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The following equation shows the conversion of aluminum oxide (from the ore bauxite) to aluminum: $$ 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=+3350 \mathrm{~kJ} / \mathrm{mol} $$ (a) Is the reaction exothermic or endothermic? (b) How many kilojoules are required to produce \(1.00 \mathrm{~mol}\) of aluminum? (c) How many kilojoules are required to produce \(10.0 \mathrm{~g}\) of aluminum?

Short Answer

Expert verified
(a) Endothermic. (b) 837.5 kJ. (c) 310.39 kJ.

Step by step solution

01

Identify Reaction Type

The given reaction has an enthalpy change \( \Delta H = +3350 \, \text{kJ/mol} \). Since the enthalpy change is positive, it indicates that the reaction requires energy to proceed. This is characteristic of an endothermic reaction.
02

Determine Energy for 1 mol of Al

The reaction shows that \( 3350 \, \text{kJ} \) is required to produce \( 4 \text{ mol of Al} \). To find how much energy is needed for \( 1 \text{ mol} \) of Al, divide the total energy by 4: \[ \text{Energy per mol of Al} = \frac{3350 \, \text{kJ}}{4} = 837.5 \, \text{kJ} \]
03

Calculate Moles of Aluminum in 10 g

To find the kilojoules required for \( 10.0 \, \text{g} \) of aluminum, first determine how many moles are in \( 10.0 \, \text{g} \). The molar mass of Al is \( 26.98 \, \text{g/mol} \). Calculate using: \[ \text{moles of Al} = \frac{10.0 \, \text{g}}{26.98 \, \text{g/mol}} \approx 0.3707 \, \text{mol} \]
04

Energy for 10 g of Aluminum

Use the moles found in Step 3 to calculate the energy needed: Multiply the moles by the energy per mole from Step 2: \[ \text{Energy required} = 0.3707 \, \text{mol} \times 837.5 \, \text{kJ/mol} = 310.39 \, \text{kJ} \]

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Key Concepts

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

Enthalpy Change
In thermochemistry, enthalpy change is a crucial concept that helps us understand how heat energy is absorbed or released during a chemical reaction. Enthalpy is represented by the symbol \( \Delta H \), and it indicates the change in heat content.

When the enthalpy change \( \Delta H \) is positive, it signifies that the system has absorbed energy, indicating an endothermic process. Conversely, a negative \( \Delta H \) means the system has released energy, characterizing an exothermic process. In the case of aluminum production, \( \Delta H = +3350 \, \text{kJ/mol} \), suggesting that the reaction requires heat to convert aluminum oxide into aluminum. This insight is crucial to determining whether a reaction is endothermic or exothermic.
Endothermic Reactions
Endothermic reactions are processes that absorb energy from their surroundings, often in the form of heat. These reactions require an input of energy to proceed, as the stored energy in the reactants is lower than that of the products.

In the aluminum production reaction given, the positive \( \Delta H \) indicates it's endothermic. This means the process of converting aluminum oxide into aluminum metal requires a significant amount of energy input, resulting in higher energy products.
  • Energy is absorbed, raising the temperature needed for the reaction.
  • Endothermic reactions are common in processes where breaking strong bonds in the reactants is necessary.
Understanding endothermic reactions provides insight into why certain processes, like aluminum extraction, necessitate specific energy investments.
Aluminum Production
Aluminum production from aluminum oxide involves chemical reactions that are notably energy-intensive due to their endothermic nature. The aluminum oxide is typically found in bauxite ore and undergoes electrolysis to separate pure aluminum metal.

The reaction provided highlights that from 2 moles of \( \text{Al}_2\text{O}_3 \), 4 moles of aluminum are produced, with an energy requirement of 3350 kJ. Modern aluminum production follows the Hall-Héroult process, an industrial method that applies a significant amount of electric energy to facilitate the reaction.

This production is vital to industries worldwide, as aluminum is a lightweight, durable metal with extensive applications, from cookware to aerospace technology.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between substances in a chemical reaction. It allows us to calculate reactants and products, ensuring balanced equations align with the law of conservation of mass.

In the context of producing aluminum, stoichiometry enables us to determine the energy required for different amounts of aluminum. For instance, to calculate energy needed for 1 mole of aluminum, we use the reaction data that showed 3350 kJ for 4 moles. Thus:
  • Energy per mole of Al = \( \frac{3350 \, \text{kJ}}{4} = 837.5 \, \text{kJ} \).
For any given mass of aluminum, say 10 g, stoichiometry involves calculating moles using its molar mass and then using this relation to find the energy required for the conversion. Understanding stoichiometry is key to solving many real-world problems in chemistry.

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Most popular questions from this chapter

The reaction between hydrogen gas and carbon to produce the gas known as ethene is: $$ \begin{array}{c} 2 \mathrm{H}_{2}(g)+2 \mathrm{C}(s) \longrightarrow \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(g) \\ \Delta G=+68.2 \mathrm{~kJ} / \mathrm{mol} \text { at } 298 \mathrm{~K} \end{array} $$ (a) Is this reaction spontaneous at \(298 \mathrm{~K} ?\) (b) Would it be reasonable to try to develop a catalyst for the reaction run at \(298 \mathrm{~K} ?\) Explain.

Classify each of the following as having potential or kinetic energy. For those identified as having potential energy, discuss how the potential energy would be realized by conversion to another form of energy. (a) gunpowder (b) a bullet in flight (c) a cell phone (lithium ion) battery (d) wind (e) a candy bar (f) spinning wind mill blades

Write the equilibrium constant expressions for the following reactions: (a) \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{CO}_{2}(g)\) (b) \(\mathrm{Mg}(s)+\mathrm{HCl}(a q) \rightleftarrows \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g)\) (c) \(\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q)\) (d) \(\mathrm{S}(s)+\mathrm{O}_{2}(g) \rightleftarrows \mathrm{SO}_{2}(g)\)

What is meant by the term "chemical equilibrium"? Must amounts of reactants and products be equal at equilibrium?

As we exercise, our bodies metabolize glucose, converting it to \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O},\) to supply the energy necessary for physical activity. The simplified reaction is: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)+2840 \mathrm{~kJ} $$ An individual weighing \(68 \mathrm{~kg}\) jogging at \(8 \mathrm{~km} / \mathrm{h}\) for 30 minutes would burn \(1138 \mathrm{~kJ}\). How many moles of glucose would need to be metabolized to generate this required energy?

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