Question

The thermite process is used for welding railway track together. In this reaction, aluminium reduces iron(III) oxide to metallic iron:

$2\text{Al}\left(s\right)+{\text{Fe}}_2{\text{O}}_3\left(s\right)\to 2\text{Fe}\left(s\right)+{\text{Al}}_2{\text{O}}_3\left(s\right)$

Igniting a small charge of barium peroxide mixed with aluminium triggers the reaction of a mixture of aluminium powder and iron.(III) oxide; the molten iron produced flows into the space between the steel rails that are to be joined.

(a) Calculate the standard enthalpy change for this reaction, using data from Appendix D.

(b) Calculate the heat given off when 3.21g iron (III) oxide is reduced by aluminium at constant pressure.

Short answer

(a) The enthalpy change value, $∆H^{°}=-851.5kJ$.

(b) The amount of heat evolved, $q_p=-17.1\text{kJ}$.

Step-by-step solution

Given data

(a) $2\text{Al}\left(s\right)+{\text{Fe}}_2{\text{O}}_3\left(s\right)\to 2\text{Fe}\left(s\right)+{\text{Al}}_2{\text{O}}_3\left(s\right)$

$\Delta H_f^o$Value of $\text{Fe}\left(s\right)$is 0.

$\Delta H_f^o$Value of $\text{Al}\left(s\right)$is 0.

$\Delta H_f^o$Value of ${\text{Al}}_2{\text{O}}_3\left(s\right)$is $-1675.7{\text{kJmol}}^{-1}$.

$\Delta H_f^o$Value of ${\text{Fe}}_2{\text{O}}_3\left(s\right)$is $-824.2{\text{kJmol}}^{-1}$.

$\Delta H_f^e$Represents the standard enthalpy value.

(b) 3.21g Iron (III) oxide is reduced by aluminium at constant pressure.

 Step 2: Concept of Enthalpy

Enthalpy change is the amount of heat absorbed or evolved in a reaction when carried out at a constant pressure. It is generally calculated as the difference in energy required for bond breaking in a chemical reaction and energy gained in forming new bonds in the chemical reaction.

It is represented by $∆H$.

Negative enthalpy change represents the exothermic reaction when energy is released from the reaction and positive enthalpy change represents an endothermic reaction in which energy is taken in from surroundings.

 Step 3: Calculate the enthalpy change∆H°

(a)

The given expression is, $2\text{Al}\left(s\right)+{\text{Fe}}_2{\text{O}}_3\left(s\right)\to 2\text{Fe}\left(s\right)+{\text{Al}}_2{\text{O}}_3\left(s\right)$.

The enthalpy change is calculated as follows:

$∆H_{}^{\circ }=n∆H_f^{\circ }\left(Products\right)-n∆H_f^{\circ }\left(Reac\tan ts\right)$

$∆H_f^{\circ }$Represent the standard enthalpy value of the molecule.

role="math" localid="1663752328895" $$\begin{gathered} ∆H^{\circ }=\left(2mol\right)∆H_f^{\circ }\left[Fe\right(s\left)\right]+\left(1mol\right)∆H_f^{\circ }\left(A{l}_2{O}_3\right(s\left)\right)-\left\{\right(2mol)∆H_f^{\circ }[Al\left(s\right)]+(1mol)∆H_f^e[F{e}_2{O}_3\left(s\right)\} \\ =(2mol)\times 0+(1mol)\times (-1675.7kJmo{l}^{-1})-(2mol)\times 0+(1mol\left)\right(-824.2kJmo{l}^{-1} \\ =-1675.7kJ+824.2kJ \\ =-851.5kJ \end{gathered}$$

Calculate the molar mass of Fe2O3

(b)

Molar mass:

$$\begin{gathered} {\text{Fe}}_2{\text{O}}_3=2\times (atomicmassof\text{Fe})+3(atomicmassofO) \\ M\left({\text{Fe}}_2{\text{O}}_3\right)=2\times \left(55.85{\text{gmol}}^{-1}\right)+3\times \left(16.00{\text{gmol}}^{-1}\right) \\ =159.7{\text{gmol}}^{-1} \end{gathered}$$

Calculate the moles of Fe2O3

Calculate moles of $3.21\text{g}{\text{Fe}}_2{\text{O}}_3$.

Molar mass of Fe₂O₃ is 159.7gmol^-1.

$$\begin{gathered} {\text{moles of Fe}}_2{\text{O}}_3=\frac{3.21{\text{gFe}}_2{\text{O}}_3}{159.7{\text{gmol}}^{-1}} \\ =0.0201\text{mol}{\text{Fe}}_2{\text{O}}_3 \end{gathered}$$

Calculate the value of qp.

$∆H^{\circ }=-851.5kJ$ When 1mol of Fe₂O₃ is reduced, so heat evolved associated with 0.0201molFe₂O₃ is:

$$\begin{gathered} \Delta H=0.0201{\text{molFe}}_2{\text{O}}_3\times \frac{-851.5\text{kJ}}{1{\text{mol of Fe}}_2{\text{O}}_3} \\ =-17.1\text{kJ} \end{gathered}$$

At constant pressure, $q_p=\Delta H$.

So, Amount of heat evolved, $q_p=-17.1\text{kJ}$$q_p=-17.1\text{kJ}$.