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MCQMediumJEE 2023Enthalpies (Bond, Combustion, Formation…)

JEE Chemistry 2023 Question with Solution

Solid fuel used in rocket is a mixture of Fe2{}_2O3{}_3 and Al (in ratio 1:21 : 2) the heat evolved (KJ) per gram of the mixture is _____ (Nearest integer) Given ΔHf∘(Al2O3)=−1700 KJ mol−1\Delta H_f^{\circ}(Al_2O_3) = -1700 \, \text{KJ mol}^{-1} ΔHf∘(Fe2O3)=−840 KJ mol−1\Delta H_f^{\circ}(Fe_2O_3) = -840 \, \text{KJ mol}^{-1}

  • A

    2 KJ2 \, \text{KJ}

  • B

    3 KJ3 \, \text{KJ}

  • C

    4 KJ4 \, \text{KJ}

  • D

    5 KJ5 \, \text{KJ}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Solid fuel is a mixture of Fe2{}_2O3{}_3 and Al in the ratio 1:21:2.

Find: Heat evolved per gram of the mixture.

The reaction is

Fe2O3+2Al→Al2O3+2FeFe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe

Using standard enthalpy of formation values,

ΔH=ΔHf∘(Al2O3)−ΔHf∘(Fe2O3)\Delta H = \Delta H_f^{\circ}(Al_2O_3) - \Delta H_f^{\circ}(Fe_2O_3)

Substituting the given values,

ΔH=(−1700)−(−840)=−860 KJ/mol\Delta H = (-1700) - (-840) = -860 \, \text{KJ/mol}

For the stoichiometric mixture, 11 mole of Fe2{}_2O3{}_3 reacts with 22 moles of Al. The solution gives the total mass of the mixture as

160+54=214 g160 + 54 = 214 \, \text{g}

Therefore, heat evolved per gram of mixture is

−860 KJ214≈−4.01 KJ/g\frac{-860 \, \text{KJ}}{214} \approx -4.01 \, \text{KJ/g}

Thus, the magnitude of heat evolved per gram is approximately 4 KJ/g4 \, \text{KJ/g}. The nearest integer corresponds to the correct option C.

Note: the solution contains a mass inconsistency because it states Fe2O3Fe_2O_3 weighs 112 g112 \, \text{g} but then uses 160 g160 \, \text{g} in the total. The final numerical answer still matches 44.

Using reaction stoichiometry

Given:

  • ΔHf∘(Al2O3)=−1700 KJ mol−1\Delta H_f^{\circ}(Al_2O_3) = -1700 \, \text{KJ mol}^{-1}
  • ΔHf∘(Fe2O3)=−840 KJ mol−1\Delta H_f^{\circ}(Fe_2O_3) = -840 \, \text{KJ mol}^{-1}

Find: Heat evolved per gram of the mixture.

First write the thermite reaction:

Fe2O3+2Al→Al2O3+2FeFe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe

Since elemental Fe and Al in standard state have zero enthalpy of formation, only Fe2O3Fe_2O_3 and Al2O3Al_2O_3 contribute.

Hence,

ΔH=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants}) ΔH=[ΔHf∘(Al2O3)+2ΔHf∘(Fe)]−[ΔHf∘(Fe2O3)+2ΔHf∘(Al)]\Delta H = \left[\Delta H_f^{\circ}(Al_2O_3) + 2\Delta H_f^{\circ}(Fe)\right] - \left[\Delta H_f^{\circ}(Fe_2O_3) + 2\Delta H_f^{\circ}(Al)\right] ΔH=(−1700)−(−840)=−860 KJ/mol\Delta H = (-1700) - (-840) = -860 \, \text{KJ/mol}

The heat released is therefore 860 KJ860 \, \text{KJ} for one stoichiometric set of reactants. Dividing by the total mass used in the solution,

860214≈4.01 KJ/g\frac{860}{214} \approx 4.01 \, \text{KJ/g}

Therefore, the nearest integer value is 44, so the correct option is C.

Common mistakes

  • Using the sign of ΔH\Delta H directly as the answer is incorrect. The question asks for heat evolved, so the required value is the magnitude of heat released per gram. Use 44, not −4-4.

  • Ignoring stoichiometry is wrong because the ratio 1:21:2 corresponds to the reaction Fe2O3+2Al→Al2O3+2FeFe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe. Always compute heat for the balanced reaction before converting to per gram value.

  • Forgetting that elements in their standard states have ΔHf∘=0\Delta H_f^{\circ} = 0 leads to an incorrect enthalpy calculation. Only Fe2O3Fe_2O_3 and Al2O3Al_2O_3 contribute here.

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