Question

An intimate mixture of ferric oxide and aluminium is used as solid fuel in rockets. Calculate the fuel value per ${\text{c}}{{\text{m}}^3}$ of the mixture. Heats of formation and densities are as follows:
${{\text{H}}_{{\text{f}}\left( {{\text{A}}{{\text{l}}_2}{{\text{O}}_3}} \right)}} = - 399{\text{kcalmo}}{{\text{l}}^{ - 1}}$; \[{{\text{H}}_{{\text{f}}\left( {{\text{F}}{{\text{e}}_2}{{\text{O}}_3}} \right)}} = - 199{\text{calmo}}{{\text{l}}^{ - 1}}\]
Density of ${\text{F}}{{\text{e}}_2}{{\text{O}}_3}$$ = 5.2{\text{gc}}{{\text{m}}^{ - 3}}$; Density of ${\text{Al}} = 2.7{\text{gc}}{{\text{m}}^{ - 3}}$

Answer 1

Hint: Fuel value is equal to the enthalpy change. It is equal to the difference between the enthalpy change of formation of products and reactants. Standard enthalpy change of reaction can be calculated from the enthalpy change of formation.

Complete step by step answer:
The data given in the question are,
Enthalpy change of formation of ${\text{A}}{{\text{l}}_2}{{\text{O}}_3}$, ${{\text{H}}_{{\text{f}}\left( {{\text{A}}{{\text{l}}_2}{{\text{O}}_3}} \right)}} = - 399{\text{kcalmo}}{{\text{l}}^{ - 1}}$
Enthalpy change of formation of ${\text{F}}{{\text{e}}_2}{{\text{O}}_3}$, \[{{\text{H}}_{{\text{f}}\left( {{\text{F}}{{\text{e}}_2}{{\text{O}}_3}} \right)}} = - 199{\text{calmo}}{{\text{l}}^{ - 1}}\]
Density of ${\text{F}}{{\text{e}}_2}{{\text{O}}_3}$$ = 5.2{\text{gc}}{{\text{m}}^{ - 3}}$
Density of ${\text{Al}} = 2.7{\text{gc}}{{\text{m}}^{ - 3}}$
The chemical reaction is given below:
$2{\text{Al}} + {\text{F}}{{\text{e}}_2}{{\text{O}}_3} \to {\text{A}}{{\text{l}}_2}{{\text{O}}_3} + 2{\text{Fe}}$

Initially we have to calculate the fuel value or the enthalpy change. Enthalpy is a thermodynamic quantity equivalent to the total heat content of a system. Standard enthalpy changes can be represented by $\Delta {{\text{H}}^ \circ }$. There are different types of standard enthalpy changes like formation, neutralization, combustion, atomization, solution, hydration etc.
Standard enthalpy change of formation is the enthalpy change when one mole of a compound is formed from its elements under standard condition. Standard condition denotes that the temperature is $298{\text{K}}$ and pressure is ${\text{1atm}}$.
Standard enthalpy change, $\Delta {{\text{H}}^ \circ }$$ = \Delta {{\text{H}}^ \circ }_{{\text{f}}\left( {{\text{products}}} \right)} - \Delta {{\text{H}}^ \circ }_{{\text{f}}\left( {{\text{reactants}}} \right)}$, where $\Delta {{\text{H}}^ \circ }_{{\text{f}}\left( {{\text{products}}} \right)}$ is the enthalpy change of formation of products.
$\Delta {{\text{H}}^ \circ }_{{\text{f}}\left( {{\text{reactants}}} \right)}$ is the enthalpy change of formation of reactants.
$\Delta H^{0} = \left [ \Delta H^{0}_{(Al_2 O_3)}+ 2\Delta H^{0}_{(Fe)}\right ] - \left [ 2\Delta H^{0}_{(Al)} + \Delta H^{0}_{(Fe_2 O_3)} \right ]$

From the values given, we can substitute the values.
$ \Rightarrow \left[ { - 399{\text{kcal}} + 2 \times 0} \right] - \left[ {2 \times 0 + \left( { - 199{\text{kcal}}} \right)} \right] = - 399{\text{kcal}} + 199{\text{kcal}} = - 200{\text{kcal}}$
It is given that density of ${\text{F}}{{\text{e}}_2}{{\text{O}}_3}$$ = 5.2{\text{gc}}{{\text{m}}^{ - 3}}$ and density of ${\text{Al}} = 2.7{\text{gc}}{{\text{m}}^{ - 3}}$
Molecular mass of ${\text{F}}{{\text{e}}_2}{{\text{O}}_3}$ is $160$.
Atomic mass of aluminium is $27$.

From the values of densities and mass, volume can be calculated from the equation given below:
Volume of reactants $ = \dfrac{{{\text{Mass of F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}}}{{{\text{Density of F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}}} + \dfrac{{{\text{Mass}}\;{\text{of}}\;{\text{aluminium}}}}{{{\text{Density}}\;{\text{of}}\;{\text{aluminium}}}}$
$ \Rightarrow \dfrac{{160{\text{g}}}}{{5.2{\text{gc}}{{\text{m}}^{ - 3}}}} + \dfrac{{2 \times 27{\text{g}}}}{{2.7\,{\text{gc}}{{\text{m}}^{ - 3}}}} = 30.77\,{\text{c}}{{\text{m}}^3} + 20\,{\text{c}}{{\text{m}}^3} = 50.77\,{\text{c}}{{\text{m}}^3}$
Now we have to find the fuel value per volume or per ${\text{c}}{{\text{m}}^3}$.
Fuel value per ${\text{c}}{{\text{m}}^3}$$ = \dfrac{{{\text{enthalpy}}\;{\text{change}}}}{{{\text{volume}}\;{\text{of}}\;{\text{reactants}}}} = \dfrac{{200{\text{kcal}}}}{{50.77{\text{c}}{{\text{m}}^3}}} = 3.94{\text{kcal}}$
Hence the fuel value per ${\text{c}}{{\text{m}}^3}$is $3.94\,{\text{kcal}}$

Additional information-Some reactions give off so much energy that they are explosive. So, it’s very useful to know how much heat a reaction will give off or absorb. This quantity is called enthalpy.


Note:
From the enthalpy change, we can decide whether a reaction is exothermic or endothermic. Most chemical reactions are accompanied by energy changes. Some absorb energy and others release it.
Endothermic reaction-absorbs energy
Exothermic reaction-releases energy