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Chapter 7: Problem 20

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.

Short Answer

Expert verified
The quantity of heat liberated from the reaction is -852000 Joules which is indeed more than enough (383846.8 Joules needed) to raise the temperature of the products to the melting point of iron.

Step by step solution

01

Calculate total reaction's heat

First part will involve the calculation of heat produced by the complete reaction. We multiply the given \(\Delta H^\circ\) with the number of moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacted. Considering reaction is exothermic, \(\Delta H^\circ\) is negative. Therefore, \(q_{reaction} = -852 \, \mathrm{kJ/mol} \times 1.00 \, \mathrm{mol} = -852 \, \mathrm{kJ}\). This is equal to -852,000 Joules.
02

Calculate total mass of the reactant

We calculate the total mass of the reactant product mixture. For this, we multiply number of moles with molar mass for each, then add them together. The molar mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is 159.69 g/mol and that of \(\mathrm{Al}\) is 26.98 g/mol. Thus, total mass ='159.69 g/mol \times 1.00 mol + 2 \times 26.98 g/mol \times 2.00 mol = 319.38 g'.
03

Calculate required heat for temperature Change

Next, we calculate the heat (\(q\)) required to raise the temperature of the resultant product to the melting point of iron using formula \(q = mc\Delta T\). Given: \(m = 319.38 g\), \(c = 0.8 \,Jg^{-1}K^{-1}\) and \(\Delta T = 1530 - 25 = 1505 K\). Substituting these values, 'Required heat = 319.38 g \times 0.8 g^{-1} K^{-1} \times 1505 K = 383645.8 Joules'.
04

Compare the heat of reaction and required heat

In the final step, we compare the total heat generated from the reaction (-852000 Joules) with the calculated heat necessary (383846.8 Joules). Since the magnitude of heat derived from the reaction is significantly higher than the required, it's sufficient to raise the temperature of the products to the melting point of iron

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Key Concepts

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

Exothermic Reactions
In exothermic reactions, energy is released in the form of heat as opposed to being absorbed. This release of energy is what makes these reactions so exciting and useful, such as in the case of the thermite reaction used for welding. When the thermite mixture combines iron oxide and aluminum, a lot of heat is liberated, which can be harnessed for practical applications like fusing metals. The enthalpy change (\( \Delta H \)) is negative in exothermic reactions because the system loses heat to its surroundings.
Calorimetry
Calorimetry is the science that measures the heat of chemical reactions or physical changes. It allows us to calculate the amount of energy released or absorbed by a substance. In the context of the thermite reaction, calorimetry can help determine how much heat is generated, which is vital for figuring out whether the heat is sufficient to raise the products' temperature to desired levels. Using calorimetry principles, the total heat from a complete reaction has been calculated to be -852,000 Joules, providing insight into the energy changes occurring during this exothermic process.
Specific Heat Capacity
Specific heat capacity is a property that indicates how much heat energy is required to change the temperature of a given amount of a substance by 1 degree Celsius. In the thermite reaction, the specific heat capacity of the products is given as 0.8 J/g掳C. This value helps in calculating the necessary heat needed to increase the temperature of the products. A lower specific heat capacity means that less heat is needed to change the temperature, making it crucial in understanding how the reaction's heat will affect the temperature change of the entire mixture.
Heat Transfer
Heat transfer refers to the movement of thermal energy from one object or substance to another. It is a fundamental concept in thermodynamics and is particularly important in understanding chemical reactions. In the context of our thermite reaction, the liberated heat is transferred from the reaction site to the surrounding products. This heat needs to be efficiently managed to ensure it accomplishes the task of raising the temperature sufficiently. Calculations show that the exothermic reaction produces enough heat to exceed the required heat uptake needed to reach the melting point of iron.
Chemical Reactions
Chemical reactions involve the breaking and forming of bonds to create new substances. They can be categorized broadly as exothermic or endothermic. In the provided exercise, the thermite reaction is a prime example of an exothermic chemical reaction, where iron oxide (\( \mathrm{Fe}_2\mathrm{O}_3 \)) reacts with aluminum (\( \mathrm{Al} \)) to form aluminum oxide (\( \mathrm{Al}_2\mathrm{O}_3 \)) and elemental iron (\( \mathrm{Fe} \)). The balanced chemical equation shows the stoichiometry of the reaction, emphasizing how the reactants combine in a specific ratio to form the products.

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Most popular questions from this chapter

A coffee-cup calorimeter contains \(100.0 \mathrm{mL}\) of \(0.300 \mathrm{M}\) HCl at \(20.3^{\circ} \mathrm{C}\). When \(1.82 \mathrm{g} \mathrm{Zn}(\mathrm{s})\) is added, the temperature rises to \(30.5^{\circ} \mathrm{C}\). What is the heat of reaction per mol Zn? Make the same assumptions as in Example \(7-4,\) and also assume that there is no heat lost with the \(\mathrm{H}_{2}(\mathrm{g})\) that escapes. $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$$

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$$C_{P, \mathrm{m}}=28.58+3.77 \times 10^{-3} T-\frac{0.5 \times 10^{5}}{T^{2}} \mathrm{JK}^{-1} \mathrm{mol}^{-1}$$

The metabolism of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) yields \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) as products. Heat released in the process is converted to useful work with about \(70 \%\) efficiency. Calculate the mass of glucose metabolized by a \(58.0 \mathrm{kg}\) person in climbing a mountain with an elevation gain of \(1450 \mathrm{m}\). Assume that the work performed in the climb is about four times that required to simply lift \(58.0 \mathrm{kg}\) by \(1450 \mathrm{m} \cdot\left(\Delta H_{\mathrm{f}}^{2} \text { of } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}) \text { is }-1273.3 \mathrm{kJ} / \mathrm{mol} .\right)\)

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?

What is the change in internal energy of a system if the system (a) absorbs \(58 \mathrm{J}\) of heat and does \(58 \mathrm{J}\) of work; (b) absorbs 125 J of heat and does 687 J of work; (c) evolves 280 cal of heat and has 1.25 kJ of work done on it?
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