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Given the following data \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-23 \mathrm{~kJ}\) \(3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-39 \mathrm{~kJ}\) \(\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=+18 \mathrm{~kJ}\) calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

Short Answer

Expert verified
The enthalpy change, ΔH, for the given reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ is approximately -25.67 kJ.

Step by step solution

01

Identify the common species in the reactions

Observe that CO is a reactant in all the given reactions and CO₂ is a product in all the given reactions.
02

Calculate the moles of the target reaction

The goal is to manipulate the given reactions to match the target reaction. So the target reaction is: $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$
03

Manipulate the given reactions

In order for the target reaction to be obtained, multiply the first given reaction by \(\frac{1}{3}\) and reverse the last given reaction.
04

Writing the modified reactions and their enthalpy changes

The modified reactions are: $$ \frac{1}{3}(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)) \quad \Delta H^{\circ}=-\frac{1}{3} \times 23 \, \mathrm{kJ}=-\frac{23}{3} \, \mathrm{kJ} $$ $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-18 \, \mathrm{kJ} $$ Notice that when we reverse the third given reaction, its enthalpy change becomes negative. Now add the first modified reaction and the second modified reaction together to get the target reaction.
05

Combine the modified reactions and find the ΔH for the target reaction

Summing both modified reactions, we get the target reaction: $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ Now, according to Hess's law, the enthalpy change for the target reaction is equal to the sum of the enthalpy changes for the modified reactions: $$ \Delta H^{\circ}_{\text{target}}=\Delta H^{\circ}_{1}+\Delta H^{\circ}_{2}= -\frac{23}{3} \, \mathrm{kJ} - 18 \, \mathrm{kJ} $$ Calculate the final enthalpy change for the target reaction: $$ \Delta H^{\circ}_{\text{target}}=-\frac{23}{3} \, \mathrm{kJ} - 18 \, \mathrm{kJ} = -\frac{77}{3} \, \mathrm{kJ} $$
06

Final answer

The enthalpy change, ΔH, for the given reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ is -25.67 kJ (approximately).

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Key Concepts

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

Enthalpy Change
Enthalpy change is a key concept in understanding Hess's Law and how heat is absorbed or released during a chemical reaction. When we talk about the enthalpy change, we're referring to the difference in energy between the reactants and the products. If a reaction releases heat, it's called exothermic, and the enthalpy change is negative, indicating that energy is leaving the system.
For most reactions, enthalpy change is expressed in kilojoules per mole, showing how much energy is gained or lost per mole of reactant. Hess's Law allows us to determine the enthalpy change for a reaction by adding up the enthalpy changes of individual steps. This is because enthalpy is a state function—it depends only on the initial and final states, not on the path taken.
In the exercise, we use the given enthalpy changes of several reactions to calculate the enthalpy change of the target reaction using this principle. By manipulating the given reactions, we can combine them to express the target reaction and sum their enthalpy changes to find the overall enthalpy change.
Chemical Reactions
Chemical reactions are processes where reactants transform into products. This transformation can involve breaking and forming chemical bonds, which affects the energy state of the molecules involved.
Reactions can be represented in a balanced chemical equation, indicating both the reactants and products along with their stoichiometric coefficients. Stoichiometric coefficients are the numbers in front of compounds that indicate the relative number of moles involved in the reaction. The exercise demonstrates this as it involves manipulating several reactions to create a new equation.
  • Reversing a reaction changes the sign of its enthalpy change, indicating the opposite energy shift when the reaction proceeds in the reverse direction.
  • Multiplying the coefficients, including the enthalpy change, by a number ensures that the equation reflects the correct stoichiometric quantities needed to express the desired reaction.
By understanding these fundamental concepts, we can compute enthalpy changes for reactions that are not directly measurable.
Thermodynamics
Thermodynamics is the study of energy changes, particularly those involving heat and work. In the context of chemical reactions, it helps us understand how and why reactions occur, often focusing on energy in the form of enthalpy.
Hess's Law is a principle derived from the first law of thermodynamics, stating that the total enthalpy change for a chemical reaction is the same regardless of the path taken, assuming the initial and final conditions are constant. This means we can add or subtract enthalpy changes from other reactions to find out the energy change of a target reaction if its direct measurement is complex or impractical.
  • Thermodynamics does not only explain how energy is transferred but also helps predict the direction a reaction will naturally proceed by considering both enthalpy and entropy.
  • The focus on enthalpy in this exercise exemplifies its importance in comprehending energy relations and efficiency within chemical processes.
This theoretical foundation aids in the rational design of chemical processes and predicting reaction feasibility.

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Most popular questions from this chapter

It takes \(585 \mathrm{~J}\) of energy to raise the temperature of \(125.6 \mathrm{~g}\) mercury from \(20.0^{\circ} \mathrm{C}\) to \(53.5^{\circ} \mathrm{C}\). Calculate the specific heat capacity and the molar heat capacity of mercury.

When \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution at \(30.0^{\circ} \mathrm{C}\) is added to \(2.00 \mathrm{~L}\) of \(0.750 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution at \(30.0^{\circ} \mathrm{C}\) in a calorimeter, a white solid \(\left(\mathrm{BaSO}_{4}\right)\) forms. The temperature of the mixture increases to \(42.0^{\circ} \mathrm{C}\). Assuming that the specific heat capacity of the solution is \(6.37 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that the density of the final solution is \(2.00 \mathrm{~g} / \mathrm{mL}\), calculate the enthalpy change per mole of \(\mathrm{BaSO}_{4}\) formed.

On Easter Sunday, April 3, 1983 , nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: \(2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) a. Calculate \(\Delta H^{\circ}\) for this reaction. Approximately \(2.0 \times 10^{4}\) gal nitric acid was spilled. Assume that the acid was an aqueous solution containing \(70.0 \% \mathrm{HNO}_{3}\) by mass with a density of \(1.42 \mathrm{~g} / \mathrm{cm}^{3}\). What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? \(\left(\Delta H_{\mathrm{f}}^{\circ}\right.\) for \(\mathrm{NaNO}_{3}(a q)=-467 \mathrm{~kJ} / \mathrm{mol}\) ) b. According to The Denver Post for April 4, 1983 , authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of \(\Delta H^{\circ}\), what was their major concern?

Quinone is an important type of molecule that is involved in photosynthesis. The transport of electrons mediated by quinone in certain enzymes allows plants to take water, carbon dioxide, and the energy of sunlight to create glucose. A \(0.1964-\mathrm{g}\) sample of quinone \(\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}\right)\) is burned in a bomb calorimeter with a heat capacity of \(1.56 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter increases by \(3.2^{\circ} \mathrm{C}\). Calculate the energy of combustion of quinone per gram and per mole.

Given the following data $$ \begin{aligned} \mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) & \longrightarrow 4 \mathrm{PCl}_{3}(g) & & \Delta H=-1225.6 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & & \Delta H=-2967.3 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) & \longrightarrow \mathrm{PCl}_{5}(g) & & \Delta H=-84.2 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \mathrm{Cl}_{3} \mathrm{PO}(g) & & \Delta H=-285.7 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) $$

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