A mixture of Fe_2O_3 and Al, present in the molar ratio 1/2, is placed in an adiabatic container at 298 K, and the Thermit reaction
2 Al+Fe_2O_3\rightarrow 2Fe+Al_2O_3
is allowed to proceed to completion. Calculate the state and the temperature of the reaction products.
From the thermochemical data
H_{Al_2O_3,298}=-1,675,700 J/mole
and
H_{Fe_2O_3,298}=-823,400 J/mole
the heat released by the Thermit reaction at 298 K is calculated as
\Delta H_{298}=-1,675,700+823,400=-852,300 J
and this heat raises the temperature of the reaction products. Assume, first, that the sensible heat raises the temperature of the products to the melting temperature of Fe, 1809 K, in which state the reactants occur as 2 moles of liquid Fe and 1 mole of solid Al_2O_3 . The molar heat capacities and molar heats of transformation are
• c_{p,Al_2O_3(s)}=117.49+10.38\times 10^{-3}T-37.11\times 10^{5}T^{-2} in the range 298-2325 K
• c_{p,Fe(\alpha )}=37.12+6.17\times 10^{-3}T-56.92T^{-0.5} J/K in the range 298-1187 K
• c_{p,Fe(\gamma )}= 24.48+8.45\times 10^{-3}T in the range 1187-1664 K
• c_{p,Fe(\delta )}= 37.12+6.17\times 10^{-3}T-56.92T^{-0.5} J/K in the range 1667-1809 K
• For Fe_{(\alpha )}\rightarrow Fe_{(\gamma )}, \Delta H_{trans}= 670 J at 1187 K
• For Fe_{(\gamma )}\rightarrow Fe_{(\delta )}, \Delta H_{trans}= 840 J at 1664 K
• For Fe_{(\delta )}\rightarrow Fe_{(l )}, \Delta H_{m}= 13,770 J at 1809 K
The heat required to raise the temperature of 1 mole of Al_2O_3 from 298 to 1809 K is
\Delta H_1 =\left[117.48\times (1809-298)\right] +\left[\frac{10.38}{2}\times 10^{-3}(1809^2-298^2) \right]+37.11\times 10^{5}\left\lgroup\frac{1}{1809}-\frac{1}{298} \right\rgroup
=183,649 J
and the heat required to raise the temperature of 2 moles of Fe from 298 to 1809 K and melt the 2 moles at 1809 K is
\Delta H_2 =\left[2\times 37.12\times (1187-298)\right]+\left[\frac{2\times 6.17}{2}\times 10^{-3}(1187^2-298^2) \right]+\left[\frac{2\times 56.92}{0.5}(1187^{0.5}-298^{0.5} ) \right] +(2\times 670)
+\left[2\times 24.28\times (1664-1187)\right]+\left[\frac{2\times 8.45}{2}\times 10^{-3}(1664^2-1187^2) \right]
+(2×840)+[2×37.12×(1809-1664)]
+\left[\frac{2\times 6.17}{2} \times 10^{-3}(1809^2-1664^2)\right]+\left[\frac{2\times 56.92}{0.5} (1809^{0.5}-1664^{0.5} )\right]
+(2×13,770)
=(78,058+1340+34,654+1680+14,268+27,540)
=157,541 J
The total heat required is thus
\Delta H_1+\Delta H_2=183,649+157,541=341,190 J
The remaining available sensible heat is 852,300 – 341,190 = 511,110 J.
Consider that the remaining sensible heat raises the temperature of the system to the melting temperature of Al_2O_3 , 2325 K, and melts the mole of Al_2O_3 . The heat required to increase the temperature of the mole of Al_2O_3 is
\Delta H_3=\left[117.49\times (2325-1809)\right]+\left[\frac{10.38}{2}\times 10^{-3}(2325^2-1809^2) \right]+37.11\times 10^{5}\left[\frac{1}{2325}-\frac{1}{1809} \right]
=71,240 J
and, with c_{p,Fe(l)}=41.84 J/K, the heat required to increase the temperature of the 2 moles of liquid Fe is
\Delta H_4=2\times41.84\times (2325-1809)=43,178 J
The molar latent heat of melting of Al_2O_3 at its melting temperature of 2325 K is 107,000 J, and thus, the sensible heat consumed is
71,240+43,178+107,000=221,418 J
which still leaves 511,110 – 221,418 = 289,692 J of sensible heat. Consider that this is sufficient to raise the temperature of the system to the boiling point of Fe,3343 K. The constant-pressure molar heat capacity of liquid Al_2O_3 is 184.1 J/K, and thus, the heat required to increase the temperature of 1 mole of liquid Al_2O_3 and 2 moles of liquid Fe from 2325 to 3343 K is
(2×41.84+184.1)×(3343-2325)=272,600 J
which leaves 289,692 – 272,600 = 17,092 J. The molar heat of boiling of Fe at its boiling temperature of 3343 K is 340,159 J, and thus, the remaining 17,092 J of sensible heat is used to convert
\frac{17,092}{340,159}=0.05moles of liquid iron to iron vapor. The final state of the system is thus 1 mole of liquid Al_2O_3 , 1.95 moles of liquid Fe, and 0.05 moles of iron vapor at 3343 K.
Suppose, now, that it is required that the increase in the temperature of the products of the Thermit reaction be limited to 1809 K to produce liquid Fe at its melting temperature. This could be achieved by including Fe in the reactants in an amount sufficient to absorb the excess sensible heat. The sensible heat remaining after the temperature of the mole of Al_2O_3 and the 2 moles of Fe has been increased to 1809 K has been calculated as 511,110 J, and the heat required to raise the temperature of 2 moles of Fe from 298 to 1809 K and melt the Fe has been calculated as \Delta H_2=157,541 J. The number of moles of Fe which must be added to the reacting mole of Fe_2O_3 and 2 moles of Al_2O_3 is thus
\frac{511,110}{0.5\times 157,541} =6.49
The required final state is thus achieved by starting with Fe, Al, and Fe_2O_3 at 298 K,
occurring in the ratio 6.49/2/1. The Thermit reaction is used to weld steel in locations which are not amenable to conventional welding equipment.