Question

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

Hint:

To calculate the heat evolved per gram, divide the total heat by the total mass of the mixture.

Answer 1

Correct Answer: 4

To solve this problem, we need to determine the heat evolved per gram of a mixture of Fe₂O₃ and Al in a 1:2 molar ratio. The reaction between these two components can be represented as follows:

\[ \text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow 2\text{Fe} + \text{Al}_2\text{O}_3 \]
This reaction involves the formation of \(\text{Al}_2\text{O}_3\) and the reduction of \(\text{Fe}_2\text{O}_3\) to \(\text{Fe}\). The enthalpy changes for the formation of \(\text{Al}_2\text{O}_3\) and \(\text{Fe}_2\text{O}_3\) are provided:

\(\Delta H_f^{\circ}(\text{Al}_2\text{O}_3) = -1700 \, \text{KJ mol}^{-1}\)
\(\Delta H_f^{\circ}(\text{Fe}_2\text{O}_3) = -840 \, \text{KJ mol}^{-1}\)

The enthalpy change for the reaction is calculated by using the equation:
\[ \Delta H_{\text{reaction}} = \Delta H_f^{\circ}(\text{Al}_2\text{O}_3) - \Delta H_f^{\circ}(\text{Fe}_2\text{O}_3) \]
Substituting the provided values:
\[ \Delta H_{\text{reaction}} = -1700 \, \text{KJ mol}^{-1} - (-840 \, \text{KJ mol}^{-1}) = -860 \, \text{KJ mol}^{-1} \]
The negative sign indicates that heat is evolved.

Next, we calculate the total mass of the mixture (Fe₂O₃ : 159.7 g/mol, Al : 26.98 g/mol):

1 mole of Fe₂O₃ = 159.7 g
2 moles of Al = 2 × 26.98 g = 53.96 g
Total mass = 159.7 g + 53.96 g = 213.66 g

Heat evolved per gram of mixture:
\[ \text{Heat per gram} = \frac{860 \, \text{KJ/mol}}{213.66 \, \text{g/mol}} \approx 4.02 \, \text{KJ/g} \]
Rounding to the nearest integer, the heat evolved is approximately 4 KJ/g.

Finally, confirming the solution fits within the provided range 4 to 4. It does, validating our computations.