Solid fuel used in rocket is a mixture of Fe_2 O_3 and Al (in ratio 1 : 2). The heat evolved (kJ) per gram of the mixture is \underline\hspace2cm . (Nearest integer) Given: \Delta H_f^\theta(Al_2 O_3)=-1700 ~kJ ~mol^-1\Delta H_f^\theta(Fe_2 O_3)=-840 ~kJ ~mol^-1 Molar mass of Fe, Al and O are 56, 27 and 16 g mol ^-1 respectively.

Thermodynamics PYQ — JEE Main 2023 · 11 Apr · Shift 1 · Q23 with Solution

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Correct answer: 4

The mixture contains $\mathrm{Fe_2O_3}$ and $\mathrm{Al}$ . The reaction is:

$\mathrm{Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe}$

So the given ratio $1:2$ is the stoichiometric ratio in moles.

Using

$\Delta H^\circ_{\text{rxn}}=\sum \Delta H_f^\circ(\text{products})-\sum \Delta H_f^\circ(\text{reactants})$

For elements in standard state,

$\Delta H_f^\circ(\mathrm{Al})=0, \qquad \Delta H_f^\circ(\mathrm{Fe})=0$

Thus,

$\Delta H^\circ_{\text{rxn}}=\Delta H_f^\circ(\mathrm{Al_2O_3})-\Delta H_f^\circ(\mathrm{Fe_2O_3})$

$= (-1700)-(-840)$

$=-860\ \text{kJ}$

So, $860\ \text{kJ}$ heat is evolved per mole of reaction.

Molar mass of $\mathrm{Fe_2O_3}$ :

$2(56)+3(16)=112+48=160\ \text{g}$

Mass of $2$ moles of Al:

$2(27)=54\ \text{g}$

Total mass of mixture:

$160+54=214\ \text{g}$

$\text{Heat per gram} = \frac{860}{214}\ \text{kJ g}^{-1}$

$\approx 4.02\ \text{kJ g}^{-1}$

Nearest integer:

$\boxed{4}$