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Calculate \(\Delta H_{\mathrm{rxn}}\) for the reaction. $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)$$ Use the following reactions and given \(\Delta H^{\prime}s\): \begin{equation} \begin{array}{ll}{2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)} & {\Delta H=-824.2 \mathrm{kJ}} \\ {\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)} & {\Delta H=-282.7 \mathrm{kJ}}\end{array} \end{equation}

Short Answer

Expert verified
\textbackslash(Delta H_\textbackslash{rxn\textbackslash}) = (-24.9 \mathrm{kJ})

Step by step solution

01

Identify the Target Reaction

Write down the target reaction that we need to find the enthalpy change for, which is \[\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2\mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)\]
02

Analyze Given Reactions

Note the two given reactions with their respective enthalpy changes:1) \[2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s), \Delta H=-824.2 \mathrm{kJ}\]2) \[\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g), \Delta H=-282.7 \mathrm{kJ}\]
03

Manipulate Given Reactions to Match Target

Reverse the first reaction and multiply the second reaction by 3 to match the reactants and products of the target reaction:1) \[\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s) + \frac{3}{2} \mathrm{O}_{2}(g), \Delta H= +824.2 \mathrm{kJ} \text{(reversed)}\]2) \[3(\mathrm{CO}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)), 3 \times \Delta H=3(-282.7) \mathrm{kJ} \text{(multiplied by 3)}\]
04

Combine the Manipulated Reactions

Add the manipulated reactions together to obtain the target reaction:\[(\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s) + \frac{3}{2} \mathrm{O}_{2}(g)) + 3(\mathrm{CO}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)) = 2\mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)\]The \(\frac{3}{2} \mathrm{O}_{2}(g)\) and \(3 \times \frac{1}{2} \mathrm{O}_{2}(g)\) will cancel out, yielding our target reaction.
05

Calculate \textbackslash(Delta H_\textbackslash{rxn\textbackslash})

Sum the enthalpy changes of the manipulated reactions to find the enthalpy change for the target reaction:\[\Delta H_{\mathrm{rxn}} = (+824.2 \mathrm{kJ}) + 3(-282.7 \mathrm{kJ})\]

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Key Concepts

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

Thermochemistry
Thermochemistry is the branch of chemistry that deals with the energy changes accompanying chemical reactions. Central to this field is the concept of enthalpy (\( H \)), which is a measure of the total energy of a thermodynamic system. It includes both the internal energy of the system (such as chemical bonds) and the product of its pressure and volume (work done on or by the system). When a reaction occurs at constant pressure, the change in enthalpy, denoted as \( \Delta H \), corresponds to the heat evolved (\( \Delta H < 0 \) for exothermic reactions) or absorbed (\( \Delta H > 0 \) for endothermic reactions) by the reaction.

Enthalpy can't be measured directly; instead, it's calculated based on changes observed in a chemical reaction. In your homework, you often look at how the system exchanges heat with the surroundings, applying concepts like heat capacity, calorimetry, and energy conservation, to understand energy flow during chemical transformations.
Enthalpy of Reaction
The enthalpy of reaction, or reaction enthalpy, represents the amount of heat absorbed or released by a chemical reaction at constant pressure. It is denoted by \( \Delta H_{\mathrm{rxn}} \) and is expressed in kilojoules per mole (kJ/mol). The value is negative for exothermic reactions, indicating that energy is released, and positive for endothermic reactions, where energy is absorbed.

To calculate the enthalpy change of a given reaction, we may use various methods like calorimetry, which involves the measurement of heat exchange. However, when direct measurement isn't possible, we turn to Hess's Law and established \( \Delta H \) values from other known reactions to calculate the enthalpy change indirectly. This process usually involves breaking down the overall reaction into a series of steps for which \( \Delta H \) values are known or can be easily calculated.
Hess's Law
Hess's Law asserts that the total enthalpy change of a chemical reaction is the same, no matter how the reaction occurs in a series of steps. This principle allows us to calculate the enthalpy change for a reaction if we have data for the enthalpy changes of related reactions, even if we cannot directly measure it for the target reaction. The law is a manifestation of the conservation of energy, indicating that the total energy change depends only on the initial and final states, not on the specific path between them.

To apply Hess's Law, we often need to reverse and/or multiply given reactions to match the target equation. Reversing a reaction changes the sign of its \( \Delta H \) value, while multiplying a reaction by a factor also multiplies its \( \Delta H \) by the same factor. Finally, we sum up the enthalpy changes of the manipulated reactions to determine the enthalpy change of the target reaction. This method is extremely useful for calculating enthalpy changes for reactions that would be difficult or impossible to measure directly.

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Most popular questions from this chapter

The lattice energy of CsF is \(-744\) kJ/mol whereas that of \(B a O\) is \(-3029\) kJ/mol. Explain this large difference in lattice energy.

Consider the thermochemical equation for the combustion of acetone \(\left( \mathrm { C } _ { 3 } \mathrm { H } _ { 6 } \mathrm { O } \right) ,\) the main ingredicnt in nail polish remover. \(\mathrm { C } _ { 3 } \mathrm { H } _ { 6 } \mathrm { O } ( l ) + 4 \mathrm { O } _ { 2 } ( g ) \longrightarrow 3 \mathrm { CO } _ { 2 } ( g ) + 3 \mathrm { H } _ { 2 } \mathrm { O } ( g )\) $$\Delta H _ { \mathrm { rmn } } ^ { \circ } = - 1790 \mathrm { kJ }$$ If a bottle of nail polish remover contains 177\(\mathrm { mL }\) of acetone, how much heat is released by its complete combustion? The density of acetone is 0.788\(\mathrm { g } / \mathrm { mL }\) .

A system releases 622\(\mathrm { kJ }\) of heat and does 105\(\mathrm { kJ }\) of work on the surroundings. What is the change in internal energy of the system?

Identify each energy exchange as primarily heat or work and determine the sign of \(\Delta E\) (positive or negative) for the system. a. A rolling billiard ball collides with another billiard ball. The first billiard ball (defined as the system) stops rolling after the collision. b. A book is dropped to the floor.(The book is the system.) c. A father pushes his daughter on a swing. (The daughter and the swing are the system.)

How much work (in \(\mathrm { J }\) ) is required to expand the volume of a pump from 0.0\(\mathrm { L }\) to 2.5\(\mathrm { L }\) against an external pressure of 1.1\(\mathrm { atm }\) ?

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