Question

"Thermite" reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is \({\bf{F}}{{\bf{e}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s) + 2Al(s)}} \to {\bf{A}}{{\bf{l}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s) + 2Fe(s)}}\). Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb \({\bf{851}}{\bf{.8\;kJ/mol}}\)of heat.

Short answer

The given reaction is spontaneous.

Step-by-step solution

Determining the spontaneous reaction.

• The definition of spontaneous reaction in science is a reaction that occurs in a specified set of conditions without interference. A spontaneous reaction is one that occurs in the absence of intervention in a given set of conditions. A spontaneous reaction is completed without the assistance of anyone else.
• A reaction is spontaneous if the overall entropy, or disorder, increases, according to the Second Law of Thermodynamics.

Analzing the reaction.

When we have to determine whether a reaction is spontaneous or not, we can use the total entropy change in the universe. If a reaction is spontaneous, then the entropy of the universe must increase. Otherwise, it is not a spontaneous reaction.

\(\begin{array}{l}\Delta S({\rm{ universe }}) = \Delta S({\rm{ system }}) + \Delta S({\rm{ surroundings }})\\\Delta S({\rm{ surroundings }}) = \frac{q}{T}\end{array}\)

where \(q\)is the heat released during the reaction and \(T\)is the thermodynamic temperature

\(\Delta S({\rm{ universe }}) = \Delta S({\rm{ system }}) + \frac{q}{T}\)

We can calculate the \(\Delta S(\)system \()\) using the formula

\(\Delta S = S({\rm{ products }}) - S({\rm{ reactants }})\)

We can look up the standard entropy values for each species in the table in AppendixG in the book.

\(\Delta S({\rm{ system }}) = S\left( {A{l_2}{O_3},s} \right) + 2S(Fe,s) - \left( {S\left( {F{e_2}{O_3},s} \right) + 2S(Al,s)} \right)\)

\(\Delta S({\rm{ system }}) = (50.92 + 2 \cdot 27.3 - (87.40 + 2 \cdot 28.3))\frac{J}{{K \cdot mol}}\)

\(\Delta S({\rm{ system }}) = - 38.48\frac{J}{{K \cdot mol}}\)

So,

\(\Delta S({\rm{ universe }}) = - 38.48\frac{J}{{K \cdot mol}} + \frac{{861800J_{mol}^{ - 1}}}{{298K}}\)

\(\Delta S = 2.9\frac{{kJ}}{{K \cdot mol}}\)

Pay attention to the units! Before we insert the digits into our calculator, we must first express the kilojoules in joules (or vice versa). Since the \(\Delta S\) (universe) is positive, which means that the entropy of the universe has increased, the reaction is spontaneous.

Therefore the given reaction is spontaneous.