Problem 106
Question
Given the following data $$\begin{array}{ll}\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g) & \Delta H^{\circ}=-23 \mathrm{kJ} \\ 3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g) & \Delta H^{\circ}=-39 \mathrm{kJ} \\ \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g) & \Delta H^{\circ}=18 \mathrm{kJ} \end{array}$$ calculate \(\Delta H^{\circ}\) for the reaction $$\mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)$$
Step-by-Step Solution
Verified Answer
The calculated enthalpy change for the desired reaction is \(\Delta H^{\circ}_{\mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)} = 9.17 \,\mathrm{kJ}\).
1Step 1: Analyzing the target reaction
Compare the given reactions with the desired reaction. We can see that all the substances in the target reaction are present in the given reactions. We will use the given reactions to extract the necessary information to find the enthalpy change for the target reaction.
2Step 2: Manipulate the given reactions
First, we have to manipulate the given reactions to match the target reaction. We can find the desired equation by adding a combination of these reactions. We notice that to cancel out the \(\mathrm{Fe}_{2}\mathrm{O}_{3}\), we need to reverse the direction of the first two reactions.
$$-\Delta H^{\circ} = -[\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)]$$
$$-\Delta H^{\circ} = -[3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)]$$
Before we add up these equations, we need to ensure the substances we don't need to cancel each other out. Thus, we can rewrite the equations as:
$$\frac{1}{3}\times(-\Delta H^{\circ} = -[\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)])$$
$$\frac{1}{2}\times(-\Delta H^{\circ} = -[3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)])$$
$$\Delta H^{\circ} = [\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g)]$$
Now we can add up these equations getting:
$$\mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)$$
3Step 3: Calculate the enthalpy change for the target reaction
Now that we have obtained the linear combination of the given reactions to get the desired reaction, we can find the enthalpy change for this reaction by summing the enthalpy change of the given reactions after manipulating them. We have:
$$\Delta H^{\circ}_{target} = \frac{-1}{3}\times(-23 \,\mathrm{kJ}) + \frac{-1}{2}\times(-39\, \mathrm{kJ}) + 18\, \mathrm{kJ}$$
$$\Delta H^{\circ}_{target} = \frac{23}{3} \,\mathrm{kJ} + \frac{39}{2}\, \mathrm{kJ} - 18\, \mathrm{kJ}$$
Calculating the enthalpy change for the target reaction:
$$\Delta H^{\circ}_{target} = 7.67 \,\mathrm{kJ} + 19.5\, \mathrm{kJ} - 18\, \mathrm{kJ} =\boldsymbol{9.17\, \mathrm{kJ}}$$
The calculated enthalpy change for the desired reaction is \(\Delta H^{\circ}_{\mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)} = 9.17 \,\mathrm{kJ}\).
Key Concepts
Enthalpy ChangeChemical ReactionsThermochemistry
Enthalpy Change
When discussing chemical reactions, one of the key concepts you'll encounter is the enthalpy change, denoted as \( \Delta H \). Enthalpy change represents the heat absorbed or released during a chemical reaction at constant pressure.
It's a crucial component because it helps us understand how energy is transferred during a reaction.Enthalpy changes can be:
It's a crucial component because it helps us understand how energy is transferred during a reaction.Enthalpy changes can be:
- Positive: indicating an endothermic reaction where heat is absorbed.
- Negative: indicating an exothermic reaction where heat is released.
Chemical Reactions
Chemical reactions are processes where substances, known as reactants, transform into different substances, called products. The transformation involves the breaking and forming of bonds between atoms.
Writing a balanced chemical equation is essential to understand the changes that occur in a reaction and to calculate the enthalpy change accurately. In the context of this exercise, we're dealing with reactions involving iron oxides and carbon monoxide. These reactions illustrate single replacement and combustion reactions:
Writing a balanced chemical equation is essential to understand the changes that occur in a reaction and to calculate the enthalpy change accurately. In the context of this exercise, we're dealing with reactions involving iron oxides and carbon monoxide. These reactions illustrate single replacement and combustion reactions:
- Single Replacement: where one element replaces another in a compound.
- Combustion: a substance combines with oxygen, releasing energy.
Thermochemistry
Thermochemistry is the branch of chemistry concerned with the heat energy involved in chemical and physical processes. It helps us quantify the energy changes taking place during chemical reactions.
This is important for practical applications like designing energy-efficient systems.A few key points about thermochemistry include:
This is important for practical applications like designing energy-efficient systems.A few key points about thermochemistry include:
- It relies heavily on the first law of thermodynamics, which states that energy cannot be created or destroyed, only transformed.
- Thermochemical equations express the energy change of reactions, often including the enthalpy change \( \Delta H \).
- Understanding the energy profiles of reactions allows chemists to predict reaction behavior and propose mechanisms.
Other exercises in this chapter
Problem 103
The bomb calorimeter in Exercise 102 is filled with 987 g water. The initial temperature of the calorimeter contents is \(23.32^{\circ} \mathrm{C} . \mathrm{A}\
View solution Problem 105
Consider the following equations: $$\begin{aligned}3 \mathrm{A}+6 \mathrm{B} \longrightarrow 3 \mathrm{D} & & \Delta H=-403 \mathrm{kJ} / \mathrm{mol} \\\\\math
View solution Problem 107
At \(298 \mathrm{K},\) the standard enthalpies of formation for \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) and \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) are \(227 \mathrm{
View solution Problem 109
A sample of nickel is heated to \(99.8^{\circ} \mathrm{C}\) and placed in a coffeecup calorimeter containing \(150.0 \mathrm{g}\) water at \(23.5^{\circ} \mathr
View solution