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.

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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