the heat capacity of the substance in question. Heat capacity is an extensive quantity and heat capacities are usually quoted as molar heat capacities \(C_{P, \mathrm{m}},\) the heat capacity of one mole of substance; an intensive property. The heat capacity at constant pressure is used to estimate the change in enthalpy due to a change in temperature. For infinitesimal changes in temperature, $$d H=C_{p} d T \quad(\text { at constant pressure })$$ To evaluate the change in enthalpy for a particular temperature change, from \(T_{1}\) to \(T_{2}\), we write $$\int_{H\left(T_{1}\right)}^{H\left(T_{2}\right)} d H=H\left(T_{2}\right)-H\left(T_{1}\right)=\int_{T_{1}}^{T_{2}} C_{P} d T$$ If we assume that \(C_{P}\) is independent of temperature, then we recover equation (7.5) $$\Delta H=C_{P} \times \Delta T$$ On the other hand, we often find that the heat capacity is a function of temperature; a convenient empirical expression is $$C_{P, \mathrm{m}}=a+b T+\frac{c}{T^{2}}$$ What is the change in molar enthalpy of \(\mathrm{N}_{2}\) when it is heated from \(25.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C} ?\) The molar heat capacity of nitrogen is given by$$C_{P, \mathrm{m}}=28.58+3.77 \times 10^{-3} T-\frac{0.5 \times 10^{5}}{T^{2}} \mathrm{JK}^{-1} \mathrm{mol}^{-1}$$
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