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Question: An intimate mixture of ferric oxide and aluminium is used as solid fuel in rockets. Calculate the fu...

An intimate mixture of ferric oxide and aluminium is used as solid fuel in rockets. Calculate the fuel value per cm3{\text{c}}{{\text{m}}^3} of the mixture. Heats of formation and densities are as follows:
Hf(Al2O3)=−399kcalmol−1{{\text{H}}_{{\text{f}}\left( {{\text{A}}{{\text{l}}_2}{{\text{O}}_3}} \right)}} = - 399{\text{kcalmo}}{{\text{l}}^{ - 1}}; Hf(Fe2O3)=−199calmol−1{{\text{H}}_{{\text{f}}\left( {{\text{F}}{{\text{e}}_2}{{\text{O}}_3}} \right)}} = - 199{\text{calmo}}{{\text{l}}^{ - 1}}
Density of Fe2O3{\text{F}}{{\text{e}}_2}{{\text{O}}_3} =5.2gcm−3 = 5.2{\text{gc}}{{\text{m}}^{ - 3}}; Density of Al=2.7gcm−3{\text{Al}} = 2.7{\text{gc}}{{\text{m}}^{ - 3}}

Explanation

Solution

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 Al2O3{\text{A}}{{\text{l}}_2}{{\text{O}}_3}, Hf(Al2O3)=−399kcalmol−1{{\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 Fe2O3{\text{F}}{{\text{e}}_2}{{\text{O}}_3}, Hf(Fe2O3)=−199calmol−1{{\text{H}}_{{\text{f}}\left( {{\text{F}}{{\text{e}}_2}{{\text{O}}_3}} \right)}} = - 199{\text{calmo}}{{\text{l}}^{ - 1}}
Density of Fe2O3{\text{F}}{{\text{e}}_2}{{\text{O}}_3} =5.2gcm−3 = 5.2{\text{gc}}{{\text{m}}^{ - 3}}
Density of Al=2.7gcm−3{\text{Al}} = 2.7{\text{gc}}{{\text{m}}^{ - 3}}
The chemical reaction is given below:
2Al+Fe2O3→Al2O3+2Fe2{\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 ΔH∘\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 298K298{\text{K}} and pressure is 1atm{\text{1atm}}.
Standard enthalpy change, ΔH∘\Delta {{\text{H}}^ \circ } =ΔH∘f(products)−ΔH∘f(reactants) = \Delta {{\text{H}}^ \circ }_{{\text{f}}\left( {{\text{products}}} \right)} - \Delta {{\text{H}}^ \circ }_{{\text{f}}\left( {{\text{reactants}}} \right)}, where ΔH∘f(products)\Delta {{\text{H}}^ \circ }_{{\text{f}}\left( {{\text{products}}} \right)} is the enthalpy change of formation of products.
ΔH∘f(reactants)\Delta {{\text{H}}^ \circ }_{{\text{f}}\left( {{\text{reactants}}} \right)} is the enthalpy change of formation of reactants.
ΔH0=[ΔH(Al2O3)0+2ΔH(Fe)0]−[2ΔH(Al)0+ΔH(Fe2O3)0]\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.
⇒[−399kcal+2×0]−[2×0+(−199kcal)]=−399kcal+199kcal=−200kcal\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 Fe2O3{\text{F}}{{\text{e}}_2}{{\text{O}}_3} =5.2gcm−3 = 5.2{\text{gc}}{{\text{m}}^{ - 3}} and density of Al=2.7gcm−3{\text{Al}} = 2.7{\text{gc}}{{\text{m}}^{ - 3}}
Molecular mass of Fe2O3{\text{F}}{{\text{e}}_2}{{\text{O}}_3} is 160160.
Atomic mass of aluminium is 2727.

From the values of densities and mass, volume can be calculated from the equation given below:
Volume of reactants =Mass of Fe2O3Density of Fe2O3+Mass  of  aluminiumDensity  of  aluminium = \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}}}}
⇒160g5.2gcm−3+2×27g2.7 gcm−3=30.77 cm3+20 cm3=50.77 cm3\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 cm3{\text{c}}{{\text{m}}^3}.
Fuel value per cm3{\text{c}}{{\text{m}}^3} =enthalpy  changevolume  of  reactants=200kcal50.77cm3=3.94kcal = \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 cm3{\text{c}}{{\text{m}}^3}is 3.94 kcal3.94\,{\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